PROBLEMS
IN REAL
ANALYSIS
A Workbook
With Solutions
2nd Edition
Charalambos D. Aliprantis
Owen Burkinshaw

PROBLEMS IN
REAL ANALYSIS
Second Edition
A Workbook with Solutions
CHARALAMBOS D. ALIPRANTIS
Departments of Economics and Mathematics
Purdue University
and
OWEN BURKINSHAW
Departments of Mathematical Sciences
Indiana University, Purdue University, Indianapolis
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International Standard Book Number 0-12-050253-4
This book accompanies the following title, catalogued by the Library of Congress:
Library of Congress Cataloging-in-Publication Data
Aliprantis, Charalambos D.
Problems in real analysis /Charalambos D. Aliprantis and Owen Burkinshaw.
p. cm.
Includes bibliographical references and index.
ISBN 0-12-050257-7 (acid-free paper)
1. Mathematical analysis. 2. Functions of real variables. I. Burkinshaw, Owen. DL Title.
QA300.A48 1998
515-dc21 98-3955
CIP
Printed in the United States of America
98 99 00 01 02 DS 9 8 7 6 5 4 3 2 1

CONTENTS
Foreword 7
CHAPTER 1. FUNDAMENTALS OF REAL ANALYSIS 1
1. Elementary Set Theory 1
2. Countable and Uncountable Sets 6
3. The Real Numbers 11
4. Sequences of Real Numbers 20
5. The Extended Real Numbers 34
6. Metric Spaces 45
7. Compactness in Metric Spaces 54
CHAPTER 2. TOPOLOGY AND CONTINUITY 65
8. Topological Spaces 65
9. Continuous Real-Valued Functions 73
10. Separation Properties of Continuous Functions 92
11. The Stone-Weierstrass Approximation Theorem 98
CHAPTER 3. THE THEORY OF MEASURE 107
12. Semirings and Algebras of Sets 107
13. Measures on Semirings 112
14. Outer Measures and Measurable Sets 116
15. The Outer Measure Generated by a Measure 122
16. Measurable Functions 133
17. Simple and Step Functions 137
18. The Lebesgue Measure 146
19. Convergence in Measure 157
20. Abstract Measurability 160
V

vi CONTENTS
CHAPTER 4. THE LEBESGUE INTEGRAL 171
21. Upper Functions 171
22. Integrable Functions 174
23. The Riemann Integral as a Lebesgue Integral 190
24. Applications of the Lebesgue Integral 206
25. Approximating Integrable Functions 220
26. Product Measures and Iterated Integrals 224
CHAPTER 5. NORMED SPACES AND L,-SPACES 239
27. Normed Spaces and Banach Spaces 239
28. Operators between Banach Spaces 245
29. Linear Functionals 251
30. Banach Lattices 259
31. Lp -Spaces 271
CHAPTER 6. HILBERT SPACES 297
32. Inner Product Spaces 297
33. Hilbert Spaces 310
34. Orthonormal Bases 325
35. Fourier Analysis 333
CHAPTER 7. SPECIAL TOPICS IN INTEGRATION 345
36. Signed Measures 345
37. Comparing Measures and the
Radon-Nikodym Theorem 353
38. The Riesz Representation Theorem 365
39. Differentiation and Integration 379
40. The Change of Variables Formula 395

FOREWORD
This book contains complete solutions to the 609 problems in the third edition
of Principles of Real Analysis, Academic Press, 1998. The problems have been
spread over forty sections which follow the format of the book.
All solutions are based on the material covered in the text with frequent
references to the results in the text. For instance, a reference to Theorem 7.3 refers to
Theorem 7.3 and a reference to Example 28.4 refers to Example 28.4, both in the
third edition of Principles of Real Analysis.
This problem book will be beneficial to students only if they use it "properly."
That is to say, if students look at a solution of a problem only after trying very
hard to solve the problem. Students will do themselves great injustice by reading
a solution without any prior attempt on the problem. It should be a real challenge
to students to produce solutions which are different from the ones presented here.
We would like to express our most sincere thanks to all the people who made
constructive recommendations and corrections regarding the text and the problems.
Special thanks are due to Professor Yuri Abramovich for his contributions and
suggestions during the writing of this problem book.
C. D. ALIPRANTIS AND O. BURKINSHAW
West Lafayette, Indiana
July, 1998
vii

CHAPTER 1
FUNDAMENTALS OF
REAL ANALYSIS
1. ELEMENTARY SET THEORY
Problem 1.1. Establish the following set theoretic relations:
1. (AUB)C\C = (AnC)U{BDC) and
(A fl B) U C = (A U C) fl (5 U C);
2. (AUfl)\C = (A\C)U(B\C) a/id
(Anfl)\C = (A \C)D(B \C);
3. A\5 = /\nfic;
4. K5 <!=> £cCAc;aw/
5. (/lU5)c = Acn5c and (AD B)c = Ac U Bc.
Also, for an arbitrary function f:X->Y, establish the following claims:
7. /(n,6/ *) c n/€/ /Wf);
8. r,(U/e/«/) = Ue//-I(B/);
9. /-l(n/6#ft) = n/e#rl(ft);^
10. r,(5c) = [/-,(5)]c
Solution. (1) We establish the first formula only. We have
a- e(AUB)HC <=> x € AUB and x eC
<=> [x e A or x e B] and x eC
«=> [x € A and x € C ] or [x <= B and x e C ]
«=> ;ce/inC orjcefinC
<==> * G(/inC)U(5nC).
(2) Again, we establish the first formula only. Observe that
jc € {A U B) \ C *=$> x e AUB and a £ C

2
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
[x € A or x e B ] and x <£ C
[x e A and x $ C ] or [x e B and x £ C ]
x e A\C or x e B \C
xe(A\C)U(B\ C).
(3) Note that
x e A\ B <=» x e A and x £ B
x e A and x e Bc <==> x e A n Bc.
(4) Let A C B. Then, jc g Bc implies x: $ B and so x <£ A (i.e., x e Ac) so that
Bc C Ac.* On the other hand, if Bc c Ac holds, then (by the preceding case) we
have A = (AC)C c (Bc)c = B.
(5) Note that
x e (A H £)c <=* ^AH5 <=> jc £ A or x £ B
<=^ x e Ac or x e Bc <=$> x e Ac U Bc.
Moreover,
jc € (A U £)c <=* x i A U £ «=» x $ A and * £ £
<=> * € /4C and jc g £c <=» jc g i4c n Bc.
(6) We have
y€/((jA/) <=> 3 a-gIJ^, with>; = /(jc)
/€/ 16/
3 / G / with x G A/ and y = /(jc)
3 i € / with y G /(A) <=> y € (J/(*/)•
/€/
(7) From the inclusion /(f]ieJ Aj) c f(Aj) for each J, we see that
/(|>)£fV(*)-
ie/ /€/
(8) We have
e f~]{UBi) *=* fM-e\jBi <=> 3 i e / with /(jc) G £,
/€/ /€/
3 / € / with jc g /"'(fl/) <=* jc e\Jf~l(Bi).
/€/

Section 1: ELEMENTARY SET THEORY
3
(9) Note that
jr e rl(f]Bi) <=* f(x) € f]B{ *=> f(x) e B, for each / € /
<=> x G /"'(B/) for each i el <=> jc e p| /"'(fl,-).
(10) Observe that
A'G/"1^0) <=> f(x)eBc *=> f(x)iB
*=> xif~\B) <=> xe[r\B)]c.
Problem 1.2. For two sets A and B show that the following statements are
equivalent:
a. A c B ;
b. A U B = B ;
c. ADB = A
Solution, (a) => (b) Clearly, B c AUB holds. On the other hand, if x e AUB,
then x e A or „r e B, and so in either case, x e B. This means A U B C B, and
hence, AUB = B.
(b) => (c) By part (1) of the preceding problem, we have
A n B = A n (A u B) = (A n A) u (A n B) = a u (a n B) = a.
(c) => (a) Clearly, A = A n B c B.
Problem 1.3. S/zcw r/zar (AAB)AC = AA(BAC) holds for eveiy triplet of
sets A,B, andC.
Solution. Note first that for any three sets X, Y, and Z we have
XAY \Z = [X\(FUZ)]U[r\(XU Z)]
and
z \ (X Ay) = [z\(xur)]u[xnrn z].
For instance, to verify the first identity, note that
x eX&Y \ Z <=* [jc 6X\yor.r € Y\X]andx £ Z
<=> [x e X,x i Yt andx $ Z] or [jc e Y, x $ X, and x $Z]
<=> xe[X \(YUZ)]U[Y \(XUZ)].

4
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Thus,
(AAB)AC = [(AAB)\C]U[C \ (AAB)]
= [A\(BUC)]U[B\(AUC)]U[C\(AUB)]U[APiBn C]
= {[ A\(B UC)]U(ADBr\C)}U {[B\(C U A)] U [C\(B U A)]}
= [A\(BAC)]U[(BAC) \ A]
= AA(BAC).
Problem 1.4. Give an example of a function f\X-±Y and two subsets A and
BofX such that f(A n B) # /(A) n /(B).
Solution. Define /: {0, 1} -+ {0, 1} by /(0) = /(l) = 0. If A = {0} and
B = (1), then f(A n B) = 0 # {0} = /(A) n /(B).
Problem 1.5. For a function f:X -» y, s/zow f/20/ the following three
statements are equivalent:
a. / w one-to-one.
b. /(A fl B) = /(A) fl /(B) holds for all A, B e V(X).
c. For every pair of disjoint subsets A and B of X,we have f(A)f) f(B) = 0.
Solution, (o) =» (£) If y € /(A) O /(B), then there exist a e A and fc € B
with j = /(a) = /(£). Since / is one-to-one, a = b € A C\ B, and so
y € f(A n B). Thus, /(A) n /(B) c f(A n B) c /(A) n /(B).
(£) => (c) Obvious.
(c) => (o) Let f(a) = /(ft). If o # 6, then the two sets A = [a] and B = {&}
satisfy AHB=0, while /(A) n /(B) = [f(a)} # 0.
Problem 1.6. Le/ /:X^Kk a function. Show that f(f~l(A)) C A /or a//
KF.WK f-\f(B))forall Kl
Solution. Clearly, a € /_1(/4) if and only if /(a) e A. Thus, f(f~\A)) c A.
Similarly, a- e /_1(/(B)) if and only if /(a) g /(B), and so B c /-*(/(B))
holds.
Problem 1.7. Show that a function f:X-+Y isontoifandonlyiff(f~\B)) = B
holds for all B <ZY.
Solution. Assume that / is onto and B C Y. If b e B, then there exists some
a 6 X with /(o) = 6; clearly, a e f~\B). Thus, b = /(a) € /(/_,(B)), and
so B c f{f~\B))<z B holds.

Section 1: ELEMENTARY SET THEORY
5
For the converse, note that the relation /(/ ]({b}))=z{b] implies / ]({b})^
0 for each b € Y so that / is onto.
Problem 1.8. Let X -U Y -^ Z. // A C Z, show that
{gof)-\A) = r\g-\A)).
Solution. Note that
a' € (gofy\A) *=> g(f{x)) eA*=> f(x) € g~\A) *=>xe f~\g-l(A)).
Problem 1.9. Show that the composition of functions satisfies the associative
f R h
law. That is, show that if X —> Y —> Z —> V, then {h o g) o / = h o (g o /).
Solution. Observe that for each x e X we have
[(hog) of ](x) = h o g(f(x)) = h(g(f(x))) = h((g o /)(*)) = [ho(go f)](x).
Therefore, (h o g) o / == h o (g o /).
Problem 1.10. Let f:X -> Y. Show that the relation 11 on X, defined by
x\7Zx2 whenever f(x\) = f(x2), is an equivalence relation.
Solution. We must show that the relation 71 is reflexive, symmetric, and
transitive.
Reflexivity: Note that f(x) = f(x) implies xlZx for each x e X.
Symmetry: Let x\7lx2. Then, f(x\) = f(x2) or f(x2) = f(x\), so that x27lx].
Transitivity: If X[7lx2 and x-iR,xi, then f(x\) = f(x2) and f(x2) = /to) both
hold. It follows that f{x\) = /(X3), and so x{Rx^.
Problem 1.11. // X and Y are sets, then show that
V(X)r\V(Y)=:V(XnY) and V(X) U V(Y) c V(X U Y).
Solution, (a) Note that
A e V(X) n V(Y) ^ACXandAcy
*=> Acxnr «=> a eV(Xnr).
(b) Clearly,
A e P(X) U P(K) => AQX or AC,Y => A<ZXUY => AeV(XUY).

6
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
If X and Y are two nonempty disjoint sets, then X UY <£ V{X) U V(Y)y and so
equality is seldom valid.
2. COUNTABLE AND UNCOUNTABLE SETS
Problem 2.1. Show that the set of all rational numbers is countable.
Solution. Let Q be the set of rational numbers and let Q+ = {/• € Q: r > 0}.
Then the function /:N x N -* Q+ defined by f(m,n) = % is onto. The
conclusion now follows from Theorems 2.7 and 2.5.
Problem 2.2. Show that the set of all finite subsets of a countable set is countable.
Solution. We can assume that A = {p\, p2,...} is the set of all prime numbers.
Let T denote the collection of all finite subsets of A. Define f\T —> IN by
/(F) = the product of the elements of F, for each F e T. Then/ is one-to-one,
and the conclusion follows from Theorem 2.5.
Problem 2.3. Show that a union of an at-most countable collection of sets, each
of which is finite, is an at-most countable set.
Solution. This follows immediately from Theorem 2.6.
Problem 2.4. Let A be an uncountable set and let B be a countable subset of
A. Show that A is equivalent to A\B.
Solution. Let B = [b\, b2,...}. Since A is uncountable, the set A \ B is also
uncountable. Let C = [c\, c-i,...} be a countable subset of A \ B. Now define
/: A \ B -* A by
/(*) =
a-, if x £ C;
cn+|f if x = c2n+\ (n = 0, 1, 2,...);
[bn, if x = c2n(n = 1,2,...).
Then / is one-to-one and onto, proving that A % A \ B.
Problem 2.5. Assume that f:A->B is a surjective (onto) function between
two sets. Establish the following:
a. card B < card A.
b. If A is countable, then B is at-most countable.

Section 2: COUNTABLE AND UNCOUNTABLE SETTS
7
Solution, (a) Consider the family [f~l(b): b e B}. Clearly, this is a family of
disjoint subsets of A. By the Axiom of Choice there exists a subset C of A such
that C H f~\b) consists precisely of one element of A for each b e B. The
conclusion now follows by observing that /: C -> B is one-to-one and onto,
(b) This follows immediately from part (a).
Problem 2.6. Show that two nonempty sets A and B are equivalent if and only
if there exists a function from A onto B and a function from B onto-A.
Solution. If A and B are equivalent, then there exists a function /: A —> B
which is one-to-one and onto. Clearly, f~l: B —> A is a surjective function.
For the converse, assume that there exists a function from A onto B and a
function from B onto A. A glance at the preceding problem guarantees that
card B < card A and card A < card B. Now, use the Schroder-Bernstein theorem
to conclude that A and B are equivalent sets.
Problem 2.7. Show that if a finite set X has n elements, then its power set
V(X) has 2" elements.
Solution. We shall use induction on n. Assume that {1,..., n] has 2" subsets.
Then the subsets of the set {1,..., ny n + 1} consist of:
a. The subsets of {1,..., /?}, which are 2" altogether; and
b. The subsets of the form AU(/i + l), where A is a subset of (1,..., n],
again 2" altogether.
Thus, the number of subsets of {1,...,/?,/? + 1} is 2" + 2" = 2"+l.
A direct proof goes as follows. Notice that the number of subsets of {1, 2,..., h]
having k elements (where 0 < k < n) is precisely (£). So, the total number of
subsets of {1, 2,..., n] is
+(:)+G)+-+CH+i>"^
where the last equality holds true by virtue of the binomial theorem.
Problem 2.8. Show that the set of all sequences with values 0 or 1 is
uncountable.
Solution. For each subset A of IN define the sequence f(A) = [xn] by xn = 1
if n e A and xn = 0 if n g A. Then / defines a function from V(1N) onto

8
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
the sequences with values 0 and 1. Since / is clearly one-to-one and onto, the
conclusion follows from Theorem 2.8.
Problem 2.9. If 2 = {0, 1}, then show that 2X % V(X) for eveiy set X.
Solution. Define f:V(X) -+ 2X by A i—> fA% where fA(x) =1 if a* € A
and fA(x) = 0 if a: £ A. Note that / is one-to-one and onto. Therefore,
2X % V(X).
Problem 2.10. Any complex number that is a root of a {nonzero) polynomial
with integer coefficients is called an algebraic number. Show that the set of all
algebraic numbers is countable.
Solution. Let Z = {..., —2, —1,0, 1,2,...}. Fix n > 1. Since every
polynomial /?(a) = ao-\-a\x-\ \-anxn is determined uniquely by (tfo> ^i» • • •»tfn),itis
easy to see that the nonzero polynomials of degree <n with integer coefficients are
in one-to-one correspondence with the countable set Zn+]\{(0, 0,..., 0)}. Let
{pii P21 •. •} be an enumeration of all these polynomials. By the Fundamental
Theorem of Algebra, the set A* = {a g C: Pa(a) = 0} is a finite set. Thus, the
set of all zeros of the polynomials {pi, pi,...} of degree < n is precisely the set
Rn = |J~ j A;, which (by Theorem 2.6) is a countable set. Now, note that the set
of all algebraic numbers is U^li ^«» which—as a countable union of countable
sets—is itself countable; see Theorem 2.6.
Problem 2.11. For an arbitrary function f: R -»■ R show that the set
A = {a e R: lim /(a) exists and lim /(a) ^ f(a)}
1 x—>a x-*a '
is at-most countable.
Solution. Let X denote the set of all open subintervals of R with rational
endpoints and note that 1 is a countable set. Also, let Q denote the countable set
of all rational numbers of R.
For each rational real number r, let
A, = \a e A: Either f{a) < r < lim /(a) or lim /(a) < r < f(a)}.
x-*>a x-+a
Clearly, A = UreQ^r holds. Thus, in order to establish that A is at most
countable, it suffices to show that each Ar is at-most countable.
So, fix some r e Q and a e A/ and assume (without loss of generality) that
f(a) < r < lim^a /(a). Then there exists as 8 > 0 such that a—8<y<a + 8
and y # a imply f(y) > r. Next, pick an open interval Ia with rational
endpoints (i.e., Ia el) such that a e Ia and Ia c (a — 8, a + 8). Since /(v) > r

Section 2: COUNTABLE AND UNCOUNTABLE SETS
9
holds for each y € Ia with y ^ a, we see that y £ Ar for each y e /a\{a}. In
particular, note that Ar Ci Ia = {a}.
Thus, we have established a mapping a i—> Ia from Ar into J (which in view
of Ar fl /a = {a} for each a G /4r) is alsd one-to-one. This implies that y4r is
at-most countable, and hence, A is likewise at-most countable.
Problem 2.12. Show that the set of real numbers is uncountable by proving the
following:
a) (0, 1) % R; and
b) (0, 1) is uncountable.
Solution, (a) The function /:(0, 1) -> 1R defined by the formula f(x) =
tan(7r;t — |) is one-to-one and onto.
(b) If (0, 1) is countable, then let [x\,X2,...} be one enumeration of (0,1).
For each n write x„ = 0.d„\d„2 • • • in its decimal expansion, where each d/y
is 0, 1,..., or 9. Now, consider the real number y of (0, 1) whose decimal
expansion y = 0.y\yi • * * satisfies yn = 1 if dnn ^ 1 and yn = 2 if dnn = 1. An
easy argument now shows (how?) that y ^ xn for each n, which is a contradiction.
Hence, the interval (0, 1) is an uncountable set.
Problem 2.13. Using mathematical induction prove the following:
a. If a > -\,then(\+a)n > l+naforn = 1,2,.. .(Bernoulli's inequality).
b. // 0 < a < 1, f/ze* 1 4- 3na > (1 + a)n for n = 1, 2,... .
c. cos(fl7r) = (-l)n/0rrt = 1,2
Solution, (a) Let a > — 1. For ai = 1 the inequality is trivially true; in fact, it is
an equality. For the induction step, assume that (1 4- a)n > 1 + na holds true for
some n. Since 1 4- a > 0 is assumed to be true, it follows that
(l+a)n+{ = (l4-a)(l+a)n >(l+a)(l+/ifl)= l+fla + a + mz2
= 1 4- (/i + \)a 4- Aza2 > 1 4- (n 4- IK
which is the desired inequality when n takes the value n + \. This completes the
induction.
(b) Assume 0 < a < 1. Since 1 4- 3a > 1 4- a, the desired inequality is true
for n = 1. For the inductive step assume 1 4- 3na > (1 4- a)n. Then, taking into
account that 0 < a < 1, we see that
(1 4- a)n+x = (1 + a)(l + a)n < (1 + fl)(l + 3na)
= 1 + 3na + a + 3na2 = 1 + (3n + 3"a 4- \)a
< 1 -I- (3" + 3" 4- 3n)a = 1 4- 3 • 3na = 1 + 3/,+la,

10
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
which is the desired inequality valid when n is replaced by n 4-1. By the Principle
of Mathematical Induction, the inequality is true for every natural number n.
(c) For n = 1, we have cos(l • n) = cos7r = —1 = (—l)1. Now, assume
that cos(/27r) = (—l)n. Then, using the trigonometric formula cos(jc 4- y) =
cos x cos y — sin a* sin y, we see that
COs[(/2 + l)7r] = COS(rt7T + 7l) = COS(rt7T) COS 7T — sin(rtJT) SHITT
= (-l)w(-l) - sin(>ur) • 0 = (-l)n+1,
and the induction is complete.
Problem 2.14. Show that the Well-Ordering Principle implies the Principle of
Mathematical Induction,
Solution. Let S c IN satisfy
a. 1 e 5, and
b. n 4- 1 € S whenever n e S.
We must show that S = IN, or equivalently that IN \ S = 0.
To this end, assume by way of contradiction that we have IN \ S ^ 0. Then,
by the Well Ordering Principle, n = min(lN \ S) exists. Clearly, \ < n eJN \ S.
Thus, h — 1 € 5, and consequently az = (n — 1)+ 1 6 5, a contradiction.
Therefore, N\5 = 0or5 = N.
Problem 2.15. Show that the Principle of Mathematical Induction implies the
Well-Ordering Principle.
Solution. Assume that the Principle of Mathematical Induction is true. Consider
the subset S of N consisting of all natural numbers n with the property: whenever
a nonempty subset A of N contains a natural number m < n, then A has a
least element. To establish the Well-Ordering Principle, we need to show that
5 = IN.
To this end, note that 1 e S. Now assume that n e S. Also, assume that a
nonempty subset A of N contains some natural number m < n + 1. If A contains
a natural number k < n + 1, then A also contains a natural number (namely k
itself) less than or equal to n, and so, in view of n e 5, A must have a least element.
On the other hand, if A does not contain any natural number strictly less that n •+• 1,
it follows that n 4-1 G A, in which case /? 4-1 is the least element of A. Therefore,
n 4- 1 € 5, and so by the validity of the Principle of Mathematical Induction, we
infer that S = IN.

Section 3: THE REAL NUMBERS
11
3. THE REAL NUMBERS
Problem 3.1. If av b = max {a, b) and a Ab = min{tf, b], then show that
a v b = \{a 4- b 4- \a - b\) and a A b = \(a 4- 6 - |a - 6|).
Solution. Since all expressions do not change their values if we interchange a
and b, we can assume a > b. Thus,
±(a+b + \a-b\) = j(a + b + a-b) = a = avb%
and
±(a 4- 6 - \a - 6|) = {[a + b - (a - b)] = & = a a 6.
Problem 3.2. Show that \\a\ -\b\\<\a + b\< \a\ 4- |fe| for all a, b e R.
Solution. From -|a| < a < |a| and -|6| < b < \b\, it follows that
~{\a\ + \b\)<a + b<\a\ + \b\.
So, \a + b\ <\a\ + \b\.
Substituting a — b in the place of a, we get \a\ < \a — b\ 4- |6| so that
N ~ l&l < |c — b\. Interchanging a and b yields —(|a| — |6|) < \a — b\, and so
||fl|-|6|| < |a-6| also holds.
Problem 3.3. Show that the real numbers V2 and \fl 4- V3 are irrational
numbers.
Solution. Assume by way of contradiction that V2 = ^ with m, n e IN. We can
suppose that m and n have no common positive divisors other than 1. Squaring,
we get m2 = 2a?2. This implies that m is even, i.e., m — 2k for some £ e JN
(otherwise /rz = 2k 4- 1 implies that m2 is odd, a contradiction). It follows that
Ak2 = 2/r, or «2 = 2£2, which in turn implies that n is even, i.e., n—2l for some
I e IN. But then, m and a have the common factor 2, which is a contradiction.
Hence, V2 is not a rational number. (This simple proof is due to Eudoxus.)
With a different and more elegant proof one can establish the following general
result:
• The square root yfk of a natural number k is a rational number if and only
ifk is a complete square, i.e., k = p2 for some p e IN.

12
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
If k = p2, then clearly «Jk = p g N. On the other hand, if v^ is a rational
number, then \fk is a rational root of the polynomial p(x) = x2 — k. But the
positive rational roots of this polynomial are of the form ^, where m e N is a
divisor of k and n e N is a divisor of 1. Thus, y/k = m € IN, and so & = /h2.
To see that V2 + V3 is not a rational number, assume by way of contradiction
that y/2 -f \/3 = /* > 0 is a rational number. Then \/3 = r — %/2 and by squaring,
we get 3 = r2 — 2/*%/2 -f 2. This implies \/2 = c~^, a rational number, contrary
to our previous conclusion. Hence, y/l + \/3 is an irrational number.
Problem 3.4. Show that between any two distinct real numbers there is an
irrational number.
Solution. Let a < b. Choose a rational number r with a < r < b, and then
select some n so that 0 < — < b—r. Note that the irrational number x = r-f^
n n
satisfies a < x < b.
Alternatively: Note that the open interval (a, b) is uncountable, while the set
of all rational numbers is countable.
Problem 3.5. This problem will introduce {by steps) the familiar process of
subtraction in the framework of the axiomatic foundation of real numbers.
a. Show that the element 0 is uniquely determined, i.e., show that ifx 4- 0* = x
for allx <= JR and some 0* e R, then 0* = 0.
b. Show that the cancellation law of addition is valid, i.e., show that x + a =
a: + b implies a = 6.
c. Use the cancellation law of addition to show that 0 • a = Ofor all a e 1R.
d. Show that for each real number a the real number —a is the unique real
number that satisfies the equation a + x =0. (The real number —a is
called the negative of a.)
e. Show that for any two given real numbers a and b, the equation a + a* = b
has a unique solution, namely x = b + (—a). The subtraction operation
— o/IR is now defined by a — b = a + (—b); the real number a —bis also
called the difference of b from a.
f. Forany real numbers a andb show that-(—a) = aand—(a+b) = —a—b.
Solution, (a) Assume than another element 0*gR satisfies 0*+a*=a*+0*=a-
for all x e JR. Letting x = 0, we get 0* + 0 = 0. Now, recalling that 0 + y =
y + 0 = y also holds for all y e IR, letting y = 0* yields 0* = 0* + 0 = 0.
(b) Let ji: + a = x 4- b. By Axiom 5 there exists some z € R such that z + a' =
x + z = 0. So,
a = 0 + a = (z+a:) + 0 = z + (a'+^) = z + (x + b) = (z + a') + 6 = 0 + 6 = *.

Section 3: THE REAL NUMBERS
13
(c) Clearly,
0-a + 0 = 0-a = (0 + 0)-a = 0-a+0.a,
and so by the cancellation law of addition, 0 • a = 0 for each aeR
(d) Assume a + x = 0. Since a + (—a) = 0, we see that a 4- * = a 4- (—a), and
so, by the cancellation law we have established in (b) above, x — —a\ the negative
of a.
(e) If a 4- z = a 4- v = fr, then by the cancellation law, we get z = y. Thus, given
a and fr, the equation a 4- * — b has at-most one solution x e R. Since
0 + [>+(-a)] = (a+b)+(-a) = (-fl) + (fl+ft) = [(-a)+fl)]+fe = 04-6 = 6,.
we see that the only solution of the equation a 4- x = b is x = b -f (—a). We
denote this number by b — a and call it the subtraction of a from b.
(f) A close look at the equation a 4- (—a) = (—a) 4- a = 0 guarantees immediately
that —(—a) = a. Moreover, from
a+bH-a-b) = fl+ft+[-fl+(-«] = [(*+«+(-a)]+(-&) = *+(-« = 0,
we easily infer that — (a 4- b) = — a — b.
Problem 3.6. This problem introduces {by steps) the familiar process of division
in the framework of the axiomatic foundation of real numbers.
a. Show that the element 1 is uniquely determined, i.e., show that ifl*-x — x
for all x e R and some V e R, then 1* = 1.
b. Show that the cancellation law of multiplication is valid, i.e., show that
x • a = x • b with x ^ 0 implies a = b.
c. Show that for each real number a ^ 0 the real number a~l is the unique
real number that satisfies the equation x • a = 1. The real number x = a ~{
is called the inverse (or the reciprocal) of a.
d. Show that for any two given real numbers a and b with a ^ 0, the equation
ax = b has a unique solution, namely x = a~xb. The division operation
-r (or I) of R is now defined by b -4- a = a~lb; as usual, the real number
b + a is also denoted by b/a or £.
e. For any two nonzero a, b e R show that (a-1)"1 = aand(ab)~{ = a~lb~{.
f. S/zow r/?af y = a/<?r each a, £ = 0/or eac/z b ^ 0, and ^ = 1 /or eac/z
Solution, (a) Assume that some real number 1* satisfies 1* • x = x • 1* = a* for
each x e R. In particular, letting x = 1, we get 1* • 1 = 1. Since y • 1 = y for
all y 6 R, letting y = 1* yields 1* = 1* - 1 = 1. So, 1 is the only real number r
which satisfies r • x = x for each a g R.

14
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
(b) Assume x • a = x • b with x ^ 0. By Axiom 7 there exists a real number
y e R such that y -x = 1. Now, observe that
a = 1 . a = (yjc)a = y(;ttf) = yCx'6) = (yjc)6 =1-6 = 6.
(c) If <z;c = ay = 1 with a ^ 0, then by (b), we must have x = y. This shows
that the reciprocal a-1 of a is uniquely determined.
(d) To see that the equation ax = b with a ^ 0 has at-most one solution jc,
notice that if ax = cry = 6, then by the cancellation law of multiplication, we
have x = y. Moreover, notice that
fl.(fl-16) = (a-fl-1)6= 1-6 = 6.
The above show that the equation ax = 6 with a ^ 0 has the unique solution
x = fl"16.
(e) If a ^ 0, then the equation a • a"1 = 1 readily says that (a~])~] = a. In
addition, from
(aft) • (6"1a-1) = a(b • 6"1)fl-1 = a • 1 . a~x = 1,
we easily obtain (ab)"1 = 6*~1<2~I.
(f) Since 1 • a = a, we obtain y = a for each a e R. The equation 6-0 = 0
also implies that jj = 0 for each 6^0. From a • 1 = a, we get immediately f = 1
for all a # 0.
Problem 3.7. Establish the following familiar properties of real numbers using
the axioms of the real numbers together with the properties established in the
previous two problems.
i. The zero product rule: ab = 0 if and only if either a = 0 or 6 = 0.
ii. The multiplication rule of signs: (—a)b = a(-b) = -{ab) and (—a)
(-b) = ab for alia, b € R.
iii. The multiplication rule for fractions: For 6, d ^ 0 and arbitrary real
numbers a, ewe have
a c ac
b'd^bd'
In particular, (ff # 0, then (f)-' = £.
iv. The cancellation law of division: If a ^ 0 andx ^ 0, f/zefl j£ = \for
each real number 6.
v. The division rule for fractions: Division by a fraction is the same as

Section 3: THE REAL NUMBERS
15
multiplication by the reciprocal of the fraction, i.e., whenever the fraction
| -T- ^ ™ defined, we have
a c ad ad
b d be be'
Solution, (i) We already know from the previous problem that 0 • b = 0 for each
b G R. On the other hand, if ab = 0 (= a • 0) and a ^ 0, the cancellation law of
multiplication shows that b = 0.
(ii) Clearly,
ab + (-a)b = [a+(-a)]b=0 - b=0 and ab+a(-b) = a[b+(-bj\ = a.0 = 0f
and so — (ab) = (—a)b = a(—b). This implies
(-*)(-« = ~[a(-b)] = -[-(aft)] = aft.
(iii)If&,d 5*0, then
«(?•§)-KM'•§)-«•
and this shows that fd = f • §. Since f • f = ^f = 1, we see that (f)"1 = ^.
(iv) If c = -, then ac — b and so (a.r)c = &a* for each -v ^ 0, which shows that
c = - = —. a
(v) Notice that the identity
c ad adc a
d be dbc b
guarantees |-f = ^ = f • f.
Problem 3.8. 77z/.y problem establishes that there exists essentially one set of
real numbers that satisfies the eleven axioms stated in Section 3. To see this, let R
be a set of real numbers (i.e., a collection of objects that satisfies all eleven axioms
stated in Section 3 of the text).
a. Show that 1 > 0.
b. A real number a satisfies a = —a if and only ifa = 0.
c. // n = 1 4-1 H hi (where the sum has "n summands" all equal to 1),
then show that these elements are all distinct; as usual, we shall call the
collection N of all these numbers the natural numbers oflR.
d. Let Z consist of IN together with its negative elements and zero; we shall
call Z, of course, the set of integers o/R. Show that Z consists of distinct
elements and that it is closed under addition and multiplication.

16
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
e. Define the set Q of rational numbers by Q = {^: m,n e Z and n ^ 0}.
Show that Q satisfies itself axioms 1 through 10 and that
a = sup{r eQ:r<fl) = inf{s e Q: a <s}
holds for each a e R.
f. Now, let R' be another set of real numbers and let Q' denote its rational
numbers. If V denotes the unit element o/R', then we write
for the sum having "n-summands" all equal toV'. Now, define the function
f:Q^Q! by
f(Ul\ 2*1
J ^ n ' ri
and extend it to all o/R via the formula
f(A) = sup{/(r): ;• < a}.
Show that R and R' essentially coincide by establishing the following:
i. a < b holds in R if and only if f(a) < f(b) holds in R'.
ii. / is one-to-one and onto.
iii. f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b)foralla,b€R.
Solution, (a) Since 1 ^ 0, we have two possibilities: either 1 > 0 or 0 > 1. If
0 > 1, then (by Axiom 9) we have 0 + (-l) > l + (-l) = 0or-l > 0, i.e.,-1
is a positive number. Now, using Axiom 10, we infer that 0 • (—1) > 1 • (—1), or
0 > — 1, contrary to — 1 > 0. Hence, 1 > 0.
(b) Since 0 4- 0 = 0, we know that —0 = 0. Conversely, assume that a real
number a satisfies a = —a. This implies a + a = (1 + \)a = 0. However, since
1 > 0, we have 1 + 1 > 1 -f 0 = 1 > 0, and so 1 + 1 ^ 0. Consequently, from
the zero product rule, (1 -f- \)a = 0 implies a = 0.
(c) As shown in part (b) above, 1 + 1^0 and in fact 1 + 1 ^ 1; otherwise
1 + 1 = 1 = 1-1-0 implies (in view of the cancellation law) 1=0, which is
impossible. Now, by induction, assume that
0<l<l + l<l + l + l<...<l + l + .-. + l.
n-summands
We claim that the real number n + 1 = l + l-f--- + l-f-l(a sum of n + 1
summands) satisfies /i + 1>/2 = 1 + 1 + --- + 1 (where the last sum has n
summands). Indeed, if n + 1 < n, then (n -f 1) + (—n) < n -f (—n) or 1 < 0,
which is a contradiction. Hence, n + 1 > n and the induction is complete.
(d) By part (c) we know that the natural numbers together with zero are all
distinct real numbers. If —m = —n with m, n e IN, then m—n, which shows that
distinct natural numbers have distinct negatives. If m = —n with m, n e IN, then

Section 3: THE REAL NUMBERS
17
m + n = 0 contradicting (c), and so no natural number can be equal to a negative
integer. It now follows that Z consists of distinct elements.
(e) Observe that if - and £ are two rational numbers, then
m p mq + np m p mp
n q nq n q nq '
and if j ^ 0, then (j)~ = ^. That is, Q is closed under addition, multiplication
and inverses. Since all real numbers satisfy axioms 1 through 10, it follows that
Q itself satisfies axioms 1 through 10 in its own right.
For the second part, fix a e 1R and let A = [r e Q: r < a}. Since there exists
a rational number between a — 1 and a (see Theorem 3.4), A is nonempty, and
clearly A is bounded from above by a. By the Completeness Axiom (Axiom 11),
sup A exists in R and satisfies sup A < a.
Now, let € > 0. By Theorem 3.4, there exists some rational number r such
that a — e < r < a. Clearly, /• e A, and so a — € < sup Ay or a < sup A + e,
holds for all 6 > 0. This implies a < sup A, and hence a = sup A. The equality,
a = inf[s e Q: a < s] can be proven in a similar manner.
(f) Notice that the mapping is well defined. That is, if ^ = £ in Q, then
f(^) = /(-). Indeed, since j = ^ is equivalent to mq = np, we see that
m'q' = riq' or ^ = ^. Now, let us verify properties (i), (ii), and (iii).
i. a < b holds in R if and only if f(a) < f(b) holds in R'.
Note first that two rational numbers r, s e Q satisfy r < s if and only if r' < s'.
Indeed, to see this it suffices to assume that r and s are positive rational numbers
(why?). We have
m p tiii i m' p' i
r = — < s=- «=> mq < np <=> mq < nq <$=> r —— < ~ = s .
n q n' q'
Now, let a < b. Then {/• € Q: r < a] c. {s e Q: s < b], and from this it
easily follows that f(a) < f(b). For the converse, assume that f(a) < f(b). If
a < b is not true, then we must have b < a. But then, by Theorem 3.4 there
exist two rational numbers r,s € Q such that b < r < s < a. This implies
f(b) < r' < s' < /(a), a contradiction,
ii. / is one-to-one and onto.
To see that / is onto, let a! e R'. Then by Theorem 3.4,
a' = sup{r e Q': t < a'}.
If, we let S = {r e Q: r' < a1}, then this set is bounded above in R (why?) and
so a = sup S exists in R. Moreover, notice that
{/ <= Q': t < a'} = {/-': r e Q and r < a).
Now, it is easy to see that f{a) — a1.

18
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
To verify that / is one-to-one assume f{a) = f(b). Then by part (i) we have
a < b and b <a, i.e., a = b.
iii. f(a +b) = f(a) + f(b) and f(ab) = f(a)f(b) for all a, b e R.
We verify the additivity property only and leave the multiplicative property
for the reader. Clearly, f(r + s) = f(r) + f(s) holds for all rational numbers
;*,s. Now, fix <2,b e R and assume r,s € Q satisfy r < a and s < b. Then
/(r) = r' < f{a) and f(s) = s' < f{b). Since r -f s € Q and r + s < a + b, we
see that /(r) + /(*), = r' + j' = /(r + s) < f(a + 6). This easily implies
/(*) + /(«</(a+ 6).
For the reverse inequality, let ef > 0 in R'. Then there exist rational numbers
r,s e Q with r < a and s < b such that f(q) - €f < f(r) and /(Z?) - e' < /(^).
Since r+s < a+ Z?, it follows that/(a)-f f(b)-2ef < /(r) 4-/(5) = /(r+s) <
/(a + 6) for each 6; > 0. This guarantees /(a) 4- f(b) < f(a + b), and therefore
/(fl + « =/(«) + /(«.
Problem 3.9. Consider a two-point set R = {0, 1} equipped with the following
operations:
a. Addition (+): 0 + 0 = 0, 0 +1 = 1 + 0 = 1 and 1 + 1 = 0,
b. Multiplication (•) : 0 • 1 = 1 • 0 = 0 and 1 • 1 = 1, and
c. Ordering: 0 > 0f 1 > 1 and 1 > 0.
Does R with the above operations satisfy all eleven axioms defining the real
numbers? Explain your answer.
Solution. It satisfies all axioms except Axiom 9, which states that:
• lfx > y and z > 0, then x -f z > y + z.
To see this, assume that Axiom 9 is valid. We distinguish two cases.
CASE I: 1 > 0.
In this case, we must have 0=1 + 1 > 0 + 1 = 1, which contradicts 1 > 0.
CASE II: 0> 1.
This implies l=0+l>l-fl=0, which again contradicts 0 > 1. Thus,
Axiom 9 does not hold in this case.
It should be noticed that Axiom 9 is the one that guarantees that 1 -f 1 (i.e, the
number 2) is distinct from 0 and 1; and, of course, it is the axiom that establishes
(as we saw in part (b) of Problem 3.8) the existence of the set of integers.
Problem 3.10. Consider the set of rational numbers Q equipped with the usual
operations of addition, multiplication, and ordering. Why doesn't Q coincide with
the set of real numbers?

Section 3: THE REAL NUMBERS
19
Solution. The set of rational numbers satisfies all the axioms of real numbers
except the completeness axiom. This was proven in part (e) of Problem 3.8. To see
that Q does not satisfy the completeness axiom, assume by way of contradiction
that it does. Consider the set
S = [0<r e Q: r2 <2}.
Then S is nonempty and bounded from above in Q (why?), and so b = sup S exists
in Q. Now, repeat the proof of Theorem 3.5 to conclude that b2 = 2, i.e., that
b = V2. However, we proved in Problem 3.3 that \fl is not a rational number,
and we have reached a contradiction. Hence, Q does not satisfy the completeness
axiom and it cannot coincide with the set of real numbers.
Problem 3.11. This problem establishes the familiar rules of(t exponents" based
on the axiomatic foundation of real numbers. To avoid u?inecessary notation,
we shall assume that all real numbers encountered here are positive—and so by
Theorem 3.5, all non-negative real numbers have unique roots. As usual, the
"integer" powers are defined by
all=a-a — -a, a0 = 1, al=a, and a~n = —.
• v ' an
n- factors
Extending this to rational numbers, for each my n e IN we define
a „/—_,_« 1 1
a" = \Jam and a » = —^ =
an yam
Establish the following properties:
a. a* ={l/a)mforallm,n e IN.
b. Ifm, ny p, q € N satisfy j = E, then a^ — at.
c. ///* and s are rational numbers, then:
i. aras =ar+s and £ = ar~s,
ii. (ab)r = arbr and"(g)r = £, and
iii. (ar)s =ars.
Solution. It should be noticed first that (an)m = (am)n = amn for all a e R and
all natural numbers m, n e IN.
(a) Notice that
[c^r]" = (^r-(^r---(^r = ^-^-^
n-factors mn-f actors
= {lfc)n • (#0" • ■ • Qjfr)" = a • a • - a = am ■
m-factors m-factors
Since the /?th-roots are unique (Theorem 3.5), we infer that (j/a)m = tfa™ = a?.

20
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
(b) Assume m,n, p,q eN satisfy j = & or pn = mq. Using part (a), we
see that
(a?)" = (tf*)K = [U/Zy]" = {■&)"" = (^5)"* = [M^]" = a",
and this shows that a < = a».
(c) The formulas can be established easily if r and J are integers. Now, let r and
s be rational numbers. We shall assume that r and s are also positive and leave
the "negative case" for the reader. By part (b), we can also suppose that r = &
and s = £. Since (^ • 7^)n = (^)n • ("s/fe)'1 = ab, we see that Vab = f/ay/b.
Now note that
i. aras = ^/a^j/aP =j/a^ = ^am+P = a^ = ar+*.
ii. (afc)r = y&bT = ^b^ = (ffi)(Vb^) = ^y.
iii. (aT = y(y/a™)P = y^/ar™ = "V^"" = a - = flr5.
We leave the remaining cases for the reader.
4. SEQUENCES OF REAL NUMBERS
Problem 4.1. Show that if\x\ < 1, then lim a" = 0.
Solution. Let xn = |jc|w for each a. Then jc,,+i = \x\xn holds for each /i, and
the assumption |a*| < 1 implies 0 < xn+\ < xn. By Theorem 4.3, a = limA„
exists. It follows that a = a\x\ (or (1 — \x\)a = 0) must hold, and from this that
A direct way of proving that lim a" = 0 goes as follows. Observe first that we
can suppose 0 < a < 1. Now, if e > 0 is given, then note that
a-" < € <=» ln(jcn) = n\nx < lne <*=> n > .
In a
Problem 4.2. Show that limx,, = x holds if and only if every subsequence of
{xn} has a subsequence that converges to x.
Solution. If lim xn = a, then every subsequence must converge to x. So, every
subsequence of a subsequence (as being itself a subsequence of [xn}) must converge
to A.
For the converse, assume that each subsequence of [x„} has a subsequence
that converges to x . Now, suppose by way of contradiction that [xn} does not
converge to a. Then for some e > 0 we must have \x — x„\ > s for an infinite
number of n. So, there exists a subsequence [yn] of [xn] such that \x — yn\ >e
for each n. However, the latter contradicts the fact that [yn] has a subsequence
that converges to x. Therefore, lim xn = a.

Section 4: SEQUENCES OF REAL NUMBERS
21
Problem 4.3. Consider two sequences {kn} and [mn] of strictly increasing
natural numbers such that for some t e IN we have
{£,£ + l,£ + 2,...} c {k{,k2,...}U{mum2,...}.
Show that a sequence of real numbers {x„} converges in R if and only if both
subsequences {x^} and {xmJ of {xn} converge in R and they satisfy Yimx^ =
limjcWrt (in which case the common limit is also the limit of the sequence).
In particular, show that a sequence of real numbers {xn} converges in R if and
only if the "even" and "odd" subsequences [x2n] and {x2n-\} both converge in R
and they satisfy limxin = lim^-i.
Solution. If xn -» *, then clearly xkn -> x and xm>t -> x. For the converse,
assume that Xk„ —> x and xnin --» x. Let € > 0. Choose some no G IN such that
|aX — x\ < € and \xnin — x\ < € for all n > ho- (*)
Put lo = max{£, £Wo, /w„0}, and we claim that
|jc„ — x\ < € for all n > £0-
To see this, let n > to. Then the assumption
{£f£ + l,£ + 2,...} C {*lf*2, ...}U{/wifm2f...}
guarantees the existence of some/* e IN such that kr = normr = n. Since r < nQ
implies kr < kno < to and mr < m,tQ < to, we see that r > no. Hence, either
xn = Xkr of xn = *Wr (with /* > no), and so from (•) it follows that \xn — x\ < e.
This shows that xn ->- jc.
The last part should be immediate from the above conclusion.
Problem 4.4. Find the lim sup and lim inf for the sequence {(— 1)"}.
Solution. We have liminf(— 1)" = —1 and limsup(— l)" = 1.
Problem 4.5. Find the lim sup and lim inf of the sequence [xn] defined by
X\ = 5, *2n = 5*2n-l. ^ X2n+[ = 5 + *2n far /I = 1, 2, . . . .
Solution. We claim that
n—1 /t
*2/i = 32 2^ F an(^ ^n+l = 3 2^ 3T
£=0 Jt=0
hold for az = 1, 2,... . The validity of the identities can be established by
induction. We shall establish the validity of the second identity and leave the

22
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
verification of the first to the reader. For n = 1, we have
A'3 = A'2-i + i = 3+-^2 =3 + 3^1 = 3 + 9 = 3 (I + 3) = 3 ^
Now, assume that xin+\ = \ ]Ca=o jt holds for some n. Then,
n n+\
— I _i_ v — lj.lv 1 _, 1 V^ 1 __ 1 V^ 1
*2(n+l)+l — 3 T *2(/i+l) — 3 + 3*2n+l — 3 + 9 2^ F ~ 3 Z^ ?'
and the induction is complete. Consequently,
00 00
^ = ^^ = ^•1 = 1 and HmA'2n+1 = !££ = i.
Now, we claim that £ and ^ are the only limit points of {a„}. To see this, let
a be a real number different from £ and 5. Pick some e > 0 such that
(a - £, a + e) n (± - e, ± + s) = 0 and (a - e, a + e) n (5 - £, 5 + e) = 0.
Next, note that there exists some k such that A2„ e (£ — e, g -f- e) and A2„+i €
(5 — e, 5 4- e) hold for all a > /:. Therefore, |a„ — a\ > e holds for all n > fc,
and this shows that a cannot be a limit point of the sequence {*„}. Consequently,
lim inf a„ = \ and limsupA„ = ^.
Problem 4.6. Lef (a„ } be a bounded sequence. Show that
lim sup(—xn) = — lim inf a„ tffl<tf lim inf(—x„) = — iim sup x„.
Solution. We shall use the fact that lim sup xn and lim inf x„ are the largest and
smallest limit points of {*„}, respectively. We shall establish the first formula.
Choose two subsequences [yn] and [zn] of {a„} such that limy,, = lim inf xn
and lim(—z„) = limsup(—a„). Then
— lim inf xn = lim(—yn)
< limsup(—a„) = lim(—z„) = — limz„
< —lira inf xn,
and so lim sup(—xn) = — lim inf xn.

Section 4: SEQUENCES OF REAL NUMBERS
23
Problem 4.7. // {xn} and [yn} are two bounded sequences, then show that
a. lim sup(A„ + y„) < lim supxn 4- lim sup yn, and
b. lim inf(A'„ 4- yn) > lim inf xn 4- lim inf y„.
Moreover, show that if one of the sequences converges, then equality holds in both
(a) and (6).
Solution, (a) By passing to a subsequence, we can assume that lim(jt„ 4- yn) =
limsup(-Y„ 4- yn)> Since [xn] is a bounded sequence, there exists a subsequence
{xk„} that converges. Let x = Yimx^. By the same reasoning, there exists a
subsequence of {y^} that converges to some y. Thus, there exists a strictly
increasing sequence {mn} of natural numbers such that x = lim;cm;i and y —
limymn. Hence,
lim sup(jr„ 4- y„) = x 4- y = lim xt„n 4- lim y„H < lim sup x„ 4- lim sup y„.
Finally, if x = Y\mxn holds, then pick a subsequence [ykn] of {yn} such that
lim ykn = lim sup yny and note that
lim sup A*„ 4- lim sup yn = x 4- lim yA/i
= lim(AA„ 4- X) < limsup(A„ 4- yn).
(b) It follows from (a) by using the preceding problem.
Problem 4.8. Prove that the lim sup and lim inf processes "preserve
inequalities!' That is, show that if two bounded sequences [xn] and {yn} of real numbers
satisfy x„ < yn for all n > no, then
lim inf a„ < lim inf y„ and lim sup xn < lim sup yn.
Solution. First, we shall show that if two sequences of real numbers {sn} and {/„}
converge in R (say sn -> s and tn —> t) and sn < tn for each n > no, then s < t.
Indeed if, s > t is true, then let € = ^ > 0 and note that for all n sufficiently
large, we must have
sne{s-€,s + €) = (!±!;l¥) and tn e (t - €, / + e) = (^, ^).
That is, r„ < ^ < j„ must hold for all n sufficiently large, which is impossible.
Hence, s < t.

24
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Now, assume that two bounded sequences of real numbers [xn] and [yn] satisfy
*n < yn for all n > hq. Put
sn = inf xk and /„ = inf yk.
k>n k>n
If n > no, then notice that for each r > n we have sn = infjt>„0 xk < xr < yr,
and so s„ < infr>„ yr = /„ for each n > no. By the discussion of the first part, we
infer that
liminfjc„ = \imsn < linw„ = liminfy,,.
The lim sup case can be established in a similar manner, or by using the formula
lim sup*,, = — liminf(—xn).
Problem 4.9. Show that lim j/n = 1 (and conclude fivm this that lim j/a = 1
for each a > 0).
Solution. Note that j/n = fflfn ) . An easy inductive argument shows that
vVw > 1 holds for each n. Thus, we can write ^PJn = 1 -f xn with a„ > 0.
Since (1 -f a)n > 1 -f na holds for each n and each a > 0 (see Problem 2.13),
we get
v^ = (J7Z)n = 0 + xnf > 1 + nxn%
and so 0 < xn < 4= — £. This implies lim xn = 0. Therefore,
^=(</>)2 = (l+^)2^ 1.
An alternate proof goes as follows: By L'HopitaTs Rule, we have lirnx_0o ^ =
0, and so lim,,-^ ^ = 0. Therefore, using that the exponential function is
continuous, we infer that
lim rfn = lim e~ = e° = 1.
/J—fOO «—*-oo
For the parenthetical part, assume first a > 1. Then it is easy to see that 1 <
j/a < tfh~ holds true for all n > a. Consequently, by the "Sandwich Theorem,"
we see that lim j/a = 1. If 0 < a < 1, then £ > 1, and so lim .//£ = lim-jW = 1,
from which it follows that lim tfa = 1 holds true in this case, too.

Section 4: SEQUENCES OF REAL NUMBERS
25
Problem 4.10. // {xn} is a sequence of strictly positive real numbers, then show
that
liminf —— < liminfrfx~n < limsup j/x^ < limsup ——.
/i—oc x„ «—oo „_¥QO n_ifQO Xn
Conclude from this that if lim ^ exists in JR, then lim j/x^ also exists and
limyx^ = lim -~i.
Solution. Let [xn} be a sequence of real numbers such that xn > 0 holds for
each n. We shall establish lim sup j/x^ < lim sup ^ and leave the similar proof
of the other inequality for the reader. Put
v OO 00
jczrhmsup -—= /\V—-,
X« n=\k=n *k
and note that if .v = oo, then there is nothing to prove. So, we can assume x < oo.
Let s > 0 be fixed. Then there exists some k such that —^ < x 4- £ holds
for all h > /:. Now, for /7 > k we have
-Y» = ^'^!t •" ^'^ <(* + e)fl~*** = (* + 0,,c,
where c = -r*(;t 4- £)~A is a constant. Therefore, ^/!x^ < {x 4- £)■</? holds for
each n > k and so, in view of lim rfc — 1 (see Problem 4.9) and Problem 4.8,
we infer that
lim sup ^/x^} < lim sup(„t 4- £)Vc = (x -f e) lim j/c = x 4- £.
/J-+CO H-^OO /l-*00
Since £ > 0 is arbitrary, we infer that lim sup j/xH < x = lim sup ^.
Problem 4.11. The sequence of averages of a sequence of real numbers {xn} is
the sequence {an} defined by an — xt+xi+-+x\ If{xn) is a bounded sequence of
real numbers, then show that
lim inf x„ < lim inf an < lim supan < lim supxn.
In particular, if xn —> x, then show that an —► jr. Does the convergence of [an]
imply the convergence of{xn}?

26
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Solution. The solution will be based upon the following properties of lim sup
and liminf:
• If {un} is a bounded sequence of real numbers, then for each e > 0 the
inequalities
w; > lim sup un 4 € and um < lim inf un — e
hold for finitely many k and finitely many m.
To see this, assume by way of contradiction that uj, > lim sup u„ 4 € holds true
for infinitely many k. Then there exists a subsequence [vn] of [un] satisfying
vn > lim sup un 4 € for each n. Since {vn} is a bounded sequence, there exists
a subsequence [wn] of {vn} (and hence of {un}) satisfying wn -* w € 1R. By
Problem 4.8, we know that w > lim sup un 4 €, i.e., w is a limit point of {un}
which is greater than the largest limit point (lim sup un) of [u„}, a contradiction.
Now, let {xn} be a bounded sequence of real numbers and fix € > 0. Put
I = limsupjcn and let K = {A- € N: jc* > £ 4- e}. By the above discussion, K is
a finite set. Put
Sn = {/ 6 IN: / € if and i < /*} and 7W = {/ € N: / $ K and / < /*},
and define the sequences {sn} and {/„} by
**» = YlXi and /fl = YlXi'
ieSn ieT„
Clearly, {sn} is an eventually constant sequence, tn < n(£ 4 e) holds for each n
and an = — 4- ^. Since s,,//? —> 0 and tn/n < i. + e for each h, it follows from
Problems 4.7 and 4.8 that
limsup<3n = limsup(^ 4 *f) = lim ^ 4 lim sup ^ = lim sup ^ < £ 4e.
Since £ > 0 is arbitrary, we get limsup<z„ < £ = lim sup a'„. Similarly, liminf
A'„ < lim inf a„. If*,, —► jc, then a = lim inf xn = lim sup xn, and so a =
lim inf an = lim sup g„. This implies an -* jc.
The convergence of the sequence [an} of averages does not imply the
convergence of {xn}. For instance, if xn = (—1)", then an —► 0 while {xn} fails to
converge.
Problem 4.12. For a sequence of real numbers [xn] establish the following:
a. // xn+\ — xn —► x in R, then xn/n ->■ jc.

Section 4: SEQUENCES OF REAL NUMBERS
27
b. If{xn] is bounded and 2xn < xn+\ +xn-\ holds for all n — 2, 3,..., then
xn+\ -xn t 0.
Solution, (a) Assume that xn+\ —xn -> .r in R. Notice that ^=1U/+i -xt) =
xn+\ — X\ for each n. By Problem 4.11, we have
n
/=1
Since x\/n -» 0, it follows that xn+\/n -» *. Now note that
a /i — 1 /?
(b) The condition 2*„ < jcn+i + Jf„_i can be rewritten as xn — xn—\ < .X/j-f i — JT/j
for each n = 2, 3,..., which implies that the bounded sequence {xn+\ — xn] is an
increasing sequence, and hence convergent. Let xn+\ — xn t x in R. By part (a),
wehaveA*n//? -* a\ But, since {.r„} is abounded sequence, jc„/a7 ~> 0. Therefore,
x = 0, and so a*„+i - jc„ | 0.
Problem 4.13. Consider the sequence {xn} defined by 0 < x\ < 1 tfrtd *„+1 =
1 — «J\ — xnfor n = 1, 2,... . S/iovv f/zof .t„ | 0. Also, show that ^f1 -> \.
Solution. We claim that
0 < xn+l < xn < 1 (*)
holds for each n — 1, 2,... . To verify this claim, we use induction. Since
0 < x\ < 1, wehaveO < 1 — x\ < 1, and soO < 1 —x\ < Vl — x\ < 1. Hence,
0 < 1 — V* — *i = *2 < jci < 1. That is, (*) is true for n = 1.
For the inductive argument, assume that (•) is true for some n. This implies
0 < 1 — xn < 1 — xn+\ < 1, and so 0 < Vl — xn < VI — xn+\ < 1, from which
it follows that
0 < Xn+2 = 1 - y/l ~ Xn+{ < 1 - y/\ -Xn = Xn+] < 1,
which shows that (•) is true for n -f 1. This completes the induction and guarantees
that (*) is true for each n.
Now, since [xn] is decreasing and bounded from below, it converges, say to
x g R. Clearly, 0 < .r < 1. Moreover, we have
x = lim xn+\ = lim (l - V1 — xn) — 1 — Vl — x.
In other words, x is the non-negative solution of the equation x = 1 — Vl —-*.
Solving the equation yields x = 0 or x = 1. Hence, x = 0, and so xn I 0.

28 Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
For the last part, notice that
xn+x i - yi - xn i \_
*n *n 1 + -v/1 - JL-„ 2'
and the solution is complete.
Problem 4.14. Show that the sequence {xn} defined by
is a convergent sequence.
Solution. From the binomial expansion:
*. = o+i)"=Eo±=i+i;o±
i=0 1 = 1
n
_ I _|_ V^ n(n-\)-(n-i+\) # J_
i = l
= »+i;Ao-i)(i-j)-o-¥)
/=i
i = 1
/i i 1 \n+]
= (i + ^+r) = *»+i.
Thus, xn t holds. Also, note that for n > 2 we have
/=! i=2 i=2
By Theorem 4.3, {xn} converges. (Of course, limx„ = e = 2.718 • • •.)
Problem 4.15. Assume that a sequence {xn} satisfies
/or /z = 2, 3,... tfrtd some fixed 0 < a < 1. Show that {xn} is a convergent
sequence.

Section 4: SEQUENCES OF REAL NUMBERS
29
Solution. Let c = |.i*2 — aj |. An easy inductive argument shows that for each n
we have |a„+i - xn\ < can~l. Thus,
p p
\x„+p-xn\ < ^\x«+i ~ *«+/-! I ^ cJ2<*nH~2 ± T=Zan~l
/=i
1=1
holds for all n and all p. Since lima" = 0, it follows that [xn] is a Cauchy
sequence, and hence, a convergent sequence.
Problem 4.16. Show that the sequence {xn}, defined by
a'i = 1 and xtt+\ = — for * = 1, 2,...,
3 + x„
converges and determine its limit.
Solution. Clearly, xn > 0 holds for each n. Now, note that
|A„+1 -A„| =
3+v„ 3+v„_,
_ \x„-.xn-\\ <• ]ir __ v I
"" (3+.v(l)(3+.r<l_,) - 9|A" xn-\\
holds for /7=2, 3 By Problem 4.15, the sequence {jc„} converges. If limA„ = a,
then a > 0 and
a = limA„+i =
1
1
3-flimA,, 3+a
Solving the equation, we get x = 3"y 13.
Problem 4.17. Consider the sequence {xn} of real numbers defined by x\ = 1
and a„+i = 1 + y—- for n = 1, 2,... . Show that {a,,} is a convergent sequence
and that lirnA,, = V2.
Solution. An easy inductive argument shows that xn > 0 for each n. This implies
that, in fact, we have 1 < xn < 2 for each n. Now, note that
\-xn+\ — *n\
1
1
1+Xn 1+An_i
_ \xn -X„-\\
~ (l+.rn)(l+-r„-i)
\Xn —Xn-\\ _ . ^
- (1 + 1)(l + 1)"4l^ ^-ll
for each n = 2, 3 By Problem 4.15, the sequence {a„} converges. Let

30
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
xn -+ x. Since xn > 1 for each /2, we see that x > 1. Then
A' = lim xn+} = lim (1 + — ) = 1 + ——.
That is, x is the positive solution of the equation x = H-y~r,orA2-f a* = 1-j-A-hl.
This implies x2 = 2, and so x = y/l.
Problem 4.18. Define the sequence [xn] by x\ = 1 and
xw+i = -(*„ + —)f w = l,2,....
Show that [x„] converges and that limxn = yfl.
Solution. Clearly, xn > 0 holds for each n. (Use induction to prove this!) Also,
holds for each n. Thus, if n > 2, then
and so 0 < xn+\ < xn holds for each n > 2. By Theorem 4.3, a = limAn exists.
Since x2 > 2 holds for each n > 2, we see that x > 0. From the recursive
formula, it follows that 2a = x -f- |, or x2 = 2. (Note also that the limit is
independent of the initial choice x\ > 0.)
Problem 4.19. Define the sequence xn = J2l=\ { for w = 1, 2,... . Show that
{xn} does not converge in 1R. (See also Problem 5.10.)
Solution. The inequality
Xin — xn = —rr -f —79 + h —rr
n+l ^ «+2
2n ^ 2n ^ " ~r In ~~ " ' 2n ~~ 2
> JL + J-4....4.J- =W.J- = I
shows that [xn} is not a Cauchy sequence, and hence, is not convergent in IR.
Problem 4.20. Let -co < a < b < 00 and 0 < X < \. Define the sequence

Section 4: SEQUENCES OF REAL NUMBERS 31
{xn} by x\ = a,xi = b and
xn+2 = Ajc„ + (1 - A.).rw+i for n = 1, 2,....
S/zow r/zar {*„} converges in R and find its limit.
Solution. Rewriting x„+2 = ^« + (1 — k)xn+\ = ^ -f *n+i — Xxn+] in the
form xn+2 — xn+\ = XU„ — x„+\), we see that
\Xn+2 ~" *n+l I = ^|-r/i+! ~ Xn \
holds for each n. Now, a glance at Problem 4.15 guarantees that [xn} is a convergent
sequence. However, we cannot get the limit of the sequence {jc„} by taking limits
in both sides of the recursive formula xn+2 = Xxn+(l—k)xn+\. We shall compute
the limit of the sequence [xn] using a different method.
For simplicity put \i = 1 — k. First, we shall verify that
X\ < x3 < • • • < x2n+\ < x2n < x2n-2 <•- <x2
holds for each n.
The proof is by induction. For n = 1, the inequalities reduce to x\ < x2 which
is obviously true. So, for the inductive step, assume X2n-\ < x2n for some n. Then
*2/i+I = ^x2n~\ + &x2n = *2n-\ + ^0*2/! ~ *2/i-l) > *2/i-l
and
*2/i+l = ^2«-l + M*2/i = *2/i — k(x2n ~ *2n-l) < *2n-
Now, note that
*2/i+l < kX2n-\ + (1 — k)x2n = *2n+2 < *2/i-
Next, if we let d„ = *2„ — *2/i-i »then it is easily to verify (see Figure 1.1) that
dn+] = A.a^„, (1)
X2n+\ = *2n-\ + Mm and (2)
*2/i+2 = *2/i -^ dn. (3)
From (1) it follows that
rfn = (X^),,-1d1=(XAi),,",(6-fl),

32
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
'dn
'Vdnm
•Mn-
-\\idn
-X2dn
FIGURE 1.1.
and so from (2) and (3), we obtain
n n
*2n+\ = X\ + ]T(*2/+1 ~ *2/-l) = ^1 + X^M
/ = !
= jci + M 2_Jkny xdx = jci + —
M(&-fl)[l-(W]
1=1
A/x
= a +
1 -A/z
and
*2i
,n+2 = X2~ ^(*2i ~ JC2/+2) = *2 - X2 ]T(Axz)' Xdi
i=\
/=!
A2[l-(AM)w]di , *2(Z? - fl)[l - (A/z)"]
= a*2 : : = b —
1 -A/x
1 -A/x
Therefore,
fi(b-a) X2a + fjib , A2(fc - a) X2a + p,b
*2n+\ t fl + -1 :— = -^ :— and x2„ I 6 -
1 — A/x 1 — A/x
and consequently, limjc„ = ^1^.
1 — A/x 1 — A/x
Problem 4.21. Lef G Z?e a nonempty subset of R, w/z/c/z w <z group under addition
(i.e., if x, y e G, then x + y e G and —x e G). Show that between any two
distinct real numbers there exists an element ofG or else there exists a € R such
that G = [na: /i = 0, ±1, ±2,...}.
Solution. Assume G ^ (0). Let a = inf G fl(0, 00). We distinguish two cases.
(1) a > 0. In this case, we shall show that G = {na: n = 0, ±1,...}.

Section 4: SEQUENCES OF REAL NUMBERS
33
To see this, note first that a e G. Indeed, if a £ G, then there exist x, y e G
with a < x < y < ^. Then, the element z = y — x e G satisfies 0 < z < |,
contradicting the definition of a. Now, if x e G, then na < x < (n + \)a must
hold for some integer n. However, x = na. must also hold, since otherwise the
element x — na € G satisfies 0 < x — na < a, which is again a contradiction.
(2) (3 = 0. In this case, we claim that between any two distinct real numbers there
is an element of G.
To see this, we only need to consider 0 < x < y. Let 8 = min{;c, y — x] > 0.
Choose some element z e G with 0 < z < 8. By the Archimedean property,
the set A = {n e IN: nz > y] is nonempty, and by the Well Ordering Principle
the element k = mm A exists. Now, note that the element b = (k — l)z e G
satisfies x < b < y.
Problem 4.22. Determine the limit points of the sequence {cos n).
Solution. We claim that the set of limit points of the sequence {cos/?} is [—1, 1].
To prove this, we shall need two facts from elementary calculus.
a) The Intermediate Value Theorem; and
b) The inequality | cos.y — cos y\ < \x — y\ for all jc, y e JR.
Let G = {n -\-2mn: n, m integers}. Clearly, G is a group under addition,
and since n is an irrational number, it is easy to see that the group G is not of
the form [na: n = 0, ±1, ±2,...}. Now, let x <= [-1, 1] and let e > 0. By the
Intermediate Value Theorem, there exists some y eHR. satisfying cos y = x. The
preceding Problem 4.21 shows that there exist two integers n and m satisfying
y < n + 2mn < y + s. Thus,
\x — cosn\ = |cosy — cos(a2 + 2mn)\ < n +2mn — y < e.
The above arguments show that given x e [—1,1] and e > 0, there exists
some non-negative integer n with \x — cosn\ < e. From this, it easily follows
(how?) that every point of [—1, 1] must be a limit point of {cos/z}.
Problem 4.23. For each n define /„: [—1, 1] -» R by fn(x) = .r'1. Determine
lim sup /„ and lim inf /„.
Solution. We have
lim sup/„(.*) =
fl, if jc = —1
0, if |*| < 1
ll, if jc = 1

34
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
and
liminf/„(*)= <
[-1,
0,
, 1,
if jc = -1
if |jc| < 1
if x= 1.
Problem 4.24. Show that every sequence of real numbers has a monotone
subsequence. Use this conclusion to provide an alternate proof of the Bolzano-
Weierstrass property of the real numbers: Every bounded sequence has a
convergent subsequence. (See Corollary 4.7.)
Solution. Let [x„} be a sequence of real numbers. We consider the set of natural
numbers
S = [k e N: xk < xm for all m > it},
and distinguish two cases.
1. S is infinite.
In this case, we can write S = [k\ , &2i • • •} with k\ < kn < • ♦ •. Now, it should
be clear that the subsequence [x^] of [x„] is increasing.
2. S is finite (and possibly empty).
In this case, if we put k\ = 1 -f max S (let max S = 0 if S = 0), then for each
k > k\ there exists some m > k such that xm < jc*. So, by induction, if kn has
been chosen, then we can select some natural number kn+\ with kn+\ > kn and
Xkn+l < •**„• This implies that [x/,n] is a strictly decreasing subsequence of {xn},
and the claim is established.
For the Bolzano-Weierstrass property, notice that if [xn} is a bounded sequence,
then, by the above, {xn} has a monotone subsequence which (by Theorem 4.3) must
be convergent in R. (We remark that this result shows that not only a bounded
sequence has a convergent subsequence but it also has a monotone convergent
subsequence.)
5. THE EXTENDED REAL NUMBERS
Problem 5.1. Let {xn} be a sequence of JR.*. Define a limit point of{xn] in 1R* to
be any element x of M.* for which there exists a subsequence of{xn} that converges
to x.
Show that
limsupx,, = inf [sup *J and liminfA-„ = sup [inf **1
n Lk>n n L*>« J
are the largest and smallest limit points of{xn] in JR.*.

Section 5: THE EXTENDED REAL NUMBERS
35
Solution. The limsup case is established. Let x — lim sup *„ e R*. Then
three cases arise:
a) x e R.
In this case, repeat the proof of Theorem 4.6.
b) x = co.
In this case, we have only to show that x is a limit point of [xn]. Note that
V^„*/ = co for each n. Choose some k\ > 1 such that .r*, > 1. Now, by
induction: If kn has been selected so that xkn > n, then use \Ji>k xx = co to
choose some kn+\ > kn + 1 > kn so that xkn+i > n + 1. Clearly, {**„} is a
subsequence of [xn] satisfying limx^ = co.
c) x = -co.
In this case, we shall show that limjc,, = —co. Let 0 < M < co. From
\/°lnXi 4 —co, it follows that V/^rr< < ~M for some az, and so xx < —M for
all / > n. That is, lim.r,, = —co.
Problem 5.2. Let {xn} be a sequence of positive real numbers such that
I = lim — exists in R. Show that:
a. if I < \, then lim xn = 0, and
b. if I > \, then limx„ = co.
Solution, (a) Assume I < 1 and fix some 8 such that I < 8 < 1; for instance,
let 8 = -^n. Since lim ^ = £, there exists some £ > 1 such that *f± < 5 holds
for all az > /:. Now, if n > k, then note that
Xn_l JCn_2 Xk
< lJ^'Xk = Sn'kxk = (^)s\
(n—k)-terms
and so if c = |h then
0 < *„ < cfl"
holds for all n > k. Since (in view of 0 < 8 < 1) 8n -> 0, we easily infer that
xn -» 0.
(b) Assume now I > 1 and choose some 5 such that 1 < 8 < i. Since
lim ^ = I, there exists some k > 1 such that ^ > <5 holds for all a* > £. Then,
as in the preceding case, there exists some constant C > 0 satisfying xn > C8n
for all n > k. From <5" -* co, it easily follows that xn -> co.
Problem 5.3. Let 0 < an>m < co for all m, n, and let a: IN x IN -» IN x N te

36
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
one-to-one and onto. Show that
OO OO OO 00
Y^ iL, Qn>m = Yl IZ fl^(n,m).
/j=l m=l n=l m=l
Solution. It follows immediately from Theorem 5.4.
Problem 5.4. Show that
OO OO i
n—\ m=l
Solution. The convergence or divergence of the series is according to the
convergence or divergence of the double integral f^f^jr^i- Now, note that
jx^h^i;
— = oo.
y
An alternate solution goes as follows. Note first that the inequality
11 1 11
> r >
n2 + m2 (n+m)2 (n + m)(/i + w + l) n+m w-f/w + 1
implies XT=1 ^ > E~=,(^b " Jsbi) = sir- ™**™>
00 OO t OO i
n=\ m=l
Problem 5.5. This problem describes the p-adic representation of a real
number in (0, 1). We assume that p is a natural number such that p > 2 and
x e (0,1).
a. Divide the interval [0, 1) into the p closed-open intervals
[°.i).[J.J) [*?.!).
and number them consecutively from Otop — l. Then x belongs precisely
to one of these intervals, say k\ (0 < k\ < p). Next, divide the interval
[^-, ^y~) into p closed-open intervals {of the same length), number them
consecutively from 0 to p — 1, and let ki be the subinterval to which x
belongs. Proceeding this way, we construct a sequence {kn} of non-negative

Section 5: TOE EXTENDED REAL NUMBERS
37
integers such that 0 < kn < pfor each n. Show that
kn
X3
fl=i
Pn
b. Apply the same process as in (a) by subdividing each interval now into p
open-closed intervals. For example, start with (0, 1] and subdivide it into
the open-closed intervals (0, ±], (±, |],..., i^y, 1].
As in ia), construct a sequence {mn} of non-negative integers such that
0 < mn < pfor each n. Show that
E°° mn
, Pn'
c. Show by an example that the two sequences constructed in (a) and (b) may
be different.
In order to make the p-adic representation of a number unique, we shall agree to
take the one determined by (a) above. As usual, it will be written asx = 0.k\ k2 • • •.
Solution. For (a) and (b) note that |jc - £-=1 ^ | < \ holds for all n.
For part (c) take, for instance, p = 2 and note that for i = | we have
and kn = 0 if n > 1, while m\ = 0 and mn = 1 for n > 1.
Problem 5.6. Show that 'P(IN) ^IRby establishing the following:
i. If A is an infinite set, and f: A -» B is one-to-one such that B \ fiA) is
at-most countable, then show that A % B.
ii. Show that the set of numbers o/(0, I) for which the dyadic ii.e., p = 2)
representation determined by ia) and ib) of the preceding exercise are
different is a countable set.
iii. For each x e (0, 1), let x = Q.k\k2 — - be the dyadic representation
determined by part ia) of the preceding exercise; clearly, each kx is either
0 or 1. Let fix) = [n e N: *„ = 1}. Show that /:(0, 1) -* P(N)
is one-to-one such that P(N) \ /((0, 1)) is countable, and conclude from
part ii) that (0, 1) % PQti).
Solution, (i) Let S = [a\, a2,...} be a countable subset of A.
(a) Assume B \ fiA) = [b\,..., bn] is a finite set. Then g: A -> 5 defined
by £(*) = /(*) if x $ S, giai) = b{ for 1 < / < n, and £(fl/+„) = /(a,-) for
/ = 1, 2,... is one-to-one and onto.
(b) Assume B \ fiA) = {b\, bi,...} is countable. Then g: A -> B defined
by gix) = /Or) if * i S, gia2n+\) = /(a„) and g(a2J = 6« for each n is
one-to-one and onto.

38
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
(ii) Let D be the set of all numbers of (0, 1) for which the two sequences [kn]
and {mn} determined by the preceding problem are different. Assume x =
0.k\k.2 - - - = 0.m\m2 • • ■ € D and define the natural number r = min{w: kn ^
mn}. We can assume kr — 1 and mr = 0. Then the inequalities
r-\ oc oo
2"
n=\ n-r n=r+\
-X> = £f^£
r-1
-1
guarantee that x = -^ + |* H h ^5f •+- ~. In particular, note that mn = 1 and
kn = 0 hold for each n > r.
On the other hand, it is not difficult to see that every x of the above type belongs
to D. It is now a routine matter to verify that D is a countable set. (It is also
interesting to observe that D consists precisely of the endpoints of the subintervals
appearing during the construction of the expansions.)
(iii) Let A e P(N). Define the sequence [mn] of {0, 1} by m„ = 1 if n e A
and mn = 0 if n £ A, and then set
2"
Note that A $ /((0, 1)) if and only if x e D. Thus, V(ti) \ /((0, 1)) is
countable, and so by part (i) and the fact that / is one-to-one, (0, 1) ^ P(N)
holds.
Problem 5.7. For a sequence {xn} of real numbers show that the following
conditions are equivalent:
a. The series Y1%L\ x* ^s rearrangement invariant in R.
b. For every permutation ao/N the series JZ^li xan converges in IR.
c. The series J^lj \xn \ converges in R.
d. For every sequence [sn] of{—1, 1}, the series J2%L\ snxn converges in R.
e. For every subsequence {x^} of [xn], the series Y1T=\ x** converges in R.
f. For every € > 0, there exists an integer k (depending on e) such that for
every finite subset SofJN with min S > k, we have \ Yln<=s x*\ < €-
(Any series Y1T=\ x" satisfying any one of the above conditions is also referred to
as an unconditionally convergent series.)

Section 5: THE EXTENDED REAL NUMBERS
39
Solution. (a)=>(b) Obvious.
(b)=>(c) Assume Y1T=\ \x"\ ~ °°- From our hypothesis it follows that xn > 0
and xn < 0 both hold for infinitely many n. Split [xn] into two subsequences
{yn} and {z„} such that yn > 0 and z„ < O-hold for all n. We can assume that
Eoo
.=1 yn = 00.
Now, use induction to construct a strictly increasing sequence of natural numbers
[kn] such that
1. k\ = 1 and z\ -f 5Z/ii ^ > ^ ^
2. Zi. + Ejl^+i y/ > 1 for/i = 1,2
Then note that
y\ )>*,, zi, y*,+i, • ••> y*2, 22. j^+i* •••
is a permutation of [xn} whose series is not convergent, contrary to our hypothesis.
(c)=>(d) Obvious.
(d)=>(e) Let {**„} be a subsequence of {xn}. Put s, = — 1 if / ^ kn for each
n, and s^t = 1. Then
OO OO 00
n=\ n=l
is a convergent series.
(e)=>-(f) If (f) is false, then there exists some e > 0 and a sequence {S„} of finite
subsets of natural numbers such that maxS„ < minS„+i and |X}/<=s ■**! — £ no^
for all n. Let
Qs„ = {*i,*2,...},
where kn \ . Then, it is easy to see that the series J^Li ■**„ does not converge in
R, contradicting (e).
(f)=Ka) Let a: IN —> IN be a permutation. By our hypothesis, it is readily
seen that the partial sums of both series J2T=\ x" an<^ XwJli xon ^orm Cauchy
sequences, and hence, both series converge in JR. Let x = Y^Li xn ^d y —
Zw/I=l X(T„'
Now, if s > 0 is given, then choose k so large such that
r r
/i=i
n=l ieS
hold for all r > k and all finite subsets S of IN with min S > k. Fix some r > k

40
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
such that for each 1 < / < k there exists 1 < j < r with jc, = xap and note
that
\x-y\<
k
x - j^l -f \^xn - YlxA + Is**"y
n=l n=l n=l n=l
<£ + £ + £ = 3s
holds for all e > 0, and so jc = y. In other words, the series Y^L\ x» ls
rearrangement invariant.
Problem 5.8. A series of the form ]££Li(~" l)n~lxn, where xn > Ofor each n, is
called an alternating series. Assume that a sequence {xn} of strictly positive real
numbers satisfies xn I 0. Then establish the following:
a. The alternating series X^Jl^— l)""1^ converges in R.
b. If X^li xn = oo, then the alternating series Y^L\("~l)n~ljcn is not
rearrangement invariant.
Solution, (a) Let sn = Y!k=\(-1)*""1**- We claim that
S2 < S4 < • • • < 52n-2 < S2n < S2n-\ < $2n-3 < * ' " < $3 < $1
holds for each n. The proof is by induction. For n = 1, we have s2 = x\ — x2 <
X\ = S\. So, assume the inequalities to be true for some n. Then taking into
account that x2n — *2/j+i > 0 and x2n+\ — x2n+2 > 0, we see that
1. $2n < S2n + (x2n+\ — X2n+2) = S2n+2 = ^2(n+l),
2. $2(n+l) = $2n+2 = $2n+l ~" x2n+2 < S2n+\ = ^2(n-f 1)-|,
3. ^2(n+l)-l = s2n+\ = $2n-l — fen — *2«+l) < $2n-l,
and our claim is established.
Now, if s2n t s and s2n-\ I t hold in R, then clearly s < /. Moreover, from
sin — S2/i-i = —x2n -> 0, we obtain s = t. But then, this implies that [sn]
converges to s in R; see Exercise 4.3 of Section 4. Consequently, the alternating
series converges and X^jC—l)*-1** = lim^„ = s.
(b) We must have either YltLi *2*-i = oo or ]T]JLi X2n = oo. Assume
]C*Li xuc-i = oo; the other case can be treated in a similar manner.
Since YltL\ x2k-\ = oo, there exist integers 0 = ko < k\ < k2 < • • • such
that [Xi/^+i^-i] — x2n > 1 holds for each h = 0, 1 Consider the
rearrangement [yn] of the sequence {(— l)"-1*,,} given by
A'i, *3, . . . , *2*,-li —*2i*2*,+li • • . ,*2*2-l' "~ *4,*2Jt2+l>
and note that ££li ^ = oo holds.

Section 5: THE EXTENDED REAL NUMBERS
41
Problem 5.9. This problem describes the integral test for the convergence of
series. Assume that f: [1, oo) —> [0, oo) is a decreasing function. We define the
sequences {an} and {t,,} by
cr,, = ][]/(*) and rn= f{x)dx.
Establish the following:
a. 0 < on - r„ < f(\)for alln.
b. the sequence [an — xn] is decreasing—and hence, convergent in R.
c. Show that the series Y?kL\ /W converges in R if and only if the improper
Riemann integral /,°° f{x)dx = linv^oo J[ f(x)dx exists in R.
Solution. Since / is decreasing, notice that for each kNwe have f(x) >
f(k +1) and f(x) < f{k) for each/: < x < k+\. So, integrating over [£, k + 1],
we get:
pk+\
I fWdx > /(*+l), and (1)
pk+\
J f(x)dx < f(k) (2)
for each k = 1, 2,... . (We remark that as a decreasing function, / is Riemann
integrable over every closed subinterval of [1, oo); see Section 23 of the text.)
(a) Using (1), we see that
n n—\
°n = /(I) + £ /(*) = /(l) + J2 M + l>
*=2 *=l
< /(l) + E f /Wrf-r = /W + f 'fMdx
= /(D + r„.
This implies ct„ — t„ < /(l) for each n. On the other hand, using (2), we see
that
'1-1 1-1 pk + ] nil
On > T f(k) > T / f(x) dx = / f(x) dx = T,„
and so cr„ — zn > 0 for each n.

42
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
(b) Using once more (1), we get
cr;H_i - t/i+i = on + f{n + 1) - r„ - / f{x)dx
Jn
= CTn-Tn-U f(x)dx - f{n + 1)J < on - r„.
This shows that [an — t„) is a decreasing sequence—and so lim(a„ — xn) exists in
R.
(c) Since {an} and {r,,} are both increasing sequences of non-negative real
numbers they both converge in IR*, and clearly
lim on = V*/(t) and lim z„ = / f{x)dx.
k=\ Jl
But from part (a), we have r„ < on < f{\) + rn for each n, and therefore (by
letting n —► co), we have
/»00 oo «oo
0</ f{x)dx<Yf{k)<f{l) + f{x)dx.
This inequality shows that YaL\ /W converges if and only the improper Riemann
integral ff°f{x)dx exists.
Problem 5.10. £/$£ the preceding problem to show that the series Y^fL\ ^n does
not converge in JRfor 0 < p < 1 and converges in JRfor all p > 1. The following
are problems related to the harmonic series Y^L\ »•
a. Prove with {at least) three different ways that Y1T=\ « = °°-
b. If a computer starting at 12 midnight on December 31, 1939, adds one
million terms of the harmonic series every second, what was the value
{within an error of \) of the sum at 12 midnight on December 31, 1997?
{Assume that each year has 365 days.)
oo
Solution. Notice that
Jj xp '-+oo J, xP * (linv-.oolnr if p = 1.
This limit is finite if p > 1 and infinity if 0 < p < 1.

Section 5: THE EXTENDED REAL NUMBERS
43
(a) We let an = Yll=\ £. Here are four proofs of the divergence of the harmonic
series.
1. Notice that aln - an = -^ + ^ +; • • 4- ^ > n • ^ = \ for each w.
This shows that {cx„} is not a Cauchy sequence, and hence, divergent.
2. As shown at the beginning of the solution of the problem, f™ ^ = oo, and
so Er=, i=oo.
3. We claim that a-i* > 1 + | for each /?. (If this inequality is established,
then clearly YltL] { = n^nc^2,, = oo.) The proof of the inequality is by
induction. For n = 1, we have o^i = on — 1 -f ^. Now, if we assume the
inequality true for some ;?, then
1 1 1
2" + 1 2" + 2 2" + 2"
H 1 rt -f 1
> 1 -h - + 2^ • ——- = 1+-t—•
2 2-2" 2
4. Note that
00
> 9 • — + 90 +900 + • • -
10 100 1000
9 9 9
= To + To + To + --- = 0°-
(b) From Problem 5.9, we know that the harmonic series is associated with
the function f(x) = ^ and that 0 < an — \nn < /(l) = 1 for each n. So, In/?
approximates an within an error of one. If the computer started adding the terms of
the harmonic series at 12 midnight on December 31, 1939, then up to 12 midnight
on December 31, 1997, there are
57(years) x 365(days) x 24(hours) x 60(minutes) x 60(seconds)
= 1,797,552 x 103 seconds.
So, if the computer adds 1, 000, 000 = 106 terms per second of the harmonic
series, the last number N added a second before midnight on December 31 of
1997 is N = 1, 797, 552 x 109. Therefore,
N 1
Y - = aN % In N = 35.12520213 ...,
*—' n
which shows that the harmonic series is a "very slow" divergent series.

44
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
a-1
(c) From Problem 5.8, we know that the alternating series YltL] ("~V is
convergent in JR. Also, from Problem 5.9, we know that lim(<7„ — In n) = y € JR. So,
if we let x„ = y — (pn — In n), then A'„ -> 0 and an = y -f- In n + xn for each n.
Now, note that
A(-l)*"1 ,11 1/11 1\
t-f k 3 5 2/1 — 1 \2 4 2n/
a-=
* 11 1 ./I 1 1 \
1 1 1
n + 1 /2 + 2 2n
= crin - crn = [y + ln(2n) + x2n] - [y + In /i + *„]
= In2 + A'2n -*„,
for each «. This implies YltL\ ~jt— = m2-
Problem 5.11 (Toeplitz). Lef [an] be a sequence of positive real numbers (i.e.,
an > Ofor each n) and put bn = £"=i a'- Assume tnat bn t ]C/^i ai === °°- //"
{a,,} w (2 sequence of real numbers such that xn -* jc //z R, then show that
1 "
lim — ; fl,A'/ = a.
Solution. Let e > 0. Choose some k such that |a-„ — jc | < e for each n > k. Put
M = max{|A-, — jc|: / = 1,..., /:}, and then select some I > k such that —^ < €
for all n > I. Now, notice that if n >l, then
1 ^ 1
i k \ n
< —Y^ai\Xi-x\ + — ]£ a/|.r/-*|
Mbk
< —— + 6 < € + € = 2€,
and the conclusion follows. (Note that this problem is a substantial generalization
of Problem 4.11.)
Problem 5.12 (Kronecker). Assume that a sequence of positive numbers {bn}
satisfies 0 < b\ < bi < b?> < • • • andbn \ co. If a series Y^L\xn of real numbers

Section 6: METRIC SPACES
45
converges in R, then show that
1 "
lim irYlbixi =0-
In particular, show that if[yn} is a sequence of real numbers such that the series
£,^Li f converges in R, then y]+'n+y" -> 0.
Solution. Let x = YlT=\xn- Put bo = 0, so = 0, and sn — x\ H h xn for
each n > 1. Now, notice that
n n n
Y^biXf = ]Tfr/te - J,_i) = bnsn - J^ j/_i(6/ - 6/_i).
/ = l /==! i=2
Therefore, ^ £/'=i fe/.v/ = s„ - £ E?=2-y/-i(6/ ~ &/-i)- since ^ ~ */-i > ° for
each / and £/Li(°i ~ °i-\) = °n t °°» ^ follows from the preceding problem that
TT E/U^-i(^ -bi-\) = x. Hence,
1 " r 1 " i
lim — y^biXi = lim U„ - — Y]^-i(^ - fy-i) = jt - .r = 0.
For the second part, notice that if Y1T=\ ^ IS convergent in R, then let bn — n
for each n and notice that (by the above)
as desired.
6. METRIC SPACES
Problem 6.1. For subsets A and B of a metric space (X, d) show that:
a. (A D B)° = A°n B°.
b. A°U£° £_(4U 5)°.
c. AUB = AUB_.
d. /infiCAnfi.
e. IfB is open, then AD B <Z AD B.
Solution, (a) From (ADB)° c A0 and (ADB)° C £°, it follows that (An5)° c
A0 O 5°. On the other hand, since A0 H 5° C A n B holds and A0 n B° is open,
it easily follows that A0 n 5° c (A n 5)°.

46
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
(b)From A C A UB, it follows that A0 C (AUB)°. Similarly, B° c G4U£)°,
and the desired inclusion follows.
(c) From S c. S and the fact that S is a closed set for any subset S, we see that
AUBQA\JBUAUB = AUBOAUB = AUB.
(d) Since /in B C A n 5, we have i4nfiCi4nfl = An 5.
(e) If x e A fl 5 and r > 0, then choose some 0 < 8 < r with B(x, 8) c £,
and note that
£(*, r) n (A n fl) 2 b(a-, <5) n £ n a = £(*, <$) n a # 0.
That is, ^ € A D £, and so ADB (ZADB holds.
Problem 6.2. S/jcw that in a Euclidean space R" with the Euclidean distance,
the closure of any open ball B(a,r) is the closed ball {x € Rn: d(x, a) < /*}. G/ve
a/? example of a complete metric space for which the corresponding statement is
false.
Solution. Let C(a, r) = {x e RA: d(a, x) < r}. Since C(a, r) is a closed set,
it follows that B(a,r) C.C(ayr).
For the other inclusion, let x e C(a, r). For each n let xn = jxa + (l — £)*.
The inequalities
d(*, A'n) = (1 - I)rf(fll x) < (l - !)/• < r and d(x. xn) = !</(*, jc) < J
imply that {*„} C B(a,r) and jc„ —> x. Consequently, a* e B(a,r), and thus
C(a, r) c £(tf,r) also holds.
For a counterexample, consider X = {0, 1) with the discrete distance, and
note that X is a complete metric space. Also, observe that B(0, 1) = {0}, while
C(0,1)={0,1}.
Problem 6.3. // A is a nonempty subset of R, then show that the set
B = [a € A: There exists some e > 0 with (a, a 4- s) D A = 0 }
/j1 at-most countable.
Solution. Foreachtf € B pick a rational number ra > a sothat(tf, ra) 0/4 = 0.
We claim that if a, b e B satisfy a ^ b, then ra ^ rb. Indeed, if a < b and

Section 6: METRIC SPACES
47
ra z=rh hold, then—since b e (a,ra)—the open interval (a,ra) is a neighborhood
of b e A> and so (ayra)C\ A ^ 0, contrary to the choice of ra.
The above show that the mapping a i—> rai from B into the set of rational
numbers, is one-to-one. Consequently, the set B is at-most countable.
Problem 6.4. Let f: (X, d) -* (Y, p) be a function. Show that f is continuous
ifandonlyiff~\B°) C [f-\B)]° for every subset B of Y.
Solution. Assume / continuous, and B C Y. Since B° is open, the set
f~l(B°) is likewise open. Thus, in view of B° C B , we have
/-'(50) = [/-,(B0)]°c[/-,(B)]0.
In the opposite direction, assume that the condition is satisfied. If B c Y is
open (i.e., if B — B° holds), then
[r\B)]° c /-i(5) = r\B°) c [r1^)]0
shows that f~](B) is open. Therefore, / is continuous.
Problem 6.5. S/icw that the boundaiy of a closed or open set in a metric space
is nowhere dense. Is this statement true for an arbitrary subset?
Solution. Since dA = dAc = A 0 Ac holds, we can assume that A is closed.
Thus,
OA)° = (In^)° = (An^)° = /i0n(^)0
c A°nA*= A0 n (A°)c = $.
Since dA is closed, this shows that 3A is nowhere dense.
An alternate proof goes as follows: If x e (3/4)°, then there exists some r > 0
such that_fl(jc, r)C.dA = AnAEC.A. This implies B(x, r) n Ac = 0, contrary
to* 6 i4c.
The boundary of an arbitrary set need not be nowhere dense. An example: Let
X = 1R with the Euclidean distance, and let A = Q (the set of rational numbers).
Note that dA = R.
Problem 6.6. Show that the set of irrational numbers is not a countable union
of closed subsets o/R.
Solution. Let / denote the set of all irrational numbers, and let {n, r2l...} be
an enumeration of the rational numbers of R.

48
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Assume by way of contradiction that there exists a sequence of closed sets [An]
of IR such that / = (J^li An. Then
R=(U*")u(lN)'
n—\ n—\
and by the Baire Category Theorem (Theorem 6.18), we must have (An)° =£ 0
for some n. Thus, some An contains an interval. However, since An c. I holds
and each interval contains rational numbers, this is impossible, and the conclusion
follows.
Problem 6.7. Let (X ,d)bea metric space. Show that if{xn} and [yn} are Cauchy
sequences ofX, then {d(xn, y„)} converges in JR.
Solution. Use the inequality
\d(xn, y„) - d{xm, ym)\ < d(xn,xm) + d(yn, ym).
(Also, see the discussion before Theorem 6.19.)
Problem 6.8. Show that in a metric space a Cauchy sequence converges if and
only if it has a convergent subsequence.
Solution. Let {*,,} be a Cauchy sequence in a metric space (X, d). If xn —> x
holds in X, then every subsequence of {xn} converges to x.
For the converse, assume that there exists a subsequence {a*^ } of [xn} such
that Xkn -» x holds in X. Let e > 0. Choose /i0 such that d(xkn, x) < € and
d(xn, xm) < ^ for n, m > no. Now, if n > «0, then kn > n > /?o, and so
d(xn,x) < d(xn,xkn) -f d(xLn,x) <€+€ =2*.
This shows that, limjc„ = x holds in X.
Problem 6.9. Prove that the closed interval [0, 1] is an uncountable set:
a. by using Cantor s Theorem 6.14, and
b. by using Baire's Theorem 6.17.
Solution, (a) Assume by way of "contradiction that [0, 1] is a countable set,
say [0, 1] = [x\,xi,...}. We consider [0,1] equipped with the usual distance
d(x, y) = |a- — y\ so that [0,1] is a complete metric space.

Section 6: METRIC SPACES
49
Subdivide [0, 1] into three closed subintervals (as in the construction of the
Cantor set) of equal length. Remove from [0, 1] the middle open subinterval and
consider the remaining two closed subintervals (here the subintervals [0, ^] and
[|, l]) and then select one of them, say /j, such that x\ £ l\. Next, repeat this
process with l\ in place of [0, 1] and select a closed subinterval h of l\ of length
equal to one-third of /j such that„t2 £ h> Inductively, assume that we have chosen
n closed intervals I\,..., In such that:
1. /„£/„_! c...c/2c/If
2. xk £ Ik for k = 1,..., az, and
3. the length of each Ik is ^.
As above, there exists a closed subinterval In+\ of /„ of length equal to one-third
of In such thatX/,+1 £ /„+i.
Thus, there exists a sequence {/„} of closed subintevals of [0, 1] such that
In+\ £ //!»-v„ £ /„ and d{In) = ^ for each n. By Theorem 6.14, we infer that
Pl^=i In consists exactly of one point. But, since xn £ I„ for each «, we see that
p|^j /„ = 0, a contradiction. Hence, [0, 1] must be uncountable.
(b) Again, assume by way of contradiction that [0, 1] = {^i, X2,...} and again
we consider [0, 1] as a complete metric space. If An — [xn], then each An is closed
and has no interior points. However, Theorem 6.17 (or Theorem 6.18) applied to
the equality [0, 1] = UJHi ^/» implies that some An must have an interior point,
which is impossible. This shows that [0, 1] cannot be countable.
Problem 6.10. Let [r\, /*2,...} be an enumeration of all rational numbers in the
intei-val [0, 1] and for each x e [0, 1] let Ax = [n e IN: rn < x}. Define the
function f: [0, 1] -> 1R by the formula
neAx
Show that f restricted to the set of irrational numbers o/[0, 1] is continuous.
Solution. Fix an irrational number a e [0, 1] and let s > 0. Pick a natural
number k such that Y^Lk ?F < s ^ ^et
8 = min{\a -/■,!, \a - r2\ \a - rk\] > 0.
We claim that if x e [0, 1] is an irrational number, then \a — x\ < 8 implies
|/(.r) — f(a)\ < s (which tells us that / is continuous when restricted to the
irrational numbers).
To see this, let x e [0, 1 ] be an arbitrary irrational number satisfying \x — a\<8.
Let /Y denote the half-open subinterval of [0, 1] which is open at left and closed

50
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
at right having endpoints a and x. If Bx = [n e IN: rn e Ix }, then note that
5v £ {*, * + 1, * + 2,...} (why?), and so
00
|/Cr)-/(«)| = 5><£±<«
holds, as claimed.
Problem 6.11. This problem concerns connected metric spaces. A metric space
{X, d) is said to be connected whenever 0 and X are the only subsets ofX that
are simultaneously open and closed. A subset A of a metric space (X, d) is said
to be connected whenever (A, d) is itself a connected metric space. Establish the
following properties regarding connected metric spaces and connected sets.
a. A metric space (X, d) is connected if and only ifeveiy continuous function
f: X —> {0,1} is constant, where the two point set {0,1} is considered to
be a metric space under the discrete metric.
b. // in a metric space (X, d) we have B C A C X, then the set B is a
connected subset of (A, d) if and only ifB is a connected subset o/(X, d).
c. ///: (X, d) -> (7, p) is a continuous function and A is a connected subset
ofX, then f(A) is a connected subset ofY.
d. // {A/}/e/ is a family of connected subsets of a metric space such that
Diet Ai # $■then U
€/ A,- is likewise a connected set.
e. // A is a subset of a metric space and a e A, then there exists a largest
(with respect to inclusion) connected subset Ca of A that contains a. (The
connected set Ca is called the component of a with respect to A.)
f. If a,b belong to a subset A of a metric space and Ca and Cf, are the
components of a and b in A, then either Ca = Cb or else CaC\Cb = 0.
Hence, the identity A = [JaeA Ca shows that A can be written as a disjoint
union of connected sets.
g. A nonempty subset of 1R with at least two elements is a connected set if and
only if it is an interval. Use this and the conclusion of(f) to infer that every
open subset of R can be written as an at-most countable union of disjoint
open intejyals.
Solution, (a) Let (X,d) be a connected space and let f:X —> {0, 1} be a
continuous function. Then the set A = /_,(0) is an open and closed subset of
X. Since X is connected either A = 0 (in which case /(a) = 1 holds for each
x e X) or A = X (in which case f(x) = 0 holds for each x e X).
For the converse, assume that every continuous function from X into {0, 1}
is constant and let A be a closed and open subset of X. Then the function

Section 6: METRIC SPACES
51
f:X -> 1R defined by
/W~"J0, ifx*A,
is continuous (why?). By our hypothesis, / must be a constant function, and this
implies that either A = 0 or A = X, i.e., X is a connected metric space.
(b) It follows immediately from (a).
(c) Assume that / and A satisfy the stated properties and consider the
continuous functions (A,d) —>• (f(A)y p) —> {0, 1}. By (a), the continuous function
g o / must be a constant function and from this, we see that g is also a constant
function. By (a), (/(/4), p) is a connected metric space.
(d) Assume the family {A,: i e 1} satisfies the stated properties. Put A =
(J/6/i4/ and let f:(A,d) —> {0, 1} be a continuous function. Then the function
/: (A,-, d)—^{0, 1} is a continuous function and so / restricted to each A-t is
constant. Since Hze/^/ ^ 0> we see mat / *s constant on A, and so—by
(a)—the set A is connected.
(e) Fix a e A and let
A = [B C A: B is connected and a e B).
Note that {a} e A and that f\BeAB ^ 0. By (d), the set Ca = Use^5 is a
connected subset of /\ that satisfies the desired properties.
(f) If Ca H C/, 7^ 0, then by (d) we infer that Cfl U C/, is a connected set
containing a. Hence. Cb C.CaUCb Q Ca. Similarly, Ca c Cb and so Ca = CV
(g) Let i4 be a connected subset of R and let a, 6 e i4 satisfy a < b. If
a < x < b and a* g A, then the set /4 fl (-co, x) is a proper and closed subset
of A (why?), a contradiction. Thus, (a, b) c A holds and this shows that A is an
interval.
For the converse, assume that / is an interval of R. Assume by way of
contradiction that there exists an onto continuous function /: / —> {0, 1}. Pick
a, b e I such that f(a) = 0 and f(b) = 1; we can suppose that a < b (and so
[a,b] C /). Now, let
c0 = sup{c<E [a,b): f(c) = 0).
By the continuity of /, we see that f(co) = 0 and that cq < b. Then /(a) = 1
holds for all cq < x < b, and so (by the continuity of / again) /(co) = 1 must
also hold, which is impossible. Therefore, every continuous function from / into
(0, 1} is constant, and so by (a) the interval / is a connected set.
Finally, note that if / is an open subset of R, then by (f) we know that / =
Uaei C<*> wnere eacn Cfl is a connected set. It easily follows (how?) that each
Ca is an open interval and that there are at-most countably many of them.

52
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Problem 6.12. Show that R" with the Euclidean distance is a connected metric
space. Use this conclusion to establish that, if the intersection of two open subsets
of JR." is a proper closed set, then the two open sets must be disjoint.
Solution. Let d denote the Euclidean distance of R", i.e., let
1 = 1
For each x e Rn, let Lx denote the line segment joining 0 and x, i.e., let
Lx = {rx: 0 < / < 1). We claim that Lx is a connected set.
To see this, note that the function /: [0,1] —> Lx, defined by /(/) = rx,
satisfies \f(t) — f(s)\ < d(x, 0)1$ — f|,andso / is (uniformly) continuous. From
parts (g) and (c) of Problem 6.11, we see that Lx is a connected set.
Now, use part (d) of the preceding problem and the identity Rn = UX€Rn^x to
infer that R" is itself a connected metric space.
For the last part of the problem, let U and V be two open subsets of Rn such
that K = U f! V is a closed set. Then K is both open and closed (and since K
is a proper subset of Rn) it must be the empty set.
Problem 6.13. Let C be a nonempty closed subset of JR. Show that a function
f\C —► R is continuous if and only if it can be extended to a continuous real-
valued function on R.
Solution. Let C be a nonempty closed subset of R and let f:C -> R be a
function. If / can be extended to a continuous real-valued function on R, then
/: C -> R is obviously continuous.
For the converse, assume that f:C -* R is a continuous function. Start
by observing that the complement Cc of C is an open set and so (by part (g)
of Problem 6.11) Cc can be written as an at-most countable union of pairwise
disjoint open intervals; say Cc = U/€/(a/»^/)» wnere / is at-most countable.
Since the open intervals {(a,-, bi): i e 1} are pairwise disjoint, it follows that all
the endpoints ax and bx belong to C. Hence, f{a{) and f(b\) are defined for
each z. Now, extend the domain of / by defining the graph of the function / on
the interval (a,-, bj) to be the straight line segment joining the points (a,-, /(fl/))
and (6/, f(bi)). In other words, for each a\ < x < bt we let f(x) = /(a,-) +
f(bi]ZfaM (* -*,-)—in case to, b,) = (-co, b{) or fa, b{) = to, oo) let f{x) =
bf or /U) = a/.
We claim that this extension of / to all of R is continuous. Clearly, / is
continuous at every point of C° and"at every point of Cc (why?). We only need
to verify that / is continuous at the boundary points of C. So, let a € SC and
let xn —► a with \xn) C Cc—if [xn] C C, then /(*„) —> f(a) is trivially

Section 6: METRIC SPACES
53
true. Also, we shall assume that a is not one of the endpoints a{ or bx. For each
n pick (the unique) in e I with aXn < xn < b\n. Note that in this case, we must
have lima/,, = \\mbln — a (why?). From
i/(*,)-/mi = |[/(«)+m£ziy u. - g/j] - /(*)|
—^ l/(fl)-/(fl)l=0,
we see that lim f(xn) = /(a). A similar conclusion holds true if a is one of the
endpoints ax or bx. This shows that / is continuous at a, as claimed.
For an alternate proof see Problem 10.11.
Problem 6.14. Show that a metric space is a Baire space if and only if the
complement of every meager set is dense.
Solution. Let X be a metric space. Assume first that X is a Baire space and
let A be a meager set. Pick a sequence (An] of nowhere dense sets such that
A = \J%L\ An. To show that Ac is dense, it suffices to show that V 0 Ac ^ 0 for
each nonempty open set V. To see this, let V be a nonempty open set, and assume
by way of contradiction that V n Ac = 0. This implies V C A, and so
v = \Jvn Aa.
Hence, V is a nonempty open meager set, a contradiction. Hence, Ac is a dense
set.
For the converse, assume that the complement of every meager set is dense,
and let V be an open meager set. Then Vc is dense. So, if V is nonempty, then
V n Vc 7^ 0, which is impossible. Thus, the empty set is the only open meager
set, and hence, X is a Baire space.
Problem 6.15. A subset of a metric space is called co-meager if its complement
is a meager set. For a subset A of a Baire space show that:
a. A is co-meager if and only if it contains a dense Gs-set.
b. A is meager if and only if it is contained in an Fa -set whose complement is
dense.
Solution. Notice that if A is a nowhere dense set in a metric space X, then from
Lemma 6.8 we see that
0 = (i)° = (ir = ([(A)cDc.

54
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
This implies that a subset A is nowhere dense if and only if the open set (A)c is
dense.
Now, assume that X is a Baire space and let A be a subset of X.
(a) Suppose first that A is a co-meager set. Then there exists a sequence [A,,] of
nowhere dense sets such that A = (U^Li ^n) • This implies
00 oo
n=\ n=l
By the above discussion, each set (A»)c is an open dense set, and since X is a
Baire space, the Ga-set E = f]^=i (An)c ls also dense (see Theorem 6.16). Now,
a glance at (•) shows that E C A.
For the converse, assume that A contains a dense G^-set B, i.e., B C A. So,
there exists a sequence {Vn] of open sets such that B = p|^=i V«- From BC|/„,
we see that each V„ is also dense. This implies
[(Vn)c]° = [(V„)c]c-C = (V*)c = Xc = 0,
and so each (V„)c is nowhere dense closed set. Now, use the inclusion
00
AcCBc = \J(Vn)c
n=l
to conclude that Ac is a meager set, i.e., A is a co-meager set.
(b) Assume first that A is a meager set, i.e., Ac is a co-meager set. By part (a),
there exists a dense G^-set E such that E c Ac. This implies A c £c, where now
£ is an FCT-set whose complement (Ec )c = E is dense.
For the converse, assume that KF holds, where F is an Fa-set with dense
complement. It follows that Fc c Ac, where now Fc is a dense G$-set. By part
(a), Ac is co-meager set, which means that A is a meager set.
7. COMPACTNESS IN METRIC SPACES
Problem 7.1. Let f: (X, d) -» (K, p) Z?^ a function. Show that f is continuous
if and only if f restricted to the compact subsets ofX is continuous.
Solution. Assume that / is continuous on every compact set. Let xn -* x.
Then the set A = {jcj , X2,...} U [x] is compact (note that every open cover of A
can be reduced to a finite cover), and xn ~> x holds in A. Since / restricted to
A is continuous, lim f(xn) = f(x) holds, which shows that / is continuous.

Section 7: COMPACTNESS IN METRIC SPACES
33
Problem 7.2. A metric space is said to be separable if it contains a countable
subset that is dense in the space. Show that every compact space (X, d) is
separable.
Solution. Foreach n choose a finite subset Fn of X such that X=|JA.6/7 B(x, £).
Let F = U^li ^n> an^ note tnat F ls at-most countable.
Now, let x e X and r > 0. Pick some n with y < r. Then there exists some
y G Fn with d{x, y) < £ < r. Thus, B(x, r) H F ^ 0, and so F is dense in
X.
Problem 7.3. Show that if(X, d) is a separable metric space {see the preceding
exercise for the definition), then cardX < C.
Solution. Let {.*i,.X2» •••} be a countable dense subset of X. Consider the
collection of open balls [B{xi, j): /, j = 1,2,...}. Clearly, this collection is
countable; let [B\, Bn,...} be one of its enumerations. Now, for each x e X
define the set Sx = [n e IN: x e Bn}. Thus, a mapping x i—> Sx from
X into P(IN) has been established that is clearly one-to-one. Consequently,
cardX < cardPON) = C. (See also Problem 5.6.)
Problem 7.4. Let {X\,d\),..., (X„, dn) be arbitrary metric spaces, and let
X = X{ x ... x Xn. Ifx = (.vi xn)andy = {yu ..., yn), define
Oi(*, )>) = Yldm(a'»'' ^) and °2{x,y)= \^ydm{xm, ym)]2J2.
a. S/zow that D\ and Dj are distances on X.
b. Show that D\ is equivalent to Dn.
c. Show that {X, D\) is complete if and only if each (X/, d{) is complete.
d. Show that (X, D\) is compact if and only if each (X,-, dt) is compact.
Solution, (a) Routine.
(b) Use the inequalities
±Dxix%y)<D2{x,y)<nDi{x,y).
(c) Assume that each Xm {m = 1,..., n) is a complete metric space. Let
{xk} be a DpCauchy sequence of X, where jc* = {x\,..., **). Clearly, each
(x^} is a Cauchy sequence of Xm, and thus there exists xm e Xm such that
limic-^oodm{x^yxnl) = 0. Hence, if x = (*i,... ,*„) e X, then we have
lim^oo D\{xky x) = 0, so that the metric space X is D\-complete.

56
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Now, let X be D\-complete. Fix an element (yi,..., yn) e X. Let {jc* } be a
Cauchy sequence of Xm. If ** e X is the element whose f1 component equals
yj for j ^ m and equals jc* if j = m, then {x/,} is a Cauchy sequence of X. If
x e X is its limit, then it is easy to see that lim^oo dm(x^,xm) = 0, so that each
Xm is complete.
(d) Assume first that each Xm is compact. Then following the proof of the
second part of Theorem 7.4, we can see that every sequence of X has a convergent
subsequence, and so X must be a compact metric space.
On the other hand, if X is a compact metric space, then the function fm: X —>
Xm, defined by fm(x\,..., xn) = xm, is continuous and onto for each 1 < m < n.
Hence, by Theorem 7.5, each Xm = /m(X) is compact.
Problem 7.5. Let {(Xn,dn)} be a sequence of metric spaces, and let X =
nSli %n- For each x = {xn} and y = {yn} in X, define
., v v^ 1 d„(xn,yn)
a. Show that d is a distance on X.
b. Show that (X, d) is a complete metric space if and only if each (X„, dn) is
complete.
c. Show that (X, d) is a compact metric space if and only if each (Xni dn) is
compact.
Solution, (a) Note first that if d is a distance on a set X, then p(x, y) = !+///*)
is likewise a distance on X, which is equivalent to d. From this observation it
easily follows that
,, , v^ 1 d„(x„,y„)
diX'y) = k¥-l+dn(Xn,yn)
is a distance on X = n^li^n-
(b) Let [xk] be a sequence of X, where xk = (jcj, jc^ ,.. .)• The proof follows
from the following two properties (whose verifications are straightforward).
1. xk —> x holds in X if and only if xf —► Xj holds in X/ for each /; and
2. {xk} is a Cauchy sequence in X if and only if [xf] is a Cauchy sequence
in X,- for each /.
(c) Assume that (X, d) is a compact metric space. Then the function f:X —>
X,, defined by f(x) = *,- for each x = (x\, JC2,...) € X, is continuous and onto.
By Theorem 7.5, each X/ is a compact metric space.

Section 7: COMPACTNESS IN METRIC SPACES
57
For the converse, assume that each X; is a compact metric space. By (b), X
is a complete metric space, and so by Theorem 7.8 it suffices to show that X is
totally bounded. To this end, let s > 0. Choose n such that 2~n < e, and note
that
Pn(x, y)= y,^i' i , At T
defines a distance on YYi=\ %i- ^ should be clear that pn is equivalent to the
distances of the preceding problem, and (fl/Li ^/» Pn) is a compact metric space.
Choose a finite subset F of YY!=\ %i sucn mat tne Pn-balls with centers at the
points of F and radii e cover YY!=\ %i- Next, extend each x € F to an element
of X (i.e., add to each x e F arbitrary components xn+\,x„+2* • • •)• Now, if
y = (y{, y2,...) € X, then pick some jc g F with pn(jc, y) < e, and note that
Thus, X = Ut€F^('r' ^) holds, and therefore, X is totally bounded.
Problem 7.6. A family of set T is said to have the finite intersection property
// every finite intersection of sets of T is nonempty. Show that a metric space
is compact if and only if every family of closed sets with the finite intersection
property has a nonempty intersection.
Solution. Let X be compact, and let {A,-: / € /} be a family of closed sets with
the finite intersection property. If f]i&IAi = 0, then X = U/e/^/ holds, and by
the compactness of X, there exist /1,..., in e I such that X = Uy=i ^/ • Thus,
fY!=1i4iy = 0, contrary to our hypothesis. Hence, p|/G/ A,- # 0.
For the converse, assume that every family of closed sets with the finite
intersection property has a nonempty intersection. Let X = U/e/^i' De an °Pen
cover. Then flie/^/0 = $» ^d smce Wf1 ** € /} is a family of closed sets, our
hypothesis guarantees the existence of a finite number of indices /1,..., in such
tnat n;=i^/c = $• T^S' x = Uj=\vij holds, so that X is a compact metric
space.
Problem 7.7. Let f:X -> X be a function from a set X into itself A point
a € X is called a fixed point/or / if f(a) = a.
Assume that (X, d) is a compact metric space and /: X -> X satisfies the
inequality d{f(x), f(y)) < d(x, y) for x ^ y. Show that f has a unique fixed
point.

58
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Solution. Note first that / has at most one fixed point. Indeed, if /(jc) = jc
and /(v) = y hold with x ^ y, then
d(x,y) = d(f(x),f(y))<d(x,y)
must hold, which is absurd.
Now, define the function g: X —► R by g(x) = d(x, /(jc)). From the inequality
\gM - g(y)\ < </(/(*), f(y)) + £/(*, y) < 2d{x, y)
(see the discussion preceding Theorem 6.19), it follows that g is continuous.
Since X is compact, g attains its minimum at some point a e X. If f(a) =£ a,
then the inequality
g(f(aj) = </(/(*), /(/(*))) < rf(fl, /(*)) = g(a)
shows that g does not attain a minimum at a. Thus, /(a) = a must hold, and so
a is a (unique) fixed point for /.
Problem 7.8. Let (X, d) be a metric space. A function f:X-+X is called a
contraction if there exists some 0 < a < 1 such that d(f(x), f(y)) < ad(x, y)
for all jc, y e X; a is called a contraction constant.
Show that every contraction f on a complete metric space (X, d) has a unique
fixed point; that is, show that there exists a unique point x e X such that /(jc) = x.
Solution. Note first that if f(x) = jc and f(y) = y hold, then the inequality
d(x,y) = d(f(x)yf{y)) < ctd(xyy) easily implies that d(x,y) = 0, and so
jc = y. That is, / has at-most one fixed point.
To see that / has a fixed point, choose some a € X, and then define the
sequence {jcn} inductively by
x\ = a and xn+\ = /(jc„) for n = 1, 2,... .
From our condition, it follows that
d(x„+i,xn) = d(f(xn), /(*„_!)) < ad(xn, xn-X)
holds for n = 2, 3,... . Thus, as in Problem 4.15, we can show that {jc„} is
a Cauchy sequence. Since X is complete, {jc„ } is a convergent sequence. Let
jc = lim jc„. Now, by observing that / is (uniformly) continuous, we obtain that
x = lim xn+i = lim f(xn) = /(*),
n—>oo • n-»-oo
and so jc is a (unique) fixed point for /.

Section 7: COMPACTNESS IN METRIC SPACES
59
Problem 7.9. A property of a metric space is called a topological property if it
is preserved in a homeomorphic metric space.
a. Show that compactness is a topological property.
b. Show that completeness, boundedness, and total boundedness are not
topological properties.
Solution, (a) It follows from Theorem 7.5.
(b) Consider (0, 1] and [1, oo) as metric spaces under the usual Euclidean
distance d(x, y) = \x — y\. Clearly, (0, 1] is not complete but it is bounded and
totally bounded. Also, [1, oo) is complete (because it is a closed subset of R), but
is neither bounded nor totally bounded. On the other hand, /: (0, 1] -> [1, oo),
defined by f(x) = |, is a homeomorphism, and the claims in (b) follow.
Problem 7.10. Let (X, d) be a metric space. Define the distance of two nonempty
subsets A and BofXby
d(A, B) = inf{d(x, y): x e A and y e B}.
a. Give an example of two closed sets A and B of some metric space with
A OB -0and such that d(A, B) = 0.
b. // A fl B = 0, A is closed, and B is compact (and, of course, both are
nonempty), then show that d(A, B) > 0.
Solution, (a) Let X = R2 with the Euclidean distance, and consider the closed
subsets of X
A = \{x,\): x> 1} and fi = {(jt,0): x > l}.
Note that A n B = 0, while d(A% B) = 0.
(b) Let A and B be as stated in the problem. If d(A, B) = 0, then pick two
sequences {xn} C A and [yn] c B with d(x,uyn) —> 0. Since B is compact,
by passing to a subsequence (if necessary), we can assume that yn —> y holds
for some y e B. The inequality
d(xn, y) < d(xny yn) + d{yn, y)
shows that d(xn, y) —► 0. Since A is closed, y e A, and hence AD B ^ 0,
contrary to our hypothesis. Therefore, d(A, B) > 0 must hold.
Problem 7.11. Let (X, d) be a compact metric space and f:X-*Xan isome-
try; that is, d(f(x), f(y)) = d(x, y) holds for all x, y e X. Then show that f is
onto. Does the conclusion remain true ifX is not assumed to be compact?

60
Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
Solution. Let y e X. Define the sequence [xn] of /(X) by
A'! = f(y) and x„+x = /(*„) for /z = 1, 2
Note that d(xn,xn+p) = d(y,xp) holds for all n and all p. Since /(X) is
compact, {a,,} must have a limit point in /(X). Let a be a limit point of [xn].
Now, let e > 0. Pick a > 1 and /? such that d(xn, a) < s and <i(A'n+p, a) < e.
Then
) < d(xn, a) •+■ d(x„+p, a) < 2e
holds for all e > 0, and so d(y, /(X)) = 0. Thus, y e J(X) = /(X), so that
/(X) = X holds.
If X is not supposed to be compact, then the conclusion is no longer true. A
counterexample: Take X = IN with d(n, m) = |/z — m| and consider the function
/: N -* N defined by /(/i) = n + 1.
Problem 7.12. S/zow r/2<ar a metric space X is compact if and only if every con-
tinuous real-valued function on X attains its maximum value.
Solution. Let (X, d) be a metric space. Assume that X is compact and that
/: X ->• R is a continuous function. By Theorem 7.5, we know that /(X) is a
compact subset of R, and so (by Theorem 7.4) /(X) is closed and bounded. The
maximum of /(X) is the maximum value of / on X.
For the converse, assume that every continuous real-valued function on X
attains a maximum value. Clearly, every continuous real-valued function on X
attains also a minimum value.
We shall establish first that X is a complete metric space. Let (X, d) denote
the completion of (X, d) and let X e X. The function /: X —>» R, defined by
f(x) = d(x,x), satisfies inf{/(A): x G X} = 0. So, there exists some xq e X
satisfying f(xo) = d(x, a*o) = 0. It follows that x = xo € X and so X = X.
This means that X is a complete metric space.
Next, we shall show that X is totally bounded. To establish this, assume
by way of contradiction that X is not totally bounded. Then an easy inductive
argument shows that there exist some e > 0 and a sequence {xn} of X such that
d(xluxm) > 3e holds for n ^ m. For each n consider the nonempty closed set
C„ = [B(xn, s)f = {xe X: d{x% xn) > *},
and then define the function /„:X^R by
/„(*) = d(x, C„) = inf{d(*f y): y e Cn).

Section 7: COMPACTNESS IN METRIC SPACES
61
So, fn is abounded function, fn(x) = 0 holds for each x e C„ and fn(xn) > 0.
Multiplying by a constant c„, we can assume that sup{/n(jt): x e X] > n holds
for each n. Now, define the function /: X —> R by
f( n_ I /"M' if * € 5 (**>*)
/W~J0, if **LGli *(*-.«).
and we claim that / is a continuous function. Clearly,/ is continuous at the points
of the balls B(xn, e). If x0 i U^=i B(xnf e), note that B(jc0, |) O £(*„, e) # 0
holds for at-most one n (why?). If B(xq, |) n £(*„, e) = 0 for each /?, then
/(„r) = 0 for each ,t in £(;co, §)> and so / is continuous at xq. Thus, we can
assume that B(xq, §) H £(*„, e) ^ 0 for some n. We distinguish two cases.
CAS£/: d(jt0,*,,)> e.
In this case, there exists some 0 < r < | such that £(;co, r) Pi B{xn, e) = 0.
Clearly, f(x) = 0 holds for each x e B(xq, r), and from this we see that / is
continuous at xq.
CASE II: d{x0,x„) = e.
Let {zk} be a sequence of X satisfying zk —> xq\ we can assume that zk
belongs to B(xq, f) for each k. Note that if zk £ B(xn, e), then f(zk) = 0. On
the other hand, if zk e 5(;cn, e), then
0 < f{zk) = crffe, C„) < cdfo, jc0).
Thus, 0 < /(z*) < cnd(zk, Xo) holds for each £. In view of
lim d(zkixo) = 0,
k-+oo
we see that lim f{zk) = 0 = /(*o) and so / is continuous at xq in this case too.
To contradict our hypothesis, note that / does not attain a maximum value.
Thus,X must also be totally bounded. By Theorem 7.8, we see that X is a compact
metric
space.
Problem 7.13. This exercise presents a converse of Theorem 7.7. Assume that
(X, d) is a metric space such that every real-valued continuous function on X is
uniformly continuous.
a. Show that X is a complete metric space.
b. Give an example of a noncompact metric space with the above property.
c. If X has a finite number of isolated points {an element a e X is said to
be an isolated point whenever there exists some positive r > 0 such that
B{a, r) fl (X \ {a}) = 0), then show that X is a compact metric space.

62
Chapter 1: FUNDAMENTALS OF REAL AN ALYSIS
Solution. Let (X, d) be a metric space such that every continuous real-valued
function on X is uniformly continuous.
(a) If x e X (the completion of X) is an element that does not belong to X, then
the function /: X -> R defined by f(x) = -j^f—-, x e X, is a continuous real-
valued function on X that fails to be uniformly continuous (why?), a contradiction.
Hence, X = X holds, which means that X is a complete metric space.
(b) Let X = {1,2,...} equipped with the discrete distance d. Then every set is
open and so every real-valued function / on X is continuous. Since d{x,y) < 1
implies x = y (and so f(x) — f(y) = 0), we see that every real-valued function
on X is uniformly continuous. Now, note that X is not a compact metric space.
(c) In view of (a), we need to establish that X is totally bounded. To this
end, assume that X is not totally bounded. Then, there exist some e > 0 and
a sequence of elements {xn} of X such that d{xn,xm) > 3s for n ^ m. From
our hypothesis, we can suppose that each xn is an accumulation point of X. For
each n pick an element yn such that 0 < d(xn, yn) < £ and let r„ = d(xn, yn).
Put
Cn = {xeX: d(x,xn)>rn]
and define the functions /„ and / as in the solution of Problem 7.12 (the open
ball B(xn, e) is now replaced by B(xn, rn)). Then / is a continuous function
and satisfies f(yn) = 0 for each n. Pick zn e B(xn,rn) such that f(xn) > n,
and note that
\f(yn) - /ten)! > n and lim d(yn% zn) = 0.
This shows that the continuous function / is not uniformly continuous, contrary
to our hypothesis. Hence, X is totally bounded, as desired.
Problem 7.14. Consider a function f:(X,d) -* (7, p) between two metric
spaces. The graph G of f is the subset ofXxY defined by
G = {(A-,y)GXxr:y = /U)}.
If(Y, p) is a compact metric space, then show that f is continuous if and only if
G is a closed subset ofXxY, where X x Y is considered to be a metric space
under the distance D((jc, j), (w, v)) = d(x, u) + p(y, v); see Problem 7.4. Does
the result hold true if(Y, p) is not assumed to be compact?
Solution. Observe that an arbitrary sequence [(xn,yn)} of X x Y satisfies
(xn,yn) -> (jc, y) in X x Y if and only if xn -> x and yn —► y both hold.

Section 7: COMPACTNESS IN METRIC SPACES
63
Assume (7, p) compact and G closed. If / is not continuous, then there exists
a sequence [xn] of X and some e > 0 such that xn -> x and p(f(xn), /(*)) >
e for all n (why?). Since (7, p) is compact, by passing to a subsequence, we can
assume that f(yn) -> )> holds in 7. Now, observe that (x„, /(*„)) e G holds
for each n and (*„,/(*„)) -> (jc, ^) holds in X x y. Since G is closed, it
follows that (x,y) e G and so y = /(x). This implies
p(/(A'„), /(*)) -» p(/U), /(a-)) = 0,
which contradicts p(f(xn), f{x)) > e for all n. Hence, / is a continuous
function.
If (Y, p) is not compact, then a function with closed graph need not be
continuous. For an example, consider the function /: 1R -> R defined by
/w=H if^0;
J 1 0 if x = 0 .
Problem 7.15. A cover {V/}/e/ o/tf sef X /j jtf/d to be a pointwise finite cover
whenever each x e X belongs at-most to a finite number of the V\.
Show that a metric space is compact if and only if every pointwise finite open
cover of the space contains a finite subcover.
Solution. Clearly, if X is compact, then every pointwise finite open cover of
X contains a finite subcover. For the converse, assume that every pointwise finite
open cover of X contains a finite subcover. To establish that the metric space X
is compact, it suffices to show that every sequence in X contains a convergence
subsequence.
Let [xn] be a sequence in X. We can suppose (why?) that the sequence
consists of distinct elements. Suppose by way of contradiction that [xn] has no
convergence subsequence. Then *i is not in the closure of the set [xn: n ^ 1} and
thus, there exists an open ball V\ = B(x\,8\) about x\ with radius 0 < 8\ < 1
and satisfying xn £ V\ for all n ^ 1. Also, X2 is not in the closure of the set
{xn: n ^ 2} and thus, there exists an open ball Vi = Bfe, 82) about xi with
radius 0 < 82 < ^ and such that xn & V2 for all n ^ 2. Proceeding inductively,
we see that for each k there exists an open ball V* = B(xk,8k) with radius
0 < 8k < jr satisfying xn & V^ for all n ^ k.
Since the set F = {a*i, xi, ...} contains no convergent subsequences, the set
F must contain all of its closure points. Thus, F is a closed set, and hence, the
set G = X \ F is an open set. Then, the collection C = {G, V\, V2,...} is an
open cover of X. In fact, the collection C is a pointwise finite open cover of X
because if a point a* belongs to an infinite number of sets in C, then x belongs to
an infinite number of the sets V,,. However, this would imply that a subsequence

64 Chapter 1: FUNDAMENTALS OF REAL ANALYSIS
of [xn] converges to the point x. Since the sequence {xn} contains no convergent
subsequences, we infer that C is a pointwise finite open cover.
Therefore, C contains a finite subcover of X, say V\,..., Vm, G. Since G
does not intersect {a'i , x2,...}, it follows that {jci , *2,...} c (J™^ Vim However,
this contradicts the fact xn & V* for n ^ k. Conclusion: The sequence {xn} must
have a convergent subsequence—and hence, the metric space X is compact.

CHAPTER 2
TOPOLOGY AND CONTINUITY
8. TOPOLOGICAL SPACES
Problem 8.1. For any subset A of a topological space show the following:
a. A° = (_AF)C.
b. dA = A\A°.
c. (A \ A0)0 = 0.
Solution, (a) Note that
x g A0 <=> there exists a neighborhood V of x with V c A
<£=>• there exists a neighborhood V of x with V n Ac = 0
<=> jc £ A^ <==* .x 6 (A^)0.
(b) Using (a), we see that dA = AnA* = A \ (A*)0 = A \ A0.
(c) If x e (A \ A0)0, then for some open set V we have
x eV QA\A° CA.
This implies x e A \ A0 and x e A°, a contradiction. Hence, (A \ A0)0 = 0.
Problem 8.2. If A and B are two arbitrary subsets of a topological space, then
show the following:
a. AUB = AU5.
b. (A U BY = A' U B'.
Solution, (a) See Problem 6.1.
(b) Clearly, A c B implies A' c £', and so A7 U B' C (A U £)'. For the reverse
inclusion, let x e (A U Z?)'. If ;c £ A' U £', then there exist two neighborhoods-
65

66
Chapter 2: TOPOLOGY AND CONTINUITY
V and W of x such that
An(V\{x}) = Bn(w\{x}) = 0.
Now, note that the neighborhood U = V n W of the point * satisfies
(AUB)n(£/\ {*}) = &
proving that a* £ (A U £)', a contradiction.
Problem 8.3. If A is an arbitrary subset of a Hausdoiff topological space, then
show that its derived set A' is a closed set.
Solution. Let A be an arbitrary subset of a Hausdorff topological space X. We
shall establish that (A')c is an open set (and this will guarantee that A' is a closed
set). To this end, let x € (A')c, i.e., let x £ A'. This means that there exists a
neighborhood V of x such that
Vn(A\{*})=0. (*)
We claim that V C (A')c holds. To see this, let y e V with y ^ x. Since X
is a Hausdorff topological space, there exist neighborhoods U and W of y and
x, respectively, such that U C)W =0. Now, note that V CiU is a neighborhood
of y with jt^KfK/ and so from (•), we see that (V (MJ)CiA = 0. The latter
shows that y £ A'. Hence, V C (A')Q holds proving that every point of (A')c is
an interior point, as desired.
Problem 8.4. Let X = R, <?fld /ef r be the topology on X defined in Example 8.4.
In other words, A € r if and only if for each x e A there exist € > 0 and an
at-most countable set B {both depending on x) such that (x — €,x + e)\BCA.
a. Show that r is a topology on X.
b. Verify that 0 e (0, 1).
c. Show that there is no sequence {xn} o/(0, 1) with \\mxn = 0.
Solution, (a) Straightforward.
(b) Since for each e > 0 and each countable set B the set (—e, e) \ B is
uncountable, we must have ((—£, £) \ B) D (0, 1) ^ 0. This easily implies that
OeOO).
(c)If {jc„} is a sequence of (0, 1), then V = (—1, 1) \ {jcj, Jt2,...} is a
neighborhood of zero, and xn £ V for all n. This shows that no sequence of (0, 1) can
converge to 0.
Problem 8.5. // A is a dense subset of a topological space, then show that
O C. ADO holds for every open set O. Generalize this conclusion as follows: If
A is open, then AD B C. AD B for each set B.

Section 8: TOPOLOGICAL SPACES
67
Solution. Let x e O and let V be a neighborhood of x. Since O is open,
V HO is a neighborhood of x, and so the denseness of A implies
v n (A n O) = (V n O) n a ^ 0,
which means that x e A n O.
For the general case, assume A is an open set and let x e A C\ B. If V is a
neighborhood of x, then V n A is also a neighborhood of x. Since x e #, it
follows that_K n(ADB) = (V r\A)C\B # 0. This shows that x e TPhB, and
hence, A 0 5 CAHB.
Problem 8.6. //{O/J/e/ w aw opew cover for a topological space X, then show
that a subset AofX is closed if and only if A O 0\ is closed in D{ for each i e I
(where 0\ is considered equipped with the relative topology).
Solution. If A is closed, then clearly A n 0{ is closed in 0\ for each i. For
the converse, assume that A n 0\ is closed in 0{ for each /. Put
V/ = 0/ \ ADOi =Oi \ A,
and note that—by our hypothesis—each Vt is open in Ox. Since each 0\ is an
open subset of X, it follows that each V/ is likewise an open subset of X. Now,
note that
A* = X \ A = (\JOi) \ A = \J(Oi \A) = \J V,
iel iel i€/
is an open subset of X, and so A is a closed set.
Problem 8.7. If(X, r) is a Hausdorff topological space, then show the
following:
a. Every finite subset ofX is closed.
b. Every sequence ofX converges to at-most one point.
Solution, (a) Let A = {x} be a one-point set. If y £ A, then (since X is a
Hausdorff space) there exists a neighborhood V of y with x ^ V, and so V C Ac.
Thus, Ac is open, and hence Ay is closed. Now, observe that every finite set is a
finite union of one-point sets.
(b) If x ^ y, then there exist neighborhoods Vx and Vy of x and y
respectively, such that Vx 0 Vy — 0. Now, a sequence of X cannot converge to x and
y at the same time simply because its terms cannot be eventually in both Vx and
vy.

68
Chapter 2: TOPOLOGY AND CONTINUITY
Problem 8.8. For a function f: (X, r) -» (7, x\)show the following:
a. // r is the discrete topology, then f is continuous.
b. // x is the indiscrete topology and X\ is a Hausdoiff topology, then f is
continuous if and only if f is a constant function.
Solution, (a) Note that every subset of X is open. Thus, f~~](A) is an open set
for every subset A of Y y and so / is continuous.
(b) Recall that the indiscrete topology is the topology r = [0, X}. If / is a
constant function, then f~l(A) is either 0 or X, and so / is continuous. For
the converse, let / be a continuous function. If for some jc, y e X we have
f(x) ^ f(y\ then there exists a neighborhood V of f(x) such that f(y) £ V.
Now note that f"l(V) is neither equal to 0 nor equal to X, and so f~l(V) is
not open, a contradiction. Thus, / must be a constant function.
Problem 8.9. Let f and g be two continuous functions from (X, r) into a
Hausdoiff topological space (Y, i[). Assume that there exists a dense subset A of X
such that f(x) = g{x)for all x e A. Show that f(x) = g(x) holds for all x e X.
Solution. Suppose that for some x e X we have f(x) i=- g(x). Pick a
neighborhood V of f(x) and another W of g(x) such that V n W = 0. Since
f~](V)Pig~](W) is a neighborhood of * and A is dense in X, there exists some
y € /_,(V) H g-^W) Pi 4. Now, note that /(v) = g(y) e V C\ W = 0 must
hold, which is absurd. Thus, f(x) = g(x) holds for each jc € X.
Problem 8.10. Let f: (X, r) -» (7, rj) fo a function. Show that f is continuous
if and only if f~\B°) C [/"'(B)]0 holds for every subset B ofY.
Solution. Repeat the solution of Problem 6.4.
Problem 8.11. If f: (X, r) -» (7, rO a/^ g: (K, tj) -> (Z, T2) are continuous
functions, show that their composition go f: (X, r) -> (Z, T2) is also continuous.
Solution. Use the identity (gof)-l(V) = /"^(g^OO). (See Problem 1.8.)
Problem 8.12. Let X be a topological space, let a e X, and let Afa denote the
collection of all neighborhoods at a. The oscillation of a function /: X —> JRat
the point a is the extended non-negative real number
tof{a)= inf/{ sup I/W-/O0I}.
v^f" x,yeV

Section 8: TOPOLOGICAL SPACES
69
Establish the following properties regarding the oscillation:
a. The function f is continuous at a if and only if a)f(a) = 0.
b. IfX is an open interval o/R and /: X -» R is a monotone function, then
o)f(a)=\ lim f{x)- lim f(x)\. .
r->a+ K-+a-
Solution. (a) Assume that / is continuous at a. Fix € > 0. Then there exists
some W £ Ma (i.e., some neighborhood W of a) such that x € W implies
l/to - f{x)\ < €. So, if x, y €W, then
l/to - f(y)\ < l/to - /tol + l/to - /tol < € + 6 = 2e,
and thus
0<^/to< sup |/to-/O0| <2e
for each € > 0. This implies a>f(a) = 0.
For the converse, assume co/(a) = 0. Let € > 0. Then from the definition
of the oscillation, we see that there exists some neighborhood V of a such that
supr yeV |/to - f(y)\ < €. In particular, we have \f(x) - f(a)\ < e for all
x e V, and this shows that / is continuous at a.
(b) We can assume that / is an increasing function. Note that we can consider
neighborhoods of a of the form (c, d) with a € (c, d). Consider first a
neighborhood (c, d) of a and assume that x, y e (c, d) satisfy x < a < y. Since / is
increasing, it follows that 0 < lim,_>a+ f(t) — lim/—fl- /(/) < f(y) - /to, and
from this, we infer that
lim /(/) - lim /(/) < cof(a).
On the other hand, if 6 > 0 is given, then there exists some 8 > 0 such that the
open interval / = (a — 5, a -f 5) satisfies / C X and
07(a) < sup |/to - /O0I < f lim f(t) - lim f(t)] + *.
This implies a)f(a) < lim,_a+ f{t) — limr_*a- /(/), and so
to/(a) = lim f(t) - lim f(t)
holds true.
Problem 8.13. Show that a finite union of nowhere dense sets is again a nowhere
dense set. Is this statement true for a countable union of nowhere dense sets?

70
Chapter 2: TOPOLOGY AND CONTINUITY
Solution. Let A and B be two nowhere dense sets. Using the identity S° = Sc c
(see Problem 8.1), we have
(AUB)° = (A U5)c"c= (A"c H B~c )"cc (a~c~ n B"c- )c
= A~c~c U B"c-C = (I)°U(5)°=0U0 = 0.
An easy induction argument can now complete the proof.
The countable union of nowhere dense sets need not be nowhere dense. An
example: Take X = R, and let En = {/*„}, where {n, r2,...} is an enumeration
of the rational numbers. Clearly, each En is nowhere dense, while (JJJLi ^n =
{;*i, ;*2,...} is not nowhere dense.
Problem 8.14. Show that the boundary of an open or closed set is nowhere
dense.
Solution. Repeat the solution of Problem 6.5.
Problem 8.15. Let f: (X, t) -> R, and let D be the set of all points ofX where
f is discontinuous. If Dc is dense in X, then show that D is a meager set.
Solution. From ~D* = X, it follows that D° = (7^)° = 0. Now, the proof can
be completed by observing that D is an /v-set (Theorem 8.10).
Problem 8.16. Show that there is no function /: R -> R having the irrational
numbers as the set of its discontinuities.
Solution. Let / denote the set of all irrational numbers of R. If / is the set of
discontinuities of a function /: R —► R, then (by Theorem 8.10) / is an FCT-set.
However, this is impossible by Problem 6.6.
Problem 8.17. Show that every closed subset of a metric space is a G&-set and
every open set is an Fa-set.
Solution. Let A be a nonempty closed subset of a metric space X. Then the
function /: X —> R, defined by
f(x) = d{x, A) = infWOt, y): y <= A],
is continuous (see the proof of Lemma 10.4) and satisfies
bo oo
A=r'((o})=/-'(n(-i.i))=n/",(H.i))-

Section 8: TOPOLOGICAL SPACES
71
(See the discussion at the end of Section 6 of the text.) Thus, A is a G^-set. By
Theorem 8.9 every open set is an FCT-set.
Problem 8.18. Let B be a collection of open sets in a topological space (X, t).
If for each x in an arbitrary open set V there exists some B e B with x e B C V,
then B is called a base/o/* x. In general, a collection B of subsets of a nonempty
set X is said to be a base //
1. \JBsBB = X.and
2. for every pair A, B € B and x e A n B, there exists some C e B with
x eC OADB.
Show that ifB is a base for a set X, then the collection
x = [V c X: V a- € V there exists B eB with x e B c V }
is a topology on X having B as a base.
Solution. Obviously, B c x holds. Clearly, {ier, and from condition (1)
it follows that X e x. Also, it should be clear that x is closed under arbitrary
unions.
Now, let V, W e x and x e V 0 W. Choose two sets Ay B e B with
x e A c V and a* g B c W. By condition (2), there exists some C e B with
x eC £ AHB CV n\Vy that is, V n W e r. Thus, r is a topology.
The verification that B is a base for x is straightforward.
Problem 8.19. Let (X, r) be a topological space, and let B be a base for the
topology x (see the preceding exercise for the definition). Show that there exists a
dense subset AofX such that card A < card B.
Solution. If B e B and B ^ 0, then fix some xB € B and consider the set
A = [xB: B e B \ {0}}. We claim that:
1. A is dense in X, and
2. card A < card B.
To see (1) let V be a nonempty open set. If a e V, then there exists some
B e B with x e B c V. It follows that a5 g V, and so V n A ^ 0. This shows
that A is dense in X.
For (2) note that the function f:B \ {0} —> A, defined by f(B) = xB, is
onto. By the Axiom of Choice there exists a subset C of B such that Cnf~\{x})
consists precisely of one point for each x e A. Then f:C —> A is one-to-one
and onto, proving that card A = card C < card B.

72
Chapter 2: TOPOLOGY AND CONTINUITY
Problem 8.20. Let /: X -> Y be a function. Ifz is a topology on X, then the
quotient topology Zf determined by f onY is defined by z/ = {OC.Y: f~x(0) ez}.
a. Show that z/ is indeed a topology on Y and that /: (X, r) -> (F, T/) is
continuous.
b. Ifg: (7, Zf) —► (Z, Ti) is afunction, then showthatthe composition function
g o /: (X, t) -> (Z, Tj) is continuous if and only ifg is continuous.
c. Assume that f:X->Y is onto and that z* is a topology on Y such that
f: (X, r) -* (7, r*) is an open mapping {i.e., it carries open sets ofX onto
open sets ofY) and continuous. Show that r* = r/.
Solution, a. (1) Since f'l(0) = 0 e z and fl(X) = X g t, we see that
0, Y G T/.
(2) If V, W € T/f then the identity /-l(V n W) = /_1(V) n /"'(WO implies
that V Pi IV G zf.
(3) If {V/: / G /} is a family of t/, then in view of the identity /_1 ((J V/) =
U f~l(Yi)9 we see that |J V- € r7.
b. Assume go/ is continuous. If V is an open subset of Z,then f~l(g~l(V)) =
(g o /)_1(^) € r shows that g'^V) € r/. That is, g is continuous.
c. Since / is continuous, it is easy to see that r* c Zf holds. On the other hand,
let V e Zf. Then f~l(V) e r, and moreover, since / is an open mapping and
onto, we have V = /(/_1(V)) e z* (see Problem 1.7). That is, zf C z* also
holds, and so Zf = r*.
Problem 8.21. T/zw exercise presents an example of a compact set whose closure
is not compact. Start by considering the interval [0, 1 ] with the topology z
generated by the metric d(x, y) = |jc — 3^ |. It should be clear that {[0, 1], z)isacompact
topological space. Next, put X = [0, 1] U N = [0, 1] U {2, 3,4,...}, and define
z* = rU{[0, 1]LM: A C1N}.
a. Show that z* is a non-Hausdorjf topology on X and that r* induces z on
[0, i].
b. Show that (X,z*) is not a compact topological space.
c. Show that [0, 1] is a compact subset o/(X, r*).
d. Show that [0, 1] is dense in X (and hence, its closure is not compact.
e. Why doesn't this contradict Theorem 8.12(1)?
Solution, a. (1) Clearly, 0, X e t*.
(2) Let V, W e z*. Then we have the following cases:
CASE I. V, W e z. In this case, V n W e z c r*.
CASEEL V e r and W £ r (and vice versa). Note that VfW = VGTCr*.

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS 73
CASE IE. V i x and W £ x. In this case, we have V n W = [0, 1] U A for
some ACAf. That is, V n W G r*.
(3) Let {V/: / g /} be a family of r*. If V, G x holds for each /, then clearly
(J V} e r C t* holds. On the other hand, if some V/ is of the form [0, 1] U A,
then (J V/ is of the same type, and hence, it belongs to r*.
Thus, r* is a topology on X that induces r on [0, 1].
b. The cover X = UJJli^ ^ u M) cannot be reduced to a finite cover.
c. Since ([0, 1], r) is a compact topological space and x* induces x on [0, 1],
it follows that [0, 1] is a compact subset of X.
d. If x G X \ [0, 1], then every neighborhood V of x is of the form V =
[0, 1] U A for some A c TV. Thus, V n [0, 1] = [0, 1] # 0 holds for every
neighborhood V of x. Therefore, [0, 1] = X holds.
e. This does not contradict Theorem 8.12(1) because (X, r*) is not a Hausdorff
topological space.
Problem 8.22. A topological space (X, x) is said to be connected if a subset of
X that is simultaneously closed and open (called a clopen set) is either empty or
else equal to X.
a. Show that (X, x) is connected if and only if the only continuous functions
from (X, r) into (0, 1} (with the discrete topology) are the constant ones.
b. Let f: (X, r) —► (Y, t*) be onto and continuous. //(X, x) is connected,
then show that (Y,x*) is also connected.
Solution, (a) If f:X —> {0, 1} is a nonconstant continuous function, then
/_1({0}) is a nonempty clopen set which is different from X, and so X is not
connected.
For the converse, assume that every continuous function from X into {0,1} is
constant. If A is a clopen subset of X different from 0 and X, then the function
f:X —> (0, 1}, defined by /(.r) = 1 if jc G A and f(x) = 0 if x i A,
is a nonconstant continuous function, a contradiction. Thus, X is a connected
topological space.
(b) Let A be a clopen subset of Y. By the continuity of /, the set /-1(^) is
a clopen subset of X. Since X is connected, f~](A) = 0 or /_1(^) = ^- Also,
since / is onto, f{f~[(A)) = A holds (Problem 1.7). Thus, A = 0 or A = Yy
proving that Y is a connected topological space.
9. CONTINUOUS REAL-VALUED FUNCTIONS
Problem 9.1. // u, v, and w are vectors in a vector lattice, then establish the
following identities:
a. MVii-fiiAu = H + ij;

74
Chapter 2: TOPOLOGY AND CONTTNUrTY
b. w — v v w = (w — v) A (w — iu);
c. w — v a w = (w — i>) v (w — tu);
d. a(w aw) = (aw) A (au) if a > 0;
e. |w — v\ = w v v — w A u;
f. uvv = \(u + v + \u- v\);
g. w a v = j(w + v - |w - u|).
Solution. We use the identities (a), (b), and (d) in Section 9 of the text.
(a) Replace w by —(w + v) in w a v + tu = (w + w) A (u ■+- w) to get
w a v — (w + u) = (—v) A (—w) = — w v v.
(b) M-uvii) = H (—v) A (—w) = (w — v) A (w — iu).
(c) M-uAiy = « + (—u) v (—w) = (w — u) v (w — w).
(d) If a > 0, then
a(w Al)) = a[-(-w) v (-t>)] = -a[(-w) v (-U)]
= —(—aw) v (—af) = (aw) A (au).
(e) Using (a), we see that
w v v - w A v = w v v + [w v v - (w + v)] = 2(w v u) - (w + v)
= (2w) v (2v) - (w + u) = (w - v) v (v - w) = |w - u|.
(f) Using (e) and (a), we get
w + v + |w - u| = (w v i> + w a u) + (w v v - w a v) = 2(w v v).
(g) As in (f), we get
W + U — |W — u|=WVl>-f-WAl> — (WVU — U Av) = 2(w A V).
Problem 9.2. Ifu and v are elements in a vector lattice, then show that:
a. |w + v\ v |w - v\ = |w| 4- \v\, and
b. |m + u|a|ii-i;| = ||ii|-|i;||.
Solution, (a) Note that
|w + v\ v |w - u| = [(w + u) v (-w - v)] v [(w - v) v (-w + u)]
= [(w + v) V (-w + u)] v [(-« - v) v (w - u)]
= [w v (-w) + v] v [(-w) v w - u]
= [mv(-m)] + [uv(-i;)] = |ii| + |i;|.

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
75
(b) Using the distributive law, we see that
\u + v\/\\u-v\ = [(u + v) v (~u - v)] A [(u -v)v (-U + v)]
= [(U + V) A (U-V)] V [(-m-u) A (U-V)] V [(w + u) A (-M + V)] V • • •
' ' • V [(-M - u) A (-M + U)]
= [U + V A {-V)] V [(-W) AU -v]v[u A (-M) + u] V [u A (-U) - m]
= (k - M) V (-M - |V|) V (V - \U\) V (-V - \u\)
= [u v (-«) - |u|] V[l)V (-v) - |u|] = (|u| - \V\) V (\v\ - |M|)
= I l«l — l"l I-
Problem 9.3. Show that \u\ a \v\ = 0 /io/<& i/awd only if\u + v\ = \u — v\
holds.
Solution. If |m|a|u| = 0, then using parts (a) and (b) and part (e) of Problem 9.1,
we get
\u + V\ A \u — w| = ||M| — |u|| = |u| v \v\ - |U| A |v| = |u| v |V|
= \u\ + |u| - \u\ a \v\ = \u\ + |v| = \u + v\ v \u - v\.
This easily implies that \u -f v| = \u — v\ holds.
For the converse, assume that \u + v\ = \u — v\. Then by parts (a) and (b) of
Problem 9.2, we have
\u\ + \v\ = \\u\-\v\\ = \u\v\v\-\u\a\v\
= (|H| + |V| - |l<| A |V|) - \U\ A \V\ = \U\ + \V\ - 2(|M| A |V|),
from which it follows that \u\ a \v\ = 0.
Problem 9.4. Show that the vector space consisting of all polynomials (with
real coefficients) on R is not a function space. Prove a similar result for the
vector space of all real-valued differentiable functions on ]R.
Solution. If p is the polynomial defined by p(x) = x, then \p\(x) = \p(x)\ =
\x\ holds. Clearly, \p\ is not differentiable (and hence, it is not a polynomial
either).
Problem 9.5. Let X be a topological space. Consider the collection L of all
real-valued functions on X defined by
L = {/ € R*: 3 [fn] c C(X) such that lim/„(jc) = /(*) V.v e X).
Show that L is a function space.

76
Chapter 2: TOPOLOGY AND CONTINUITY
Solution. Clearly, L is a vector space. Now, let /, g € L. Choose two
sequences {/„} and [gn] of C(X) with lim/,,(*) = f(x) and \imgn(x) = g(x)
for all x. Then /„vg„€ C(X) for each rc and
lim /„ v gn(x) = lim -[/„(*) + gn(x) + |/„(.x) - gn(x)\ ]
= \[fW + gW + I/to - *(*)l] = / v *(*),
so that /vg g L. Similarly, / a g e L, so that L is a function space.
Problem 9.6. Lef L be a vector space of real-valued functions defined on a set
X. If for every function f € L the function \f\ {defined by \f\(x) = \f(x)\for
each x e X) belongs to L, then show that L is a function space.
Solution. Use the identities
/v* = ^(/ + * + |/-*|) and /As = i(/+s-|/-*|).
Problem 9.7. Consider each rational number written in the form j, where n >
0, and m and n are integers without any common factors other than ±1. Clearly,
such a representation is unique. Now, define f:JR. ->• JR by f(x) — 0 if x is
irrational and f(x) = £ if x = ^ as above. Show that f is continuous at every
irrational number and discontinuous at every rational number.
Solution. The proof will be based upon the following property: Let {rn} be a
bounded sequence of distinct rational numbers. If rn = j1 (where kn > 0, and
mn and kn do not have common factors), then lim/:,, = oo.
To see this, pick some number M > 0 such that \rn\ < M for each n, and so
\mn\ < Mkn. Now, if for some C > 0 we have \kn\ < C for infinitely many
a, then \mn\ < MC must also hold for the same infinitely many n. However,
this contradicts the fact that there is a finite number of rational numbers ^ with
\m\ < MC and |rz| < C.
Now, let x be an irrational number. If [xn} is a sequence of irrational numbers
with xn —> x, then 0 = f(xn) —> 0 = /(jc). Thus, if / is not continuous
at jc, then there exists a sequence {/„} of rational numbers with rn —► x and
lim f(r„) ^ 0. Since x is irrational, we can assume rn ^ rm whenever n ^ m.
Write r„ = j2-, and note that f(rn) = ~ -f± 0 implies kn -/> oo, a contradiction.
Therefore, / is continuous at every irrational number.
Now, let /* be a rational number. Choose a sequence {/„} of distinct rational
numbers with rn = j°- —y r. Now, note that lim/(r„) = lim — = 0 # f(r)
holds, which shows that / is not continuous at r. That is, / is discontinuous at
every rational number.

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
77
Problem 9.8. Let f: [a, b] -+ IRbeincreasing, i.e.,x < y implies f(x) < f(y).
Show that the set of points where f is discontinuous is at-most countable.
Solution. Let /: [a, b] —► IR be increasing, and let D be the set of
discontinuities of /. For each x e D choose a rational number rx such that
lim,t.r/(f) < rx < limf|.r/(r). Since x, y £ D with x < y implies
rx < lim/(f) < lim/(0 < ryi
it follows that rx ^ ry whenever x ^ y. Thus, x i—> rx is a one-to-one function
from D into the set of rational numbers, and so D is at-most countable.
Problem 9.9. Give an example of a strictly increasing function /: [0, 1] -> R
which is continuous at every irrational number and discontinuous at every rational
number.
Solution. For each t e [0,1], let /,: [0, 1] -> [0, 1] be a strictly increasing
function which is continuous everywhere except at * = t. For instance, for
0 < t < 1 let
r, , (0.5* if 0 <x < r,
Mx)=\x if *<*<!,
and /0(jc) = 0.5 + 0.5* if 0 < x < 1 and /0(0) = 0.
If {/*i, r2,...} is an enumeration of the rational numbers of [0, 1], then define
the function /: [0, 1] -> [0, 1] by
00 ,
and note that / satisfies the desired properties.
Problem 9.10. Recall that a function /:(X, r) -> (7, t\) is called an open
mapping if f(V) is open whenever V is open. Prove that z//:R -> R is a
continuous open mapping, then f is a strictly monotone function—and hence, a
homeomorphism.
Solution. Let (a, b) be a finite open interval of R. Since / attains a maximum
value on [a, b] and /((tf, b)) is an open set, it is easy to see that the extrema of
/ on [a, b] take place at the end points. In particular, this implies f(a) ^ f(b).
(If f(a) = /(£), then /((fl, b)) must be a one-point set, contradicting the fact
that / is an open mapping.) Next, we claim that / is strictly monotone on
(a, b). To see this, assume f(a) < f(b), and a < x < y < b. Then note first that

78
Chapter 2: TOPOLOGY AND CONTINUITY
/(tf) < f(x) < fib) must hold. Indeed, if /(a) < f{a) holds, then / attains its
minimum on [ay b] at some interior point. Similarly, if /(a) > f(b) holds, then
/ attains its maximum value on [a,b] at some interior point. However, (since /
is an open mapping) both cases are impossible, and so f(a) < /(a) < f(b) holds.
By the same arguments, /(a) < f(y) < f(b). Thus, / is strictly increasing on
(a, b). Similarly, if f(a) > f(b) holds, then / is strictly decreasing on (a, b).
Now, assume that / is strictly increasing on (0, 1), and let a* < y. Choose
some n with (0, 1) c (—«, n) and a\ y e (—n, n). Since / is strictly monotone
on (—h, h), and strictly increasing on (0,1), it is easy to see that / must be
strictly increasing on (—w, n). Thus, /(a) < f(y) holds, and this shows that /
is strictly increasing on R. (We remark that the function / need not be onto.
However, the mapping /:R—> f(M) is a homeomorphism.)
Problem 9.11. Let X be a nonempty set, and for any two functions f,ge IR*
let
d(f, g) = sup
xeX l + \fW-g(x)\
Establish the following:
a. (JRX, d) is a metric space.
b. A sequence {/„} C R* satisfies d(fn, f) -+ Ofor some f € R* if and
only if{fn] converges uniformly to f.
Solution, (a) Clearly, d(/, g) > 0 for all /, g e Rx and d(f, g) = 0 if and only
if / = g. Moreover, it should be clear that d(f, g) = d(g, f) for all /, g e R*.
What needs verification is the triangle inequality. To do this, we need the following
two properties:
1. 0 < x < y implies y~ < j^- , and
2- T^<TTI + T^f°ra»*.^0-
Property (1) follows from the fact that the function f{t) = j~ [t > 0) is strictly
increasing on [0, co); notice that f\t) = (1 -f t)~2 > 0 for each / > — 1. For (2)
fix a, y > 0, and note that
(A + y)(l + A)(l + y) = A(l + A)(l +y) + y(\+x){l+y)
< [a(1 +X)(l+y) + xy(l+y)] + [y(l +x)(l + y) + av(1 +a)]
= a(1 + y)(l + a + y) + v(l + a)(1 + a + y).
Dividing across by (1 +a)(1 +y)(l +x 4-y), the validity of (2) can be established.

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
79
Now, let f,g,he Rx and x € X. From
l/Cx) ~ g(x)\ < |/(jc) - A(jr)| + |/i(.r) - g(x)\
and (1) and (2), we get
\f(x)-g(x)\ ^ \nx)-h(x)\ + \h(x)-g{x)\
1 + l/to - g(x)\ ~ 1 + \f(x) - h(x)\ + \h(x) - g(x)\
^ \f(x)-h(x)\ \h(x)-g(x)\
- 1 + |/(jc) - h(x)\ 1 + \h(x) - g(x)\
<d(f,h) + d(h,g).
for all x e X. This implies
Jit \ \fW-g(x)\ ^j,,,,., .,. ,
d{f'8) = ™U + \f(x)-g(x)^d{f'h) + d{h>8)-
(b)Let{/„} c R*. Assume first that {/„} converges uniformly to some function
/ e Rx, and let € > 0. So, there exists no such that |/„(jc) — f(x)\ < e holds for
all n > n0 and all x € X, and hence, 1^J()r^^)|_ <jfn(x)-_fj&l^jjti™ft
n > no and all x € X. It follows that
</(/m /) = SUp —- < €
reX 1 + l/nU)-/C*)l
for all n > no. This shows that d(fni f) -» 0.
For the converse, assume d(/„, /) -> 0, and let € > 0. Then there exists some
azq such that
Alt K \fnW~fM\
d(fn, f) = SUp ,,,,,. 77TT <
xex 1 + !/«(*)-/tol 1+6
for all n > no, and hence, {^rX)~J{f)L < -^ for all n > no and all x e X. This
implies |/„Cv) — f(x)\ < € for all n > no and all jceX, which means that {/„}
converges uniformly to /.
Problem 9.12. Let /, /i, fa ... be real-valued functions defined on a
compact metric space (X, d) such thatxn -> xinX implies fn (xn) -> f(x) in R. /// is
continuous, then show that the sequence of functions {/„} converges uniformly to f.
Solution. Assume that the functions /, /i, fa ... satisfy the stated
properties and that the function /: X —> R is continuous. Also, assume by way of

80
Chapter 2: TOPOLOGY AND CONTINUITY
contradiction that the sequence {/„} does not converge uniformly to /.Then an
easy argument shows (how?) that there exist e > 0, a subsequence {gn} of {/„},
and a sequence {xn} of X such that
\gn(x„) - /(a„)| > s for each n. (•)
Since X is compact, the sequence [xn] has a convergent subsequence in X,
say XLn —> a. By the continuity of /, we see that /(a%,) —> /(a). Also, from
our hypothesis, it follows that gkn(xO —► /(*)> and so
\gkM - /(a*J| —> |/(a) - /(a)| = 0,
contrary to (•). Therefore, the sequence {/„} converges uniformly to /.
Problem 9.13. For a sequence {/„} of real-valued functions defined on a
topological space X that converges uniformly to a real-valued function f onX establish
the following.
a. // xn —► x and f is continuous at x, then fn(xn) —► /(a).
b. If each fn is continuous at some point xq g X, then f is also continuous at
the point xo and
lim lim /„(*) = lim lim fn(x) = /(a0).
Solution, (a) Assume / is continuous at a, xn -» x and let e > 0. Choose
some k with |/„()0 — f(y)\ < s for all n > k and all y e X. By the continuity
of / at A', there exists some m > k with |/(a„) — /(a)| < £ for all n > m.
Thus,
\Mx„) - /(a)| < |/„(a„) - /(a„)| + |/(a„) - /(a)| < 2e
holds for all n > m, so that lim /„(*„) = /(a).
(b) Assume that each /„ is continuous at Ao € X and let € > 0. Since {/„}
converges uniformly to / on X, there exists some k satisfying |/a(a) — /(a)| < €
for all a e X. Now, the continuity of fk at Ao guarantees the existence of a
neighborhood V of Ao such that |/*(*) — /*(ao)| < € for all a € V. Then
l/U) - /Uo)l < I/O*) - /*(*)! + \fkW - A(*o)l + l/*(*o) - /(*o)l
< € + €+€= 3€

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
81
holds for all x e V, which shows that / is continuous at *o- For the equality, note
that
while
lim lim fn(x) = lim f(x) = f(x0),
lim lim fn(x) = lim fn(x0) = f(x0).
n —oo jc-Kto /i-voo
Problem 9.14. Let fn: [0, 1] -> R be defined by fn(x) = xn for x e [0, 1].
Show that {/„} converges pointwise and find its limit function. Is the convergence
uniform?
Solution. Clearly,
0, if 0 <x < 1;
Since / is not continuous, the convergence cannot be uniform; see Theorem 9.2.
Problem 9.15. Let g:[0, \] -> IRbea continuous function with g(l) = 0. Show
that the sequence of functions {/„) defined by fn(x) = xng(x) for x e [0, 1],
converges uniformly to the constant zero function.
Solution. Let s > 0. Choose some 0 < 8 < 1 with \g(x)\ < £ whenever
8 < x < 1. Now, pick some M > 0 with |gCO| < M for all x e [0, 1], and
then select some k with M8n < £ whenever n > k. Thus, for each n > k we
have \x"g(x)\ < M8n < £ for 0 < x < 8 and |jcw^(jc)| < |g(jc)| < £ for all
5 < x < 1. That is, the sequence {/„} converges uniformly to the constant zero
function.
Problem 9.16. Let {/„} be a sequence of continuous real-valued functions
defined on [a,b], and let [an] and {bn} be two sequences of [a,b] such that
\iman = a and \imbn = b. //{/„} converges uniformly to f on [a, b], then show
that
;i-*oo
Jrbn rb
' Mx)dx= /
an J a
rb
lim / fn(x)dx= I f(x)dx.
Solution. Let £ > 0. Pick some k such that for all n > k we have:
1. an — a < £ and b — bn < e; and
2. \fn(x)-f{x)\<£ forall*e[a,6].

82
Chapter 2: TOPOLOGY AND CONTINUITY
Also, since / is continuous (Theorem 9.2), there exists some M > 0 satisfying
I/OOI < M for all x e [a, b]. Thus,
\fnfn(x)dx-j f(x)dx\
= I f "fnWdx - [ "f(x)dx - P f{x)dx - /" /(*)</*
'»/a„ Ja„ J a Jbn
< I \Mx)-m\dx+ P\fw\dx+ [\fw\dx
J a Ja Jbn
< s(b - a) + M{an -a) + M(b - bn) < s(2M + b-a)
holds for all n > k, and our conclusion follows.
Problem 9.17. Let {/„} be a sequence of continuous real-valued functions on a
metric space X such that {/„} converges uniformly to some function f on every
compact subset ofX. Show that f is a continuous function.
Solution. Let xn —> x in X. Put K = [x\,x2,...} U {a*}, and note that K
is a compact set—every open cover of K can be reduced to a finite cover. Since
{/„} is a sequence of continuous functions that converges uniformly to / on K,
it follows from Theorem 9.2 that / is continuous on K. Since xn —> x holds
in K, we get f(xn) —> f(x). That is, / is a continuous function.
Problem 9.18. Let {/„} and [gn] be two uniformly bounded sequences of real-
valued functions on a set X. If both {/„} and [gn] converge uniformly on X, then
show that {fngn} also converges uniformly on X.
Solution. Assume that [fn] and {#„} converge uniformly to / and g,
respectively. Let s > 0. Choose some k with \fn(x)—f(x)\ < £ and \gn(x)—g(x)\ < e
for all n > k and all x e X. Also, pick some M > 0 so that \fn(x)\ < M and
\gn(x)\ < M hold for all n and all x. Now, note that
\fnWgn{x)-f{x)g{x)\
< \fnW\ • \g„(x) - g(x)\ + |*(*)| • \fn(x) - f(x)\ < 2Mb
holds for all n > k and all x e X.
Problem 9.19. Suppose that {/„} is a sequence of monotone real-valued
functions defined on [a, b] and not necessarily all increasing or decreasing. Show

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
83
that if [fn] converges pointwise to a continuous function f on [a,b], then {/„}
converges uniformly to f on[a,b].
Solution. Let e > 0. Since / is uniformly continuous (Theorem 7.7), there
exists some 8 > 0 so that \f(x) — f(y)\ < e holds whenever |,t — y\ < 8. Fix
a finite number of points a = xq < x\ < • • • < xk = b with jc,- — Jt/_i < 8 for
1 < / < k, and then pick some m such that |/„(*,•) — f(x{)\ < s holds for each
0 < / < k and all n > m.
Now, let n > m. Assume that fn is decreasing. If x € [a,b], then jc,_i <
x < Xi holds for some 1 < / < k, and so
\fnW - fn(Xi)\ = fn(x) - fn(x{) < /„fc_,) - fn{Xi)
= Un(Xi-l) - /(*/-!)] + t/te-l) ~ f(Xi)] + [/to) - /n(X/)]
< £ + £ + £ < 3£.
A similar inequality holds true if /„ is increasing. Therefore,
\fn(x) - f(x)\ < \fn(x) - fn(Xi)\ + \fn(Xi) - f(Xi)\ + |/U0 - /U)|
< 3e + e -f e = 5f
holds for all x e [a,b] and all n > m. That is, {/„} converges uniformly to /.
Problem 9.20. Let X be a topological space and let {/„} be a sequence of real-
valued continuous functions defined on X. Suppose that there is a function f: X ~>
R such that f{x) = lim fn(x) holds for all x e X. Show that f is continuous at
a point a if and only if for each € > 0 and each m there exist a neighborhood V
of a and some k > m such that \f(x) — fk(x)\ < € holds for all x G V.
Solution. Assume that / is continuous at some point a. Let e > 0 and an
integer m be given. Pick a neighborhood U of a such that \f{x) — f(a)\ < s
holds for all x € U. Since lim fn(a) = f(a) holds, there exists an integer r > m
such that | f(a) — fn(a)\ < e holds for all n > r. Fix any integer k > r and
note that k > m. Since /* is a continuous function, there exists a neighborhood
W of a such that \fda) - fk(x)\ < e holds for all x € W. Now, note that if
x € V = f/nU^,then
\f(x) - fk(x)\ < \f(x) - f(a)\ + \f(a) - fk(a)\ + \fk(a) - fk(x)\ < 3e.
For the converse, assume that / satisfies the stated condition at the point a
and let e > 0. Since f{a) = lim/„(a) holds, there exists an integer m such

84
Chapter 2: TOPOLOGY AND CONTINUITY
that |/(<2) — fn(a)\ < e holds for all n > m. By the hypothesis, there exist a
neighborhood V of a and an integer k > m such that |/(a) — fk(x)\ < e holds
for all x e V. By the continuity of /*, there exists another neighborhood U of
a such that \fk(a) - fk(x)\ < £ holds for all x e U. Now, note that x € U HV
implies
\f(a) - f(x)\ < \f(a) - fk(a)\ + \fk(a) - /*(*)| + \fk(x) - f(x)\ < 3e,
which shows that / is continuous at the point a.
Problem 9.21. Let {/„} be a uniformly bounded sequence of continuous real-
valued functions on a closed interval [a,b]. Show that the sequence of functions
{(/>„} defined by (pn(x) = f*fn(t)dt for each x € [a,b], contains a uniformly
convergent subsequence on [a,b].
Solution. Since the sequence {/„} is uniformly bounded, there is some M > 0
such that |/„(a*)| < M holds for all x e [a, b] and all n. Clearly,
\4>nW\ = \f*M)dt
< M(b - a)
holds for all x e [a, b] and all n. So, the sequence {<pn} is uniformly bounded
and we claim that it is an equicontinuous sequence.
To see this, let e > 0 and put 8 = s/M. Now, note that jc, y e [a, b] and
\x — y\ < 8 imply
|*«(*) - 0nOO| = \ffnO)dt - fyfn(t)dt
[y\fnQ)\dt\<\[yMdt
= M\x — y\ < e.
Thus, the set A = {0i, 021»• • •} is equicontinuous. If >4 denotes the (uniform)
closure of A in C[a, ft], then A is bounded, closed, and equicontinuous (why?).
By the Ascoli-Arzela theorem (Theorem 9.10), the set A is a compact set. Since
[$„} is a sequence of A, it follows that (<pn} has a subsequence that converges
uniformly on [a, b].
Problem 9.22. For each n let /„: R -* R be a monotone (either increasing or
decreasing) function. If there exists a dense subset AofR such that lim/n(jc)
exists in R for each x € A, then show that lim fn(x) exists in R at-mostfor all
but countably many x.

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
85
Solution. Assume that the functions /„ and the dense subset A of R satisfy
the properties of the problem. Also, assume at the beginning that all but a finite
number of the /„ are increasing functions.
Define the function /: R —> R by
/(*) = lim sup/„(*), x €R.
Note that f{x) is a real number for each x e R. Indeed, if x e R, then there
exist a, b e A with a < x < b, and so fn(a) < fn(x) < fn(b) holds for all
sufficiently large n. Consequently,
-co < lim/„(fl) = limsup/„(a)
< lim sup/„(*) = f(x)
< lim sup fn(b) = lim fn{b) < co
Clearly, f(x) = lim fn(x) holds for each x e A. Next, note that / is an
increasing function. Indeed, if x < y holds, then from fn(x) < fn(y) for all sufficiently
large n, we see that f(x) = limsup/„Cr) < limsup/„(;y) = f(y). By
Problem 9.8, we know that / has at-most countably many discontinuities in every
closed subinterval of R. Hence, / has at-most countably many discontinuities
(why?). Now, we claim that
\imfn(x) = f(x)
holds at every point of continuity of /. To see this, let xo be a point of continuity
of / and let e > 0. Pick some 8 > 0 such that xq — 8 < x < xo 4- 8 implies
\f(x) — f(xo)\ < £, and then choose a, b € A with xq — 8 < a < xo < b <
Xq + 8. Also, pick some no such that for each n > no the function /„ is increasing
and satisfies
\fn(b) - f(b)\ < 6 and \fn(a) - f(a)\ < s.
Now, note that for n > /2o, we have
/(*<>)-/nUo) < /tro) -/„(*)
= [/(*o) - /(«)] + If ia) - fn(a)] < e + 6 = 2s
and
/«U0)-/C*0) < fn(b)-f{x0)
= 1Mb) - /(«] -I- [/(&) - /(xo)] < e + e = 2e.

86
Chapter 2: TOPOLOGY AND CONTINUITY
Thus, |/„(a*o) — /(*o)| < 2s holds for all n > no, proving that lim/„(jto) =
/Uo).
For the general case, assume that there are infinitely many increasing and
infinitely many decreasing /„. Split the sequence {/„} into two subsequences {gn)
and [hn] such that each gn is increasing and each hn is decreasing. Put
g(x) = limsupg„(jt) and h(x) = liminf/zn(x) = — limsup[—h„(x)],
and note that g{a) = h(a) holds for each a e A. By the above conclusion, g and
h are continuous except possibly at the points of an at-most countable subset C
of R, and for each point x £ C we have
lim gn(x) = g(x) and lim hn(x) = h(x).
Now, let c £ C and fix e > 0. Pick some 5 > 0 such that \x — c| < 8 implies
IgU) — £(c)| < e and |/z(jc) — h(c)\ < s. Pick a e A with |a - c\ < 8 and note
that from g{a) = //(a), it follows that
\g(c) - A(c)| < \g(c) - g(a)\ + \h(a) - A(c)| < £ + e = 2e.
Since e > 0 is arbitrary, we see that g(c) = h(c) holds for each c £ C. This
implies (how?) that lim fn(c) exists in R for each c £ C.
Problem 9.23. Consider a continuous function f: [0, oo) -» R. For eac/z a
de/z/ie f/*e continuous function fn: [0, oo) -> R by f„(x) = /(a"). STzovv that the
set of continuous functions {/i, /i,...} w equicontinuous at x = 1 //and o/i/y i/
/ w a constant function.
Solution. Let / e C[0, oo), let /„:[0, oo) —► R be defined by fn(x) =
/(a"), and let E = {/i, /i, ...}. If / is a constant function, then it should be
clear that the set E is equicontinuous at x = 1.
For the converse, assume that the set £ is equicontinuous at a* = 1. Fix a > 0
and let £ > 0. The equicontinuity of £ at jc = 1 guarantees the existence of some
0 < 8 < 1 such that |jc — 11 < 8 implies \fn(x) — /„(1)| < e for each n. From
liml/a = 1 (why?), we see that there exists some no such that \j/a — 1| < 8
holds for each n > no. Thus, if n > no, then we have
\fW - /(1)| = |/(M^)n) - /(r)|= |/„(^) - /„(1)| < e.
Since s > 0 is arbitrary, it follows that f(a) = /(l) holds for each a > 0. By
continuity, we see that f(a) = /(l) for each a > 0, and so / is a constant
function.

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
87
Problem 9.24. Let (X, d) be a compact metric space and let A be an
equicontinuous subset ofC(X). Show that A is uniformly equicontinuous, i.e., show that
for each e > 0 there exists some 8 > 0 such that x, y e X and d(x, y) < 8 imply
\f{x)-f(y)\<eforallfzA.
Solution. Let (X, d) be a compact metric space, let A be an equicontinuous
subset of C(X), and let s > 0. For each x e X there exists (by the equicontinuity
of A) some 8X > 0 such that d(x, y) < 8X implies \f(x) — f(y)\ < s for all
f e A. From X = {JxeX ^(JC> if) ^^ me compactness of X, we see that there
exist xu...,xneX such that X = |J?=i *(*/. %)•
Let 5 = ^ min{6r,,..., 8Xn] > 0 and let x,y € X satisfy d(x, y) < 8. Now,
pick some 1 < i < n with d(x,X{) < -^ and observe that \f(x) — /(jc/)| < s
for all / G A. In addition, from
rf(y,-t/)<rf(y^) + d(Jc,Jr/)<^- + ^L=«r,.
we see that |/(y) — /U/)| < £ holds for all / e A. Therefore, from the above,
if d(x, y) < 8, then
|/C*) - /()0| < \f(x) - f(Xi)\ + |/(a-;) - /(y)| < £ + e = 2e
holds for all / e A That is, A is a uniformly equicontinuous subset of C(X).
Problem 9.25. Let X be a connected topological space (see Problem 8.22 of
Section 8 for the definition) and let A be an equicontinuous subset ofC{X). If for
some xq e X, the set of real numbers {f(xo): f e A] is bounded, then show that
{f(x): f € A] is also bounded for each x e X.
Solution. Let X be a connected topological space, let A be an equicontinuous
subset of C(X), and let xq £ X be a point such that the collection of real numbers
{/(*o)*- / € -4} is bounded. Consider the set
E = {x e X: The set {/(*): / e A] is bounded }.
Since xo e £, we see that E is nonempty. We claim that E is both open and
closed. If this is the case, then by the connectedness of X we must have E = X,
and the desired conclusion follows.
We shall show first that £ is a closed set. To this end, let y e E. By the
equicontinuity of A, there exists a neighborhood V of y such that
l/to-/O0l<l
holds for all x e V and all / e A. From y e £, we see that V D E ^ 0. Fix
some z e V H E, and then pick some M > 0 such that |/(z)| < M holds for

88
Chapter 2: TOPOLOGY AND CONTINUITY
each / 6 A. In particular, we have
|/(y)|<|/(y)-/(z)| + |/(z)|<i+M
for all / e A. This means that y € E, and so E = £, i.e., £ is a closed set.
Next, we shall establish that E is an open set. To this end, let y e E. Pick
some C > 0 such that 1/001 < C holds for each f e A. By the equicontinuity
of A, there exists a neighborhood W of v such that |/(jc) — /OOI < 1 holds
for each x e W and all / e A. In particular, if x e W, then
|/(*)|<|/to-/O0| + |/O0|<l+c
holds for all / e A, and so x e E. That is, W c E holds, which shows that v
is an interior point of E. Therefore, E is also an open set.
Problem 9.26. Let [fn}be an equicontinuous sequence in C(X), where X is not
necessarily compact. If for some function f:X -> R we have lim f„(x) = /(a)
for each x e X, then show that f € C(X).
Solution. Let x e X and let e > 0. Since {/„} is an equicontinuous sequence,
there exists a neighborhood V of the point x such that \fn(y) — fnW\ < £ holds
for all n and each y € V.
Now, let y eV. Pick some k with |/a(jc) — /(*)| < e and |/a()0 — f(y)\ < £,
and note that
|/(a0 - f{y)\ < |/M - /*(*)| + |/*(x) - fdy)\ + |/aO0 - /O0| < 3*.
That is, / is continuous at the arbitrary point x.
Problem 9.27. Let Xbea compact topological space, and let [fn}be an
equicontinuous sequence ofC(X). Assume that there exists some f € C(X) and some
dense subset A ofX such that lim f„(x) = f(x) holds for each x € A. Then show
that {fn} converges uniformly to /.
Solution. Let e > 0. By the equicontinuity of {/„} and the continuity of /, for
each x e X, there exists some neighborhood Vx of x such that
1- 1/nOO - fn(x)\ < £ holds for all y eVx and all n; and
2. 1/00 ~ /Ml < e holds for all y e Vx.
By the compactness of X, there exist a*i ,..., jc* e X such that X = (J-=1 Vx..

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
89
Now, let y e VXr Choose some x e A n VXn and then pick some m,- with
\fn(x) - f(x)\ < e for all n > mh Clearly, \f(x) - f(y)\ < 2e. Thus,
|/i.O0-/(30|
< \fn(y) - AU/)| + |/nU/)-/nW| + \fnW - f(x)\ + \f(x) - /(y)|
< 5e
holds for all y e VXi and all n > mi.
Finally, put m — max{/72/: 1 < / < k }, and note that |/„(y) — f{y)\ < 5e for
all y e X and all n > m.
Problem 9.28. Show that for any fixed integer n > 1 the set of functions f in
C[0, 1] such that there is some x e [0, 1 — j;]for which
\f(x + h) - f(x)\ < nh whenever 0 < h < £,
/j nowhere dense in C[0, 1] (with the uniform metric).
Use the above conclusion and Baire's theorem to prove that there exists a con-
timious real-valued function defined on [0, 1 ] that is not differentiable at any point
of[0, 1].
Solution. Let D(f,g) = ||/ - gW^ = sup{|/(jc) - g(x)\: x e [0, 1]}. For
n > 2 define
An = {/ 6 C[0, 1]: 3 * € [0, 1 - ±] with |/(jc + A) - /(x)| < */t
whenever 0 < h < £ }.
We claim that:
1. Each /\„ is closed; and
2. Each /4„ is nowhere dense in C[0, 1] (i.e., (An)° = 0).
To see that each A„ is closed, let [fk] c A,, satisfy lim £>(/*, /) = 0 (i.e., {/*}
converges uniformly to / on [0, 1]). For each k choose some jt* E [0, 1 — -]
with \fd*k + A) - /*(**) I < /i/z for all 0 < h < £. Since [0, 1 — ^] is compact,
there exists a subsequence of {x^} that converges to some x e [0, 1 — £]. We
can assume that limx; = x. By Problem 9.13, lim /*(** -f /z) = /(* + /z) and
lim/*(**) = /(.r), and so |/(x + A) - /(*)| < «A holds for all 0 < h < £.
Thus, /6A„, and hence, /4„ is a closed subset of C[0, 1].
Now, let f e An and let e > 0. Consider the function g e C[0, 1] whose
graph is shown in Figure 2.1. Note that for each x e [0, 1) we have \g(x 4- h) —
g(x)\ = 3/2/2 for all sufficiently small h > 0. Put f\ — f -f g, and note that

90
Chapter 2: TOPOLOGY AND CONTINUITY
>'=g(x)
FIGURE 2.1. The Construction of a Nowhere Differentiable Function
D(/, fi) = HgHoo = e. On the other hand, if x e [0,1) is fixed, then for all
sufficiently small h > 0 we have
nh < 2nh = 3nh - nh < \g{x + h) - g(x)\ - \f(x + h) - /(jc)|
< \g(x + h) - g(x) - [f(x) - f(x + A)]| = |/,(jc + h) - /,(a-)|.
Thus, /i i An, and so £(/, 2e) 2 A„ for all e > 0. This shows that (A„)° = 0.
Now, for each n > 2 let
*« = {/€ C[0, 1]: 3 a- € [£, 1] with |/(jc - h) - /(*)l < nA
whenever 0 </*<-}.
By the same arguments, each Bn is closed and nowhere dense. Consequently,
from Baire's Theorem 6.17, we have
C[0,l]/(Qi4B)u((jBfl).
n=2 n=2
In particular, note that every / € C[0, 1] \ {{JZLiAn) u (U^2B«) does not
have any one-sided derivative at any point of [0, 1].
Problem 9.29. Establish the following result regarding differentiability and
uniform convergence. Let [fn] be a sequence of differentiable real-valued functions
defined on a bounded open interval (a, b) such that:
a. for some xq e (a, b) the sequence of real numbers {fn(xo)} converges in R,
and

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS
91
b. the sequence of derivatives [f'n) converges uniformly to a function
g:(£if6)-^R.
Then the sequence {/„) converges uniformly to a function /: (a, b) —> R which is
differentiable at xq and satisfies f'(xo) = g(xo).
Solution. First, we shall show that {/„} is a uniformly Cauchy sequence. To this
end, let e > 0 and pick some M > 0 such that |.v — xq\ < M for each x e (a, b).
Next, choose some k such that
\f'n(x) ~ f'm{*)\ < € for all m,n>k and all x e (a, b) (•)
and
|//i(*o) ~ /mUo)| < € for all m,n>k. (**)
Using the Mean Value Theorem, (•) and (••), we see that for each a' g (a, b) and
each pair n, m > k there exists some / e (ayb) such that
|/„(-V) - fm(x)\ < I [f„(X) - fM)] ~ [fn(XQ) ~ fm(x0)] | + \fn(x0) - fm(Xo)\
= \fn(0 ~ f,'n(0\ • |.r - Aol + |/„Uo) - /m(jC0)|
< Me + € = (1 + M)€.
This shows that {/„} is a uniformly Cauchy sequence, and hence, {/„} converges
uniformly to a function /: (a, b) -> R.
Next, for each n we consider the continuous function 0„: (a, b) —> R defined
by 0„(a) = M*lz'fo) if a* ^ a0 and 0„Uo) = /„'(ao). Using the Mean Value
Theorem and (•), we see that for each x e (a, b) there exists some cv e (a, b)
such that
|0„(.v) - 0„(.r)| = | I^W-WM/.(»)-f.(*)l | = l/^c.) - £(ct)| < 6,
for all n, m > k. This shows that {<£„} is a uniformly Cauchy sequence, and hence,
it converges uniformly to the function 0: (a, b) -> R defined by 0(a) = ^(v)~^(To)
if a + xq and </>{x0) = g(x0).
Finally, from Problem 9.13, we obtain
g(x0) = lim /„'(*(>) = Hm lim 0„(a) = lim lim 0„(a)
n->-co n—>-oo r->\0 r—>-.to «-*oo
.. r /*(-*)-/(*) r /GO - /fro)
= lim lim = lim .
v— v0 /i->oo X — Aq -*-*"*<) A — Ao
This shows that / is differentiable at xq and that /'(a'o) = g(-vo)-

92
Chapter 2: TOPOLOGY AND CONTINUITY
10. SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS
Problem 10.1. Let (X, d) be a metric space and let Abe a nonempty subset of
X. The distance function of A is the function d(-, A): X ->• IR defined by
d(x, A) = \nf[d(x, a): a e A).
Show that d(x, A) = 0 if and only ifx € A.
Solution. Clearly, d(x, A) > 0 for each a e X. Assume that a g A and let e > 0.
Then£(A, e)HA ^ 0, and so there exists some y e A such that J(a, y) < €. From
the definition of the distance function, we see that 0 < d(x, A) < d(x, y) < €.
Since € > 0 is arbitrary, this implies d(x, A) = 0.
For the converse, assume that d(x, A) = 0. If e > 0, then it follows from
d(x, A) = inf{d(x, a): a e A] < e that there exists some a e A with d(x, a) < €.
Hence, B(x, e) f! A ^ 0 for each 6 > 0, and this implies that x e A.
Problem 10.2. Let (X, d) be a metric space, let A and B be two nonempty
disjoint closed sets and consider the function f:X -> [0, 1] defined by
fM =*%&*>■ Sho» that:
a. / is a continuous function,
b. f-\{0}) = A and /-,({1}) = B, and
c. // d(A, B) = inf{d(#, b): a e A and b e B] > 0, then f is uniformly
continuous.
Solution. Let C be an arbitrary nonempty subset of X. We shall show first that
the function a h+ d(x, C) is uniformly continuous. To see this, fix a, y e X.
Choosing some c e C, we see that
d{x, C) < d(x, c) < </(*, y) + d(y, c) < d(x% y) + d(y% C),
or d(x, C) — d(y, C) < d(x, y). Exchanging the roles of a and y in the last
inequality, we get d(y, C) — d(x, C) < d(x, y). Therefore,
\d{xX)-d{yX)\<d(x,y\
and the uniform continuity of x h> ^(a, C) follows.
(a) Observe that since A and B are disjoint closed sets, it follows from the
Problem 10.1 that d(x, A) + ^(a, B) > 0 for each x e X. This, in connection
with the (uniform) continuity of the functions </(•, A) and d(-, B), guarantees that
/ is a continuous function.
(b) Note that /(a) = 0 if and only_if d(x, A) = 0. Now, by Problem 10.1, we
have d(x, A) = 0 if and only if x e A = A. In other words, we have /(a) = 0 if
and only if a e A. This means /_1({0}) = A.

Section 10: SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS 93
Similarly, notice that f(x) = d{x d^ = 1 if and only if d(x, B) = 0. As
above, this shows that /_,({1}) = B.
(c) Fix some e > 0 such that d{u, v) > € for all u e A and v e B. If a e A
and b € B are arbitrary, then for each z e X'v/e have
€ < d(a, b) < d(z, a) + d(z, b) < d(z, A) + d(z, B).
Now, \fx,y e X, then the inequalities
d(x, A) d(y, A)
\fw-m\ =
I d(x, A) 4- d(jcf 5) d(yf A) + d(y, B) I
|[rf(y, A) + d(y, g)]dU, A) - [d(x, A) + £f(,Yt B)]d(y9 A)\
[d(x, A) + d(xy B)][d(y, A) + d(y, B)]
\[d{x. A) - rffr, A)]d{x, B) + [rf(y, g) - d(x, B)]d{x, A)\
[d(x% A) + d(x% fi)][d(y, 4) + d(y, B)]
[d(jr,B) + £/(.r,i4)]d(JC,y)
[d(.v, A) + d(x% B)][d{y, A) + rf(y, £)]
guarantee that / is uniformly continuous.
Problem 10.3. Let A and B be two nonempty subsets of a metric space X such
that AC\B — AC\B=0. Show that there exist two open disjoint set U and V
such that A C.U and B C V.
Solution. From the solution of Problem 10.2, we know that for each nonempty
subset C of X the function x \-> d(x,C) is (uniformly) continuous. Now, consider
the function /: X -> 1R defined by
f{x) = d(x,A)-d{x,B).
By the above, / is a continuous function. From A n B — 0 and Problem 10.1, we
see that f(x) = —d(x, B) < 0 holds for each x e A. Similarly, f(x) > 0 holds
for each x € B. Consequently, the two disjoint open sets U — f~x ((—oo, 0))
and V = /_I((0, oo)) satisfy A C U and B C V.
Problem 10.4. Show that a closed set of a normal space is itself a normal space.
Solution. Let C be a closed subset of a normal space X. We consider C equipped
with the topology induced by X. Now, assume that A and B are two disjoint closed
subsets of C. Since C is closed, it is easy to see that A and B are also closed subsets

94
Chapter 2: TOPOLOGY AND CONTINUITY
of X. Pick two open subsets V\ and W\ of X satisfying A c. V\9 B Q W\ and
VxnWi = 0. Now, if V = C fl V\ and W = C n V^, then V and W are two
disjoint open subsets of C satisfying A C. V and B c IV. This shows that C
equipped with the relative topology is a normal space.
Problem 10.5. Let X bea normal space and let A and B be two disjoint closed
subsets ofX. Show that there exist open sets V and W such that A C.V, B C.W
andVnW = 0.
Solution. Assume that A and B are two disjoint closed subsets of a normal space
X. Pick two disjoint open sets V and W\ satisfying A c V and B C W\. We
claim that V D W\ = 0. Indeed, if jc e V H W\, then on one hand W\ is a
neighborhood of x, and on the other hand, x belongs to the closure of V, which
imply W\ H V ^ 0, a contradiction.
Now, since V fl B = 0 and X is normal, there exist two disjoint open sets V\
and W such that V" C Vx and 5 C W. As before, Vi 0 W = 0, and clearly the
open sets V and W satisfy the desired properties.
Alternatively: If a continuous function f\X-+ [0, 1] satisfies A C f~l([0})
and 5 c /^({l}), then the open sets V = /_I([0, \)) and W = /_1((|, l])
satisfy A c V, B c F, and Vn TV = 0.
Problem 10.6. Show that a topological space is normal if and only if for each
closed set A and each open set V with A C V, there exists an open set W such
that KH/cfci/.
Solution. Let X be a topological space. Assume first that X is a normal space
and let a closed set A and an open set V satisfy A c V. Then A H Vc = 0
and Vc is a closed set. Pick two disjoint open sets W and U such that A C. W
and Kc C {]. In particular, W H U = 0. This implies W H Uc = 0, and so
Wc V.
For the converse, assume that the property is satisfied and let A and B be two
disjoint nonempty closed sets. If V = Z?c, then V is an open set such that A C V.
By our hypothesis, there exists an open set W such that AC.WC.WC.V = BC.
If (/ = W , then £/ is an open set disjoint from W and satisfies B c. U. This
shows that X is a normal space.
Problem 10.7. For a closed subset A of a normal topological space X, establish
the following:
a. There exists a continuous function f:X-+ [0, 1] satisfying /_1({0}) = A
if and only if A is a Gs-set.

Section 10: SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS 95
b. If A is a Gs-set and B is another closed set satisfying A D B = 0, then
there exists a continuous function g:X -> [0, 1] such that g-1({0}) = A
and g(b) — 1 for each b e B.
Solution. Let A be a closed subset of a normal topological space X.
(a) If there exists a continuous function f:X-> [0, 1] such that /_1((0}) = A,
then the identity
CO 00
A=/-'((o))=ri(n[o^))=n/-,([o.i))
n—\ //=1
shows that A is a Gs-set.
For the converse, assume that A is a Gs-set. Pick a sequence {Vn} of open sets
such that A = C\T=\ ^»- Since A n Vj; = 0, it follows from Uryson's lemma that
there exists a continuous function f„:X -> [0, 1] satisfying fn(a) = 0 for each
a e A and /„(.v) = 1 for all x e V*. Now, consider the function /: X -> [0, 1]
defined by
CO
«=1
From the Weierstrass' M-test (Theorem 9.5) and Theorem 9.2, it is easy to see that
/ is a continuous function, and we claim that f~] ({0}) = A. Clearly, f(x) = 0
for each x e A. Now, assume /(a) = 0. Then fn(x) = 0 for all /?, and so (in view
of fn(v) = 1 for each v e V*) we have .r e Vn for each n, i.e., x e fl^li Vn = A.
Therefore, /-,({0}) = A.
(b) Assume now that A is a closed Gs-set and B is another closed set such that
A fl B = 0. So, there exist two disjoint open set V and W such that A C V and
5 C W. This implies that the sequence {V,,} introduced in part (a) can be assumed
to satisfy Vn c V for each a?. In particular, each /„ satisfies fn{b) = 1 for each
b e B. Now, it is easy to see that the continuous function / constructed in the
preceding part satisfies the desired property.
Problem 10.8. Show that a compact subset A of a Hausdorff locally compact
topological space is a Gs-set if and only if there exists a continuous function
/: X -* [0, 1] such that A = /"'({O}).
Solution. If A = /-1({0)), then—as in the solution of part (a) of the preceding
problem—the set A is a Gs-set. For the converse, assume that A = fXfL] V^„,where
each Vn is an open set. By Theorem 10.8, for each n there exists a continuous

96
Chapter 2: TOPOLOGY AND CONTINUITY
function /„: X -* [0, 1] such that f„(x) = 1 for each x e A and fn(x) = 0 for
each a* ^ Vn. Now, as in the the solution of part (a) of the preceding problem,
notice that the function /: X -> [0, 1] defined by f(x) = Y1T=\ 4r/n0O satisfies
the desired properties.
Problem 10.9. A topological space X is said to be perfectly normal if for every
pair of disjoint closed sets A and B, there is a continuous function f: X -> [0, 1 ]
such that A = /-1({0}) and B = /_1({1}). {Part (b) of Problem 102 shows that
eveiy metric space is perfectly normal.)
Show that a Hausdoiff normal topological space is perfectly normal if and only
if eveiy closed set is a Gs-set.
Solution. Let X be a Hausdorff normal topological space. Assume first that X
is perfectly normal and let A be a proper closed subset of X. If a € X satisfies
a £ A, then A PI {a} =0 and {a} is a closed set. So, there exists a continuous
function f:X-+ [0, 1] with /_1({0}) = A. This implies (as in the solution of
Problem 10.7), that A is a Gs-set.
For the converse, assume that every closed set is a G^-set. Let A and B be
two closed disjoint sets. By Problem 10.7 there exist two continuous functions
gJv.X -> [0, 1] such that:
i. g-1 <{Q}\ = A and g(b) = 1 for each b e £, and
ii. h-l({0}) = B and h{a) = 1 for each a e A.
Now, let / = Ig + 1(1 - h\ and note that f\X -+ [0, 1], A = /_1({0}) and
B = f-\[l}).~
Problem 10.10. Show that a nonempty connected normal space is either a
singleton or uncountable.
Solution. Let X be a (nonempty) Hausdorff connected normal space. If X is
not a singleton, then there exist a,b e X with a ^ b. Since X is Hausdorff,
singletons are closed sets, and we have {a} n {b} = 0. Now, pick a continuous
function /: X —► [0, 1] such that f(a) = 0 and f(b) = 1. The assumption that
X is connected guarantees (according to Problem 6.11(g)) that f(X) is an interval
and so f(X) = [0, 1], This easily implies that X is uncountable—in fact, it has
cardinality greater than or equal to the cardinality of the continuum.
Problem 10.11. Let X be a normal space, let C be a closed subset ofX, and let
I be a nonempty interval—with the possibility I = (-co, oo). ///: C —► / is a
continuous function, then show that f has a continuous extension to all ofX with
values in I.

Section 10: SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS 97
Solution. Assume that C is a closed subset of a normal space X and that /: X —>
I is a continuous function, where / is an interval. The interval / must be one of the
following type: (a, b), [a, b]y [a, b), (a, b]. So, we shall establish the continuous
extension of / by steps.
STEP I: / is either of the form [a, b) or (b, a].
In this case, there exists a homeomorphism h: I —> [0, 1). For instance, if
-co < a < b < co, then h(x) = |^ ls a homeomorphism between (a, b] and
[0, 1). Likewise, if a e R, then /z(a) = ^"f r defines a homeomorphism between
(-co, a] and [0, 1).
Fix a homeomorphism h: / —> [0, 1) and consider the continuous (composition)
function h o f: C -> [0, 1) c [0, 1]. By Tietze's extension theorem, there exists a
continuous function g: X -* [0, 1] satisfying g(x) = h[f(x)) for all a* e C. The
continuity of g guarantees that the set A = g~l({l}) is a closed subset of X. Also,
since for each x e C, we have g(x) = h[f(x)) e [0, 1), we see that C D A = 0.
By Uryson's lemma, there exists a continuous function 9: X -> [0, 1] such that
9(a) = 0 for each a e A and 9(c) = 1 for each c e C.
Now, consider the function 0:X —> [0, 1] defined by 0(a) = 9(x)g(x). We
claim that (p(X) c [0, 1). To see this, let a e X. If* e A,then 0(a) = 6(x)g(x) =
0-1 =0, and if a £ /\, then 0 < #(a) < 1 and so 0(a) = 9(x)g(x) < 1 is also true.
Next, define the function /: X -> / by
/(a) = (/T1 o0)(a) = h-l(9(x)g(x)).
If a G C, then 6(x)g(x) = #(a) = h(f(x))1 and hence,
/(a) = /z-,(/z(/(a))) = /(a).
This shows that /: X -» / is a continuous extension of / to all of X.
STEP II: / = [a, 6] w/r/z -co < a < b < co.
The function /*: [a, b] -» [0, 1], defined by /z(a) = 'j^J, is a homeomorphism.
By Tietze's extension theorem, there exists a continuous function g: X -> [0, 1]
satisfying g(x) = (/io /)(a) for each x e C. Then the continuous function
/ = h~l o g: X -> [a, b] satisfies f(c) = f(c) for each ceC.
STEP III: Assume I = (a, fr) w/Y/i -co < a < b < co.
In this case, there exists a homeomorphism h: (<?, b) —► (— 1, 1). (For instance,
for —co < a < b < co let h(x) = 2{^~^ — 1 and if (a, b) = (-co, co) take
/z(a) = ^ arctanA.) Now, consider the continuous function ho f:C -» (— 1, 1) c
[—1, 1] and note that by STEP II there exists a continuous function g:X -> [—1, 1]
satisfying g(c) = (h o f)(c) for each c eC.

98
Chapter 2: TOPOLOGY AND CONTINUITY
Next, let B = g~l({-\, 1}). Then B is closed and B n C = 0. By Uryson's
lemma, there exists a continuous function 6: X -> [0, 1] satisfying 0(6) = 0 for
each 6 € 5 and 0(c) = 1 for each c e C. As before, define the continuous function
0: X -> [-1, 1] by 0(jc) = 9(x)g(x). Then it is easy to see that 0(X) c (-1, 1)
and the function f:X —> (a, b), defined by / = h~l o0, is a continuous extension
of/.
11. THE STONE-WEEERSTRASS APPROXIMATION THEOREM
Problem 11.1. Let X be a compact topological space. For a subset L ofC(X),
let L denote the uniform closure ofL in C{X). Show the following:
a. IfL is a function space, then so is L.
b. IfL is an algebra, then so is L.
Solution. Let f, g e L. Pick two sequences {/,,} and {gn} of L that converge
uniformly to / and ^respectively. Also, pick some M > 0 so that ||/„||oo < M
and llgJIoo < M hold for all n.
(a) The inequality ||/„| — |/|| < |/« — f\ shows that {|/„|} converges uniformly
to |/|. Since \fn\ e L for each /?, it follows that \f\eL. This implies that L
is a function space.
(b) From the inequalities
Wfngn ~ /Slloo < llSlloo ' II/„ - /Hoc + il/.lloo - Wgn ~ g\\oo
< M(\\ftt ~f Woo+ \\gn-g\\oo).
it follows that the sequence {/„£„} of L converges uniformly to fg. Thus,
fgeL, and so L is an algebra.
Problem 11.2. Let L be the collection of all continuous piecewise linear
functions defined on [0, 1]. That is, f e L if and only iff e C[0t 1] and there exists a
finite number of points 0 = xq < x\ < • • • < xn = 1 (depending on f) such that
f is linear on each interval [jcm_i, xm]. Show that L is a function space but not
an algebra. Moreover, show that L is dense in C[0, 1] with respect to the uniform
metric.
Solution. The verification that L is a function space is routine. Since the
function f(x) = x satisfies / € L and f2 £ L, it follows that L is not an algebra
of functions.
To see that L is dense, let / e C[0, 1] and let e > 0. By the uniform continuity
of /, there exists some 8 > 0 such that |jc — jy| < <5 implies \f(x) — f(y)\ < e.
Let 0 = a*o < a'i < • • • < xn = 1 be a finite collection of points with jc,—jc/—i < $

Section 11: THE STONE-WEIERSTRASS APPROXIMATION THEOREM 99
for each 1 < / < n. The function g, defined on each subinterval [a'/_i , •*/] by
g(t) = /te-O+^^I^-'^/-^.!),
belongs to L and satisfies ||/ — glloo < e-
An alternate way of proving the denseness of L is the following: Note that 1 e
L and L separates the points of [0, 1] (why?). Thus, by the Stone-Weierstrass
theorem, L is dense in C[0, 1].
Problem 11.3. Show that a continuous function f: (0, 1) -> R is the uniform
limit of a sequence of polynomials on (0, 1) if and only if it admits a continuous
extension to [0, 1].
Solution. Let /: (0, 1) —► 1R be a continuous function. Assume first that /
has a continuous extension to [0, 1]—which we denote by /. Then, by
Corollary 11.6, the function / is the uniform limit of a sequence of polynomials
on [0, 1], and consequently /: (0, 1) —> R is likewise the uniform limit of a
sequence of polynomials on (0, 1).
For the converse, assume that there exists a sequence of polynomials {/?„} that
converges uniformly to / on (0, 1). Let £ > 0 and then pick some no such that
l/?/i(-v) — /U")l < s holds for all x e (0, 1) and all n > no. From the triangle
inequality, we see that
\pnW ~ PmW\ < \pn(x) - /(-V)| + \p„,{x) - f(x)\ < E + S = 2e
for all x € (0, 1) and all n > no. By continuity, we infer that
\pnW ~ PmM\ < 2s
holds for all x e [0, 1] and all n > azq. The above show that {/?„} is a Cauchy
sequence of C[0, 1], and so (by Theorem 9.3) the sequence (/?„} converges in
C[0, 1], say to g e C[0, 1]. It follows that f(x) = g(x) for all x e (0, 1), and so
g is a continuous extension of /: (0, 1) —> R to [0, 1].
Problem 11.4. /// is a continuous function on[Q, 1 ] such that fQ x" f(x) dx = 0
for n = 0, 1 then show that f(x) = Ofor all x e [0, 1].
Solution. By Corollary 11.6, there exists a sequence of polynomials {/?„} that
converges uniformly to /. It easily follows that {pnf} also converges uniformly
to /2, and by our hypothesis we see that f0 pn(x)fW dx = 0 holds for each n.
Now, invoke Problem 9.16 to infer that f*f2(x)dx = lim J* pn(x)f(x)dx = 0.
The latter easily implies that f(x) = 0 holds for each x e [0, 1].

100
Chapter 2: TOPOLOGY AND CONTINUITY
Problem 11.5. Show that the algebra generated by the set {1, a2} is dense in
C[0, 1] but fails to be dense in C[-l, 1].
Solution. Since the function f{x) = x2 separates the points of [0, 1], the
algebra generated by {1,a2} also separates the points of [0, 1]. Thus, by the
Stone-Weierstrass, this algebra must be dense in C[0, 1].
To see that the algebra generated by {1, a*2} is not dense in C[—1, 1], note that
for every / in the closure of this algebra, we have /(—1) = /(l). Thus, this
algebra is not dense in C[—1, 1].
Problem 11.6. Let us say that a polynomial is odd {resp. even) whenever it
does not contain any monomial of even {resp. odd) degree.
Show that a continuous function /: [0, 1] -» R vanishes at zero {i.e., /(0) = 0)
if and only if it is the uniform limit of a sequence of odd polynomials on [0,1].
Solution. If / is the uniform limit of a sequence of odd polynomials, then it
should be clear that / vanishes at zero. For the converse, assume that / e C [0, 1 ]
satisfies /(0) = 0 and let e > 0. Define the function g: [— 1, 1] -> R by
gM=\fW< if0<x<l;
-/(-a), if -1 <x <0,
and note that g e C[—1, 1]. By the Stone-Weierstrass theorem there exists a
polynomial p such that \g{x) — p{x)\ < s for each a* e [—1, 1].
Next, write p = q+r, where q is the odd polynomial consisting of the sum of
all odd terms of p and r is the even polynomial consisting of the sum of all even
terms (including the constant term) of p. In particular, note that q{—x) — —q{x)
and r{—x) = r{x) hold for each a. Thus, if 0 < x < 1, then
|/(a) - q{x) - r{x)\ = \g{x) - p{x)\ < e, '
and g{— x) = —/(a) implies
|/(a) - <7(a) + r{x)\ = I/K-a) - g(-A)| < s.
from which it follows that |/(a) — ^(a)| < e. (Here we use the elementary
property: if\a+b\ < e and \a-b\ < e,then \a\ = |^-f^| < I^M^I <
e and \b\ < s.) In other words, thexxid polynomial q is £ -uniformly close to /
on [0, 1], and the desired conclusion follows.

Section 11: THE STONE-WEEERSTRASS APPROXIMATION THEOREM 101
Problem 11.7. ///: [0, 1] -* R is a continuous function such that J* /(2"X/x)
dx = Ofom = 0, 1, 2,..., then show that f(x) = 0 for all x <= [0, 1]. Does the
same conclusion hold true if the interval [0, 1] is replaced by the interval [— 1, 1]?
Solution. Assume that a continuous function / e C[0, 1] satisfies fQf(2"+y/x)
dx = 0 for each n = 0, 1,2, Then the change of variable u = 2nX/x
(or a- = m2,i+i) yields
f fC-*>/x)dx = (2/2 + 1) f u2nf(u)du = 0,
Jo Jo
and so fQx2n f(x)dx = 0 holds for all n = 0, 1, 2, The conclusion now
follows immediately from Problem 11.5.
The conclusion is not valid if we replace the interval [0, 1] by the
interval [—1, 1]. For instance, if f(x) = x for all x e [—1, 1], then note that
f-lfC-"+<yx)dx = 0 holds for all n = 0, 1, 2
Problem 11.8. Assume that a function /: [0, oo) ->• R /s either a polynomial or
else a continuous bounded function. Then show that f is identically equal to zero
(i.e., show that / = 0) if and only iff™ f{x)e~nx dx = Ofor all n = 1,2,3,....
Solution. Let /: [0, oo) —> R be a continuous bounded function. If / = 0,
then clearly f™ f(x)e~nx dx=0 holds for all n = 1, 2, 3,....
For the converse, assume that
i
f(x)e-ttx'dx = 0 holds for all «= 1,2,3,.... (*)
o
Using the change of variable u = e~\ it follows from (•) that
/•CO p\
/ f(x)e-"x dx = / /(- In //)"""l du = 0, w = 1, 2 (**)
Jo Jo+
In particular, f0+g(u)un du = 0 holds for each /? = 0, 1,..., where g(u) =
uf(—\nu). Since / is bounded, note that lim;,_vo+ g(u) = 0 holds, and so g
defines a continuous function on [0, 1]. From (•+), we see that /„ g(.x)x" dx = 0
holds for all n = 0, 1,2, Problem 11.4 implies that g = 0, and consequently
/ = 0.
A closer look at the above arguments reveals that we have actually proven the
following result.

102
Chapter 2: TOPOLOGY AND CONTINUITY
• Assume that f: [0, co) -> IR is a continuous function such that
/»oo
/ f(x)e~nx dx = 0 for all n = Jfc, k + 1, Jt + 2,...,
w/2ere /: is a positive integer. If limM_*o+ wm/(— In w) = 0 /or 5ome natural
number m, then the function f is identically equal to zero.
Indeed, replacing n by n 4- k 4- m + 1 in (••), we get
f um+kf(-\nu)undu=0, n = 0,1,2,...,
Jo+
which implies (as above) that / = 0. The reader can verify easily that any function
/ that satisfies \f(x)\ < Ceax for some C > 0 and a > 0 and all * > a*o also
satisfies limw-H.o+ wm/(— In u) = 0 for some natural number m. In particular, the
reader should notice that every polynomial p satisfies an estimate of the form
\p(x)\ < Cea\
One more comment regarding the above discussion is in order. Recall that if
/: [0, oo) —> R is a "nice" function, then the formula
/•OO
£(/)(*)= / e~stf(t)dt
is called the Laplace transform of /. The Laplace transform is a linear operator
and plays an important role in a wide range of applications. The reader should
notice that in actuality property (•) asserts that the Laplace transform is a one-to-
one operator when defined on an appropriate linear space of functions. (See also
Example 30 of Chapter 5 in the text.)
Problem 11.9. Show that a continuous bounded function /: [1, co) —> R is
identically equal to zero if and only if ff° x n f(x)dx = 0 for each n = 8, 9,
10
Solution. The "if" part only needs verification. Therefore, assume that the
function /: [1, co) —► R satisfies f™x~nf(x)dx = 0 for each n = 8, 9, 10,....
Using the change of variable u = jc""1, we see that
noo /»1 /»1
/ x-fWdx= u"-2f(±)du= un-8g(u)du = 0, (***)
where g(u) = w6/(£). Since / is bounded, we see that limw_o- g(u) = 0, and so
g defines a continuous function on [0, 1]. In addition, from (• • •), we see that

Section 11: THE STONE-WEIERSTRASS APPROXIMATION THEOREM 103
f*xng(x)dx = 0 holds for each n = 0, 1, 2 By Problem 11.4, it follows
that g = 0, and consequently, / = 0.
Problem 11.10. Let A be an algebra of continuous real-valued functions defined
on a compact topological space X and separating the points ofX. Show that the
closure A of A in C(X) with respect to the uniform metric is either all ofC(X) or
else that there exists a e X such that A= [f e C(X): f(a) = 0).
Solution. Let .4 c C{X) be an algebra, where X is compact. Now, consider
the sequence of polynomials {P„(x)} on [0, 1] defined by
P,(a) = 0 and Pn+l(x) = Pn(x) + {[x - (P„{x))2] for n = 1, 2
An easy inductive argument shows that each polynomial Pn{x) has a constant
term equal to zero. This guarantees that if / e A, then P„(f) e A for each n.
Also, by Lemma 11.4, we know that the sequence [P„(x)} converges uniformly
to y/x on [0, 1]. Thus, if / e A is non-zero, then put c = ||/||oo, and note that:
1. The sequence {^(fr)} £ A converges uniformly to ^. Hence, |/| e A.
2. Since {^wC—)} £ A converges uniformly to J^-, we see that J^ e A.
Thus, if / G A then both |/| and VTTT belong to A. In particular, .4 is an
algebra and a function space.
Now, suppose that A is not of the form {/ e C(X): f(a) = 0} for some
a e X. This implies that for each a e X, there exists_some f e A with
/(a*) 7^ 0. Thus, for each x e X, there exists some /v e A and a neighborhood
Vv of x with fx(y) i=- 0 for all y e V\.. By the compactness of X, there exist
Aj,..., x„ e X with X = (J"=1 V\... Note that the function g = /2 H + /£
of ^4 satisfies #(a) > 0 for each a € X. Multiplying by an appropriate constant,
we can assume that g(x) > 1 holds for all x. Put hn = ~yg, and note that
hn e A and that hn(x) I 1 for each x e X. ByJDini's theorem, {hn} converges
uniformly to the constant function 1, and so 1 6 A Theorem 11.5 now guarantees
that A = C(X) must hold.
Problem 11.11. Let A be the vector space generated by the functions
1, sin A', sin2 a, sin3 a, ...
definedon[0, 1]. That is, f e Aifand only ifthere is a non-negative integer k and
real numbers a$, ct\ , ak (all depending on f) such that /(a) = X^=0Q?„ sin" a

104
Chapter 2: TOPOLOGY AND CONTINUITY
for each a* € [0, 1]. Show that A is an algebra and that A is dense in C[0, 1] with
respect to the uniform metric.
Solution. Clearly, A is an algebra of functions that contains the constant function
1. Also, since the function f(x) = sin a separates the points of [0, 1], the algebra
A likewise separates the points of [0, 1]. By the Stone-Weierstrass theorem, A
is dense in C[0, 1].
Problem 11.12. Let X be a compact subset of JR. Show thatC(X) is a separable
metric space (with respect to the uniform metric).
Solution. The polynomials with rational coefficients form a countable set (why?).
By Corollary 11.6, this set is dense in C(X).
Problem 11.13. Generalize the previous exercise as follows: Show that if
(X, d) is a compact metric space, then C(X) is a separable metric space.
Solution. By Problem 7.2, we know that X is a separable metric space. Fix a
countable dense subset {aj, a'2, ...} of X and for each n let /„: X —> IR be
the function defined by fn(t) = d(t, xn) for each t e X.
Now, let a, y e X satisfy a ^ y. Put d(x% y) = 28 > 0. Choose some n
with d(x, a„) < 5, and note that
My) = d(y% xn) > d(x% y) - d(x, xn) >28-8 = 8> </(*, x„) = /„(*),
so that Mx) ¥" My)- This implies that the algebra generated by {1, /i, /i,...}
separates the points of X. By the Stone-Weierstrass theorem (Theorem 11.5), this
algebra must be dense in C(X).
Next, consider the collection C of ail finite products of the countable
collection {1, f\, /2,...} and note that C is a countable set, say C = (gj, g2,. • •}• To
complete the proof note that the finite linear combinations of {1, g\, gi,...} with
rational coefficients form a countable dense subset of C(X).
Problem 11.14. Let X and Y be two compact metric spaces. Consider the
Cartesian product X x Y equipped with the distance D\ given in Problem 7.4, so
that X xY is a compact metric space. Show that iffeC(XxY) and € > 0, then
there exist functions [f\,..., fn) C C(X) and [g\,..., gn) C C(Y) such that
n
\f{x,y)-Ydffc)toW\<e
holds for all (a, y) € X x Y.

Section 11: THE STONE-WEIERSTRASS APPROXIMATION THEOREM 105
Solution. Consider the set
A= [heC(XxY): 3 {/,,...,/„} c C(X\ {g] g„]QC(Y)
with A(.r, y) = J2 fiW&W V (jc, y) € X x y ).
Then, A is an algebra of functions of C(X x Y) and 1 € A. On the other hand, if
(x\, y\) ^ U2, yi), then either x\ ^ ,r2 or y\ ^ yi* If jci ^ xi, then select some
/ € C(X) with /(*,) ^ /Cv2), and let F(a, y) = f(x) for all (a, j) e X x Y. If
y\ # ^2» then pick some g e C(K) with g(y\) # g0>2), and put F(x, y) = g(;y).
In either case, F e A and F(a*i, y\) ^ F{xi, y{) holds, so that A separates the
points of X x Y. Now, by the Stone-Weierstrass theorem (Theorem 11.5), we
have A = C(X x Y), and the desired conclusion follows.

CHAPTER 3
THE THEORY OF MEASURE
12. SEMIRINGS AND ALGEBRAS OF SETS
Problem 12.1. // X is a topological space, then show that the collection
S = {C HO: C closed and 0 open] = {d \C2: C\, C2 closed sets]
is a semiring of subsets of X
Solution. From 0 = 0D0 and X = X f) X, we see that 0, X e S. Next,
notice that C\ n Olf C2 n <92 e 5 imply
(Ci n Oi) n (C2 n o2) = (Ci n c2) n (o, n o2) e s.
Now, if Ci n Oj, C2 n 02 € 5, then
Ci n o, \ c2 n o2 = {Cx n o2) n (C2 n o2)c
= (Cinooncciuoj)
= (Clno,)n[c|u(0|nc2)]
= [c, n(d nc|)]u [(Ci nc2n o\)n o,]
= /1U5,
where A = Ci n (0j fl C|) 6 5 and 5 = (Cj n C2 n 0£) n 0, e S satisfy
A n 5 = 0.
Problem 12.2. Let S be a semiring of subsets of a set X, and let Y c X. Show
that Sy = [Y n A: A e <S) w a semiring ofY (called the restriction semiring 0/
5 f0 Y).
107

108
Chapter 3: THE THEORY OF MEASURE
Solution. The conclusion follows from the identities:
a. Y PI 0 = 0;
b. (Y DA)n(Y DB) = Y n(AD B); and
c. Y DA\Y DB = Yn(A\B).
Problem 12.3. Let S be the collection of all subsets of[0, 1) that can be written
as finite unions of subsets of[0, 1) of the form [a, b). Show that S is an algebra of
sets but not a a-algebra.
Solution. Let A = U/U to. */) and B = U7=i to* dj)- Then. we nave
a. AUB eS\
b. A n B = U;=i Uy=i to. */) n lchdj) e S; and
c. [0, 1) \ A = fl/Li ([°« *) \ to. */)) € 5, where the last membership holds
since each [0, 1) \ [#/, £,) can be written as a finite union of sets of the
form [a, b).
To see that S is not a a-algebra note that P|~ , [0, £) = {0} £ 5.
Problem 12.4. Prove that the a-sets of the semiring
S=[[a,b): a,beR]
form a topology for the real numbers.
Solution. Let r be the collection of all a-sets of S. Clearly, 0 e r and R =
U^L,[—«, ?i) € r. It should be clear that r is closed under finite intersections.
Thus, in order to establish that r is a topology, we need to show that r is closed
under arbitrary unions. That is, if {[fl/,6/): / e /} is a collection of nonempty
members of S, then we must show that A = (J/€/ to. M belongs to r (i.e., that
i4 is a a-set).
To see this, let V = IJ/e/to. &/)• Then, V is an open set, and thus, there exists
an at-most countable collection ofpairwise disjoint open interval {(Cj,dj): j e J)
(see part (g) of Problem 6.11) such that V = {jj€j(Cj, dj). For each j e J, let
Aj = [cj, dj) if Cj = a\ for some / e 1 and let Aj = (c,, dj) if C; ^ 0/ for ail
i e I. Clearly, each Aj is a a-set. Moreover, it is easy to see that A = {JjeJ Aj
holds, which shows that A is a a-set.
Problem 12.5. Let S be a semiring of subsets of a nonempty set X. What
additional requirements must be satisfied for S to be a base for a topology on X ? (For
the definition of a base see Problem 8.18.) Prove that if such is the case, then each
member ofS is both open and closed in this topology.

Section 12: SEMIRINGS AND ALGEBRAS OF SETS
109
Solution. Since S is already closed under finite intersections, it follows from
the definition of a base that S will be a base if and only if (J/\eS^ = %.
Now, assume that LUgs^ = X holds. Note first that if A, B e S, then (since
S is a semiring) A \ B can be written as.a finite union of (disjoint) members
of S. It follows that A \ B belongs to the topology generated by S. Thus, if
B € S, then the relation
B* = X \ B = (\Ja) \ B = \J(A\B),
AeS AeS
shows that Bc belongs to the topology generated by S. That is, in this case, every
B € S is a closed and open set.
Problem 12.6. Let A be a fixed subset of a set X. Determine the two a-algebras
of subsets ofX generated by
a. {A}, and
b. [B\ A<ZB CX|.
Solution, (a) (0, A, Ac, X] and (b) {B: A c B or A c Bc}.
Problem 12.7. Let X be an uncountable set, and let
S == [E C X: E or Ec is at-most countable).
Show that S is the a-algebra generated by the one-point subsets ofX.
Solution. Clearly, S contains the one-point subsets of X, and every member of
S must be a member of the a -algebra generated by the one-point sets. Thus, it
remains to be shown that S is a a -algebra.
Clearly, 0, X e S and S is closed under complementation. Then let
[An} c S. If each An is at-most countable, then IJ^li ^« ls at-most countable,
and consequently (J^Li ^« e ^- ^n tne other hand, if some A* is uncountable,
then (Atf is at-most countable and the inclusion (U^li An)C Q (Atf shows
that |Xi A„ e 5.
Problem 12.8. Characterize the metric spaces whose open sets form a a -algebra.
Solution. We shall show that the open sets of a metric space X form a a -algebra
if and only if X is a discrete metric space (i.e., if and only if every subset of X
is open).
Let r be the collection of all open sets. If x ='P(X), then clearly x is
a a -algebra. On the other hand, if r is a a-algebra and x G X, then

110
Chapter 3: THE THEORY OF MEASURE
[x] = n^Li ^(A'» n^ sriows that {x} is an open set. This easily implies that every
subset of X is open (i.e., r = V(X) holds).
Problem 12.9. Determine the a -algebra generated by the nowhere dense subsets
of a topological space.
Solution. Let X be a topological space. Define
A= [A O X: A is meager or Ac is meager}.
Recall that a set is called meager if it can be written as a countable union of nowhere
dense sets—a set A is nowhere dense if (A)0 = 0. We claim that A is the a-
algebra generated by the nowhere dense sets of X. Clearly, every nowhere dense
set belongs to A, and every member of A belongs to the a -algebra generated by
the nowhere dense sets. So, it suffices to establish that A is a a -algebra of sets.
Clearly, 0, X £ A. Also, it should be obvious that A is closed under
complementation. Now, let {An} c A. If each An is meager, then clearly U^li An € A.
On the other hand, if (A*)c is a meager set for some £, then the set inclusion
(IXii An)C £ (Ak)c implies U^=i An € A. Therefore, A is aa-algebra.
Problem 12.10. Let Xbea nonempty set, and let T be an uncountable collection
of subsets ofX. Show that any element of the a-algebra generated by T belongs
to the o -algebra generated by some countable subcollection of T.
Solution. Assume T to be uncountable. Let A be the a-algebra generated by
T. Denote by \A[\ i G I } the family of alia -algebras each of which is generated
by a countable subset of T. It suffices to show that B = (J/e/ A 1S a <?-algebra
(because if this is the case, then A = B must hold, and the conclusion follows).
Clearly, 0 e B. Also, if A e B, then it is easy to see that Ac e B likewise
holds. Now, let [An] C B. Since each An belongs to a a-algebra generated
by a countable subset of T, it easily follows that there exists some / e I with
{An} C Ai. Thus, (J^lj An e Ai c B. That is, B is a cr-algebra, as required.
Problem 12.11. Show that eveiy Fa - and every G&-subset of a topological space
is a Borel set.
Solution. The Borel sets are the members of the a -algebra generated by the open
sets. So, a countable intersection of open sets (or a countable union of closed sets)
is always a Borel set.
Problem 12.12. Show that every infinite o -algebra of sets has uncountably many
sets.

Section 12: SEMIRINGS AND ALGEBRAS OF SETS
111
Solution. Let A be an infinite a-algebra of subsets of a set X. If A contains a
sequence [An] of nonempty pairwise disjoint sets, then A has uncountably many
members. Indeed, if this is the case, then for each subset s of natural numbers
let As= \JnGS An e A and note that As^At if s^t. By Problem 5.6, the
collection [As: s € 'P(IN)} has uncountably many members, and so A must
likewise have uncountably many members.
Next, we shall show that there exists a sequence [Bn] C A with Bn+\ C Bn
and Bn+\ i=- Bn for all n. If this is done, then put An = Bn \ Bn+\, and use the
above arguments to see that A is an uncountable set.
Using induction, we shall establish the existence of a sequence {Btl} such that:
1. Bn+\ c Bn and Bn+\ ^ Bn for all n, and
2. {£nri/4: A e A) is an infinite set.
The basic step of the induction is the following: Assume that Bn e A has been
chosen so that JBnn/l: Aei) has infinitely many members. Choose C e A
so that 0 C Bn fl C C Bn is a proper inclusion at both ends. In view of
Bn n A = [(£„ n C) n a] u [(£„ \ C) n a],
we see that either {(BnC\C)nA: A G A] or {(£„ \ C)DA: A € A] is infinite. If
{(Bnr\C)nA: AeA} is infinite, put Bn+l = BnC\C. If {(Bnr\C)nA: A e A}
is finite, put Bn+\ = Bn \ C.
Start the induction with B\ = X.
Problem 12.13. Let (X, t) be a topological space, let B be the a-algebra of its
Borel sets, and let Y be an arbitrary subset ofX. If Y is considered equipped with
the induced topology and By denotes the a-algebra of Borel sets of(Y, r), then
show that
BY = {AC\Y: A e B).
Solution. Let (X, r), Y, and By be as in the problem, and let
A={AnY: AeB).
We have to show that By c A and A c By both hold.
Clearly, A is a a -algebra of subsets of Y and O fl Y e A holds for each
0 € r. Thus, A contains the open sets of 7, and so By C A. Now, consider the
collection of sets
C={AeB: ADY eBY}.

112
Chapter 3: THE THEORY OF MEASURE
It is easy to see that C is a a-algebra of subsets of X satisfying r c C. Hence,
C — fi, and this implies that Ac. By. Therefore, By = A as claimed.
Problem 12.14. Let A\,..., An be sets in some semiring S. Show that there
exists a finite number ofpau-wise disjoint sets B\,...,BmofS such that each At
can be written as a union of sets from the B\,..., Bm.
Solution. We use induction on n. For n = 1 the result is trivial. Thus, assume
that the result is true for some n, and let A\,..., Ant An+\ be members of <S.
Pick a finite number of pairwise disjoint members B\,..., Bm of S such that
each A/, 1 < / < /?, can be written as a union of sets from B\,..., Bm. Clearly,
U/Li Ai c U7=i Bj- Tne sets &\ ^ ^/i+i» '..,Bmn An+\ are pairwise disjoint
members of <S. On the other hand, for each 1 < i <m there exists a finite pairwise
disjoint collection T\ C S with B\ \ An+\ = Uce^^ (by the definition of the
semiring). Thus, the collection
T = ?x U • • • U Tm U \BX H An+l,..., Bm O Afl+i} c S
is finite and pairwise disjoint. Moreover, each Aj (\ < i < n) can be written as a
union of members of T. Now, observe that
/ m \
A„+x = \A„+{ \ \J Bjj U (B, H An+l) U • • • U (Bm O A/J+1).
y=i
By Theorem 12.2(1) there exist pairwise disjoint sets D\,..., D^ in S such that
4/i+i \ Uy=i fiy = Ur=i A-. Finally, the collection T U {D,,..., Dk] C 5 is
finite and pairwise disjoint, and each set /\, (1 < / < n + 1) can be written as a
union from these sets.
13. MEASURES ON SEMIRINGS
Problem 13.1. Let [an} be a sequence of non-negative real numbers. Let fi(0) =
0, and for every nonempty subset A of N put fi(A) = YneAan- Show that
fi: V(SN) -> [0, oo] is a measure.
Solution. If {A,,} is a sequence of pairwise disjoint subsets of IN and A =
\JZL\ 4/m then note that
oo oo
«A) = £^ = £(£ **) = £>m„).
Aev4 n=l A-e/4w «=1

Section 13: MEASURES ON SEMIRINGS
113
Problem 13.2. Let S be a semiring, and let /i:S -> [0, oo] be a set function
such that fi(A) < oo for some A eS. If fi is a-additive, then show that fi is a
measure.
Solution. Write A = AU0U0U-- . Then,
fJL(A) = fJL(A) + /JL(0) + jLl(0) + • • • .
If fi(0) > 0, then ix{A) — oo, contrary to our hypothesis. Thus, fi{0) = 0, and
so /x is a measure.
Problem 13.3. Let X be an uncountable set, and let the o -algebra
S — [E C X: E or Ec is at-most countable];
see also Problem 12.7. Show that \x\ S -> [0, oo), defined by n(E) = 0 if E is
at-most countable and n(E) = 1 if Ec is at-most countable, is a measure on S.
Solution. Clearly, fi{0) — 0. For the a-additivity of \x let {£„} c S be a
pairwise disjoint sequence. Let E — (J!!Li £/i- If each En is at-most countable,
then E itself is at-most countable, and so /x(£) = J2T=\ M^m) = 0 holds. On the
other hand, if £Jr is at-most countable for some k, then (in view of En D £/. = 0
for n ^ k) we must have En C ££ for n ^ k, and so En is at-most countable
for each n ^ k. Thus,
oo
1 = M£) = m(£*) = X>(£«).
It is interesting to observe that if X = [0,1], then <S is a a-subalgebra of the
Lebesgue measurable subsets of [0, 1], and \i is the restriction of the Lebesgue
measure to S.
Problem 13.4. Let X be a nonempty set, and let /: X -> [0, oo] be a function.
Define p.:V(X) -> [0, oo] by /z(A) = £r6/4 f(x) if A ^ 0 and is at-most
countable, fi(A) = oo if A is uncountable, and fi{0) = 0. Show that \x is a
measure.
Solution. For the cr-additivity of /x, let {A,,} be a pairwise disjoint sequence of
subsets of X. Let A = U^li ^«- If some An is uncountable, then A is likewise
uncountable, and hence, in this case fu.(A) = Y1T=\ M^/i) = °° holds. On the

114
Chapter 3: THE THEORY OF MEASURE
other hand, if each An is at-most countable, then A is also at-most countable,
and so
xeA n—\ xeA„ n=\
also holds.
Problem 13.5. Let She a semiring, and let fi: S -> [0, oo] be a finitely additive
measure. Show that iffi is a-subadditive, then /j, is a measure.
Solution. Let {An} C S be a pairwise disjoint such that A = |J£=i An €$• By
hypothesis, fi(A) < Y^L\ ^An) holds. On the other hand, if k is fixed, then there
exist pairwise disjoint sets B\, ...,Bm e S such that A \ Un=i A* = U/Li ^
(see Theorem 12.2). Since A = [U£=i An] U [U/li B>] is a finite union of
pairwise disjoint members of 5, the finite additivity of fi implies
k k m
£ fi{A„) < £ H(A«) + £ M(B/) = H(A).
W=l « = 1 1 = 1
Since k is arbitrary, 5Z^=i M(^/i) 5: M(^) also holds, and so /x is a measure.
Problem 13.6. Let {/z„} be an increasing sequence of measures on a semiring
S;thatis,fJLn(A) < iin+\(A) holds for all A e S andalln. Define ii:S -* [0, oo]
by fji(A) = sup[fin(A)} for each A e S. Show that [i is a measure.
Solution. Clearly,/x(0) = 0. Now, let (A„)c5 be a pairwise disjoint sequence
such that |J5!=i An = A e S. Since each /z,- is a measure,
00 oo
M/04) = £/X/Wn) < £/4>U
n=l n=l
holds, and so /z(A) < YlT^i ^An). On the other hand, for each k we have
k k k
T]fj,(An) = lim Y]fii(An) = Jim /z,(U A„) < /x(i4).
x—• i-^oo*—• /-»-oo \ , /
/?=1 n=l /i=l
Thus, 5Z^=i MW«) ^ M(^) also holds, which shows that the measure /x is <x-
additive.

Section 13: MEASURES ON SEMIRINGS
115
Problem 13.7. Consider the semiring S = [A C IR: A is at-most countable),
and define fx:S -* [0, oo] by /x(A) = 0 if A is finite and /x(A) = oo if A is
countable. Show that \i is a finitely additive measure that is not a measure.
Solution. Let A\,..., A„ be pairwise disjoint members of S. Put A = (J/Li ^f-
If each A/ is a finite set, then A is likewise a finite set, and Xw=i M(A,) = M(^) =
0 holds. On the other hand, if one of the A/ is countable, then A itself is also
countable, and Y^l=\ m(^/) = M(^) = °° holds. Thus, /x is a finitely additive
measure.
To see that /x is not a-additive, note that IN = (J^li M* wn^e
PC
0 = £/x({k})</x(1N) = oo.
Problem 13.8. S/?0w that every finitely additive measure is monotone.
Solution. Assume that /x: S —► [0, oo] is a finitely additive measure. Let
A, B € S satisfy AC j5. Choose a finite collection of disjoint sets C\,..., Cn
of S such that B \ A = (JJU C,-. Then,
5 = A U C, U • • • U C„
is a finite union of pairwise disjoint sets of S. Thus, by the finite additivity of /x,
we have
/i(A) < /x(A) + /z(Ci) + • • • + m(C„) = /x(5).
Problem 13.9. Consider the set function \x defined in Example 13.6. That is,
consider a nondecreasing and left-continuous function /: IR -> R and then define
the set function /x:«S -> [0, oo) by fi([a,b)) = f(b) — f(a), where S is the
semiring S = {[a, b): -oo < a < b < oo}. Prove alternately the fact that /jl is a
measure.
Solution. An alternate way of proving the a-additivity of \x is as follows. Let
a < b and let [a,b) = \J^=i[ani bn) with the sequence {[an,bn)} pairwise
disjoint. For each a < x < b let
Si = £[/to) - /(«,)],
i
where the sum (possibly a series) extends over all / for which [a,-, b,) C [a, x)
holds; we let sx = 0 if there is no such interval. Since / is nondecreasing, we

116
Chapter 3: THE THEORY OF MEASURE
have sx < f(x) — f(a). Next, note that the set
A = [xe(a,b]: sx = f(x) - f(a)}
is nonempty. Let / = sup A, and note that a < t <b. Now, for x e A, we have
/(*) - /(*) = sx <st< f(t) - f{a),
and so, by the left-continuity of /, we get st = /(/) — f(a). That is, t e A.
Our objective is to establish that t = b holds. Assume by way of contradiction
that a < t < b. Then a^ < / < fa must hold for some k. Since the sequence
{[tf/M bn)} is pairwise disjoint, observe that [at, bi) c [a, t) holds if and only if
[a,-, fa) c [a, aC). Thus, st = sai holds. In particular, the relation
/(/) " f(fl) = * = sak < f(flk) - f(a) < f(t) - f(a)
guarantees that a^eA. However, this implies fa e A, which is impossible.
Therefore, / = b holds, which guarantees that
oo
m([*,/>)) = £/x([*„,W).
14. OUTER MEASURES AND MEASURABLE SETS
Problem 14.1. Show that a countable union of null sets is again a null set.
Solution. The conclusion follows from the inequality
oo oo
n—\ n—\
Problem 14.2. If \i is an outer measure on a set X and A is a null set, then show
that
fi(B) = ix(AUB) = fji(B\A)
holds for every subset BofX.

Section 14: OUTER MEASURES AND MEASURABLE SETS 117
Solution. The conclusion follows from the inequalities:
fi(B) < fi(B UA) = ix{(B \ A) U A)
< fi(B \ A) + n{A) = vl{B \ A) < /i(B).
Problem 14.3. Let /i be an outer measure on a set X. If a sequence {An} of
subsets ofX satisfies Y1T=\ ^A,,) < oo, then show that the set
E = [x e X: x belongs to Anfor infinitely many n }
is a null set.
Solution. Assume that a sequence [An] satisfies X^Li^C^/i) < oo. For each n
let En = U/^/j ^/' anc* note tnat ^ — &n holds for each n. Therefore,
00
0 < /*(£) < lx{En) < Y^ii{Ai) —► 0,
i—n
and hence fi(E) = 0.
Problem 14.4. // E is a measurable subset ofX, then show that for every subset
AofX the following equality holds:
fji(E H A) + ix{E U A) = m(£) + jLt(A).
Solution. The measurability of E gives
fj.(E U A) = /z((£ U A) H £) + /x((£ U A) fl £c) = /z(£) + /x(i4 fl £c).
Consequently, we have
/z(£ U/l) + fx(E HA) = fi(E) + fi(A D Ec) 4- ix(A D E) = fi(E) + /i{A).
Problem 14.5. Let [i be an outer measure on a set X. If A is a nonmeasur-
able subset of X and E is a measurable set such that A C E, then show that
fx(E\A)>0.
Solution. If n(E \ A) = 0 holds, then E \ A e A. Thus, A = E \ (E \ A) e
A, which is a contradiction. Therefore, fi(E \ A) > 0.

118
Chapter 3: THE THEORY OF MEASURE
Problem 14.6. Let A be a subset ofX, and let {£„} be a disjoint sequence of
measurable sets. Show that
OO 00
n=\ n=\
Solution. From the cr-subadditivity of /z, we see that
n=\ n=l n=\
On the other hand, Lemma 14.5 implies
£>04 DEn) = il(a n [U£-]) ^ KA ° [U£«])
n=l /i=l n=\
for each *, and so ]T~ , /x(i4 H £„) < /x(A n [|J£Li £*]) also holds.
Problem 14.7. Let {An} be a sequence of subsets ofX. Assume that there exists
a disjoint sequence {Bn} of measurable sets such that An C Bn holds for each n.
Show that
^(ua«)=£/x(a")-
n=\ n=\
Solution. Put A = U^Li An and note that A Pi Bn = An holds for each n.
Thus, using the preceding problem, we see that
00 / r °° "i\ °° °°
m(U An) = H\A H [(J B„J) = ^/x(Afl £„) = J]/z(4w).
Problem 14.8. Let /1 be an outer measure on a set X. Show that a subset E of
X is measurable if and only if for each e > 0 there exists a measurable set F such
that F c E, and //,(£ \ F) < e.
Solution. If E is measurable, then F = E satisfies the condition for each e > 0.
For the converse, assume that the condition is satisfied.
Start by choosing for each n a measurable set Fn with Fn c £and/z(£ \Fn)< £.
Put £ = UnLi ^n £ £» and note that F is measurable. Consequently,
M(£ \ £) < M(£ \ Fn) < £ for each w implies /x(£ \ £) = 0, and so
£ \ F is measurable. The measurability of £ now follows from the identity
£ = £U(£ \ F).

Section 14: OUTER MEASURES AND MEASURABLE SETS
119
An alternate proof of the preceding part goes as follows. Let A be a subset of
X with p(A) < oo. If € > 0 is given, pick a measurable subset F with FC£
and p(E \ F) < e. Then
p(A C\E) = fi(A n[fU(£\ F)])
< p{A n F) + /x(/\ Pi (F \ F)) < p(A C\ F) + e
implies p(A fl F) — p(A fl E) > — €, and so
ii(A) = /iWnf) + M(/inFc)
> M(/\nF) + M(/\nEc)
= fi(A n F) + +m(A n fc) + [mm n F) - /x(4 n E)]
> fj.(A n F) + n(A n Fc) - €
for all 6 > 0. This implies /x(/4) > /x(/4 n F) + M(/4 fl Fc), which shows that F
is a measurable set.
Problem 14.9. Let p be an outer measure on a set X. Assume that a subset E
ofX has the property that for each € > 0, there exists a measurable set F such
that fi(EAF) < e. Show that E is a measurable set.
Solution. Let s > 0. According to the preceding problem, it suffices to show
that p(E \ G) < £ holds for some measurable set G with G c F.
For each n choose F„ <= A with p(E&Fn) < 2~ns. Put F = f^Li F„ € A.
Since F \ E C. Fn \ E holds, we have
p(F \ F) < /x(F„ \ F) < 2-ne
for each /2, and so p(F \ F) = 0. Thus, F \ F e A, and hence F fl F =
F \ (F \ F) is also a measurable set. Now, note that F C\ E C. E holds and
CO oo
p(E\EnF) = m(£ \ Z7) = m(|J(£ \ ^)) < X>(£ \ F"} < *•
Problem 14.10. Lef X = {1, 2, 3}, T = {0, {1}, {1, 2}} awd consider the set
function p: .F -» [0, oo] cfe/zwed fry p(0) = 0, /x({ 1}) = 2 a«d ^({ 1, 2}) = 1.
a. Describe the outer measure /x* generated by the set function p..
b. Describe the a-algebra of all p*-measurable subsets ofX (and conclude
that the set {1} € T is not a measurable set).

120
Chapter 3: THE THEORY OF MEASURE
Solution, (a) The outer measure /z*: V(X) -► [0, oo] is given by
/z*(0) = O, /x*({1})=/z*({2}) = 1, M*({3})=oo,
/x*((l, 2}) = 1, m*(U, 3}) = /*'({2, 3}) = /z*({l, 2, 3}) = oo.
(b) The a-algebra of all measurable sets is A = {0, {3}, {1, 2}, X).
Problem 14.11. Let v: V(X) -» [0, oo] be a set function. Show that v is an outer
measure if and only if there exist a collection T of subsets ofX containing the empty
set and a set function /a: T —> [0, oo] with fi(0) = 0 satisfying v(A) = /x*(A)
for all A eV(X).
Solution. Assume first that v:V(X) -> [0, oo] is an outer measure. Let T =
V(X) and \x = v. We claim that v(A) = ft*(A) holds for each A e V(X), where
OO 00
/z*(A) = inf X>(4„): MnJC^and A c (J An ,
and inf 0 = oo. To see this, let A e V(X). From A = A U 0 U 0 U 0 • • •, we see
that /i*(y4) < /x(A) = v(A). On the other hand, if A C (J^1 A„ holds true, then
from the a-subadditivity of v, we see that
oo oo
and so u(A) < fA*(A) is also true. Hence, v(A) = /x*(A) for each subset A of X.
For the converse, assume that the outer measure fi* generated by a set function
fi: T -* [0, oo] satisfies /x(0) = 0 and v(A) = /z*(A) for each A e V(X). We
shall show that v is an outer measure by verifying the three properties required to
be satisfied by v in order to be a measure.
(1) From 0 < v(0) = /z*(0) < /z(0) + /x(0) + fi(0) + • • • = 0, we see that
v(0) = 0.
(2) (Monotonicity) Let A c. B and let {An} be a sequence of T with £ c
U~ ! A„. Then, A c (J~ i An* and so n\A) < £~ , MW»). Therefore,
oo oo
u(A) = /x*(A) < inf(J] /z(A„): {An} c T and 5 c |J An ) = M'(B) = v(B).
(If there is no sequence {A,,} of T with 5 C |J^=i An, then £i*(£) = oo, and
v(A) < v(£) = fi*(B) is trivially true.)

Section 14: OUTER MEASURES AND MEASURABLE SETS
121
(3) (a-Subadditivity) Let {£„) be a sequence of subsets of X and let E =
IX|£«- If £,T=i^(£„) = oo, then v(E) = n*(E) < £~, M*(£/.) =
X^i u(£«) is trivially true. So, assume X!^=i ^(En) < oo and let £ > 0.
For each n pick a sequence {Akn} of T with £„ C (J^j Akn and
PC
£ M(/\f;) < n*{En) + 2~ne = v(En) + 2-"e.
Clearly, £ C (J,*, (J^i ^ holds' and so
3C OO OO OO
v(E) = /**(£) < X] L ^^ < E[u(£n)+2_"el = £ w<£»>+£-
Since £ > 0 is arbitrary, v(E) < YjT=\ v(En), and we are done.
Problem 14.12. Consider an outer measure fi on a set X and let A be the
collection of all measurable subsets of X of finite measure. That is, consider the
family A = [A e A: fi(A) < oo}.
a. Show that A is a semiring.
b. Define a relation ^ on A by A ~ B if p,{AAB) = 0. Show that 2^ is an
equivalence relation on A.
c. Let D denote the set of all equivalence classes of A. For A e A let
A denote the equivalence class of A in D. Now, for A, B e D define
d(A, B) = p.{AAB). Show that d is well defined and that (D, d) is a
complete metric space.
Solution. Note that if A, B, and C are three arbitrary sets, then
AAC C:(AAB)\J(BAC).
(a) Straightforward. (Note that in actuality A is a ring of sets.)
(b) If A, £, and C in A satisfy A 2^ B and B ~ C, then the relation
fi(AAC) < fi({AAB) U (BAC)) < fji(AAB) + /jl(BAC) = 0
shows that A ^ C.
(c)If A ~ A\ and B ~ 51, then
/x(AA£) < Ai((AAA1)U(A1A5i)U(5IAJ5))
< /i(/\A/\i)H-/x(i4iAfi,) + )Lt(5,AB) = iLt(i41AB,).

122
Chapter 3: THE THEORY OF MEASURE
Similarly, fi(A\AB\) < fi(AAB)y and so fi(AAB) = fjL(A\AB\). This shows
that d(A, B) = fi(AAB) is well defined.
For the triangle inequality, note that
d(A, B) = fi(AAB) < ii(AAC) + /x(CAB) = d(A, C) + d(C, B).
Thus, (D, d) is a metric space. What remains to be shown is that (D, d) is a
complete metric space.
To this end, let {An} be a Cauchy sequence of D. By passing to a subsequence,
we can assume that
d(An+u An) = fji(An+lAAn) < 2-"-1
holds for each n. Set A = HaLi U/^ji A,- € A. Now, let /j be fixed and note that
A C U~n ^/ = A*. U OXnOWl \ A/)) holds. Thus,
oo
fi(A) < ii(A„) + J^ ^A<+' \ *') < ^A«) + 2~" < °°-
and so A € D. Moreover, we have
OO 00
M(A \ An) < /x((J(A,+1 \ */)) < £>04/+i \ A/) < 2~".
On the other hand, if x e An \ A, then x e An and a: $ A = P|!u=i U~a Ai-
Consequently, there exists some k > n with x $. A,- for each / > k. This implies
An \ A c (JSnM/ \ A/+I), and so /x(A„ \ 4) < ££„ mW/ \ */+i) < 2"»
also holds. Therefore,
</(A„, A) = /x(AnAA) = M(Afl \ 4) + At(i4 \ A„) < 21""
holds for each /2. This shows that lim d(An, A) = 0, and so (D, d) is a complete
metric space. (For an alternate proof of this part, see Problem 31.3.)
15. THE OUTER MEASURE GENERATED BY A MEASURE
Problem 15.1. Let (X, <S, /x) be a measure space, and let E be a measurable
subset ofX. Put SE = {EDA: A e S], the restriction of S to E. Show that
(E,Se, M*) w a measure space.

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE 123
Solution. Let £ be a measurable subset of X and let [An] be a sequence of
A such that
a. [An fl E] is a pairwise disjoint sequence; and
b. there exists some A e S such that A n E = IJ^li A„ n £•
Using the fact that /x*: A —> [0, oo] is a measure, we see that
CO 00
and so fi* is a measure when restricted to the semiring iSf.
Problem 15.2. Let (X, S, /x) be a measure space. Show that
H*{A) = M{n*(B): B is a a-set such that A <Z B]
holds for every subset AofX.
Solution. Let [An] c S and let B = |J~ , A„. By Theorem 12.2(3), there
exists a pairwise disjoint sequence {Bn} of S such that B = UULi ^"- Thus,
for A C X, there exists a sequence {A,,} of 5 with A C (J^li ^ if and only if
there exists a cr-set B with A C. B. The desired equality now follows from the
relation
00 oo
H*(A) < £>(B„) = H'(B) < £>(A„).
Problem 15.3. Show that every interval I of IR is Lebesgue measurable and
X*U) = \l\(=thelengthofI).
Solution. In Example 15.5, we established that the intervals / of the form [a, b]
and [a, oo) are Lebesgue measurable and that X*(I) = |/| holds for these cases.
We shall consider the other cases separately. Assume — oo < a < b < oo.
a. / = {a,b\. Choose a sequence [x„] of real numbers with xn I a and
a < xn < b for each n. Thus, by Example 15.5, we have
X*((a,b]) = lim A.* ([*„,&]) = lim(b-xn) = b-a = \I\.
v ' /»-*oo x ' n—►oo
b. / = (a, b). Pick a < xn < b with *„ I a and observe that [xn, b) f (a, £)•

124
Chapter 3: THE THEORY OF MEASURE
c. / = (—00, a). Note that [a — n, a) f (—00, a) and so
>.*((-oo, a)) = lim k([a - n,a)) = lim /; = 00 = |/|.
d. / = (—00, a]. Note that (—00, a) C (—00, a] and so from the inequality
00 = |(-oo, a)\ = A* ((-00, a)) < r ((-00, a]),
we see that |/1 = X*(I) = 00.
e. / = (a, 00). The conclusion follows immediately from the obvious
inclusion [a -f 1, 00) c (#, 00).
f. / = (—00, 00). Note that [0, 00) c (—00, 00).
Problem 15.4. Show that every countable subset of R has Lebesgue measure
zero.
Solution. Let a e R. Then, [a] c [a — e, a + e) holds for each £ > 0 and so
X*({a}) < X*([a -e,a + E))= 2e for all e > 0. Therefore, r((a}) = 0 holds
for all a e R. If A = {a\, <?2, • • •} = U/^li (an) is a countable set, then note that
*•(*) < ESLi ^((fln)) = 0 so that ^04) = 0.
Problem 15.5. For a subset AoflR and real numbers a and b, define the set
a A -f ft = {ax -f b: x e A). Show that
a. \*{aA+b) = \a\k*(A), and
b. if A is Lebesgue measurable, then so is a A -f b
Solution. Let A c Rand fix two real numbers a and b. Since A c (J^l, [a„, bn )
holds if and only if A 4- b c (J^li fa/i + ^» ^« + *) holds, it is easy to see that
X*(A + b) = k*(A). The identities
£ Pi (6 + A) = 6 + (£ - 6) n ,4 and £ O (ft + Af = ft 4- (£ - 6) n Ac
imply
r(£ n (ft + A)) + r(£ n (ft + Af) = r ((£ - ft) n a) + r ((£ - ft) n >4C),
which shows that A is measurable if and only if ft -f A is measurable for each
fteR.
Next, note that k*(c(s, t)) = |c[A.*((j, t)) = \c\(t-s) holds. On the other hand,
since A C U^Liton W holds ifand only if tfA C |J^=i a(an, b„) holds for each

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE 125
a e R and since k*([an, b„)) = A.*((a„, /?„)), it follows that V(aA) = |a|A.*(i4)
for each agIR. Now, the identities
ED a A =a((a-]E)nA) and E n {aAf = a((a~] E) C\ Ac) (a ^ 0),
imply
X*(£ O a A) + r (£ H (<M)C) = \a\[k*((a'lE) n A) + A* ((a-1 £) 0 /\c)],
which shows that A is measurable if and only if a A is measurable for each
a e R.
Now, (a) and (b) follow from the preceding discussion.
Problem 15.6. Let S be a semiring of subsets of a set X, and let fi:S -> [0, oo]
be a finitely additive measure that is not a measure. For each A c X define (as
usual)
oo oo
li*(A) = inf X>M„): {A„} C 5 a/id A C |J A„ .
5/70W fry a counterexample that it is possible to have p, ^ fi* on S. Why does this
not contradict Theorem 15.1?
Solution. Consider the finitely additive measure /x of Problem 13.7. Clearly,
At(N) = oo. Since N = (J^i ("1 € S, we have /x*(lN) < £™ , /x({n}) = 0, and
so 0 = /z*QN) < /i(N) = oo.
This conclusion does not contradict Theorem 15.1, since the cr-additivity of the
measure was essential for its proof.
Problem 15.7. Let E be an arbitrary measurable subset of a measure space
(X, <S, n) and consider the measure space {E, Se, v), where Se = [EdA: A e S]
and v(E D A) = //*(£ D A); see Problem 15.1. Establish the following properties
regarding the measure space (E,Se, v):
a. The outer measure v¥ is the restriction ofp,* on E, i.e., v*(B) = p.*(B)for
each B <ZE.
b. The v-measurable sets of the measure space (£, e>£, v) are precisely the
sets of the form E n A where A is a pi-measurable subset ofX, i.e.,
AV = (FC£: F € AM}.

126
Chapter 3: THE THEORY OF MEASURE
Solution. Let (X, «S, /z), £, and v be as defined in the problem.
(a) Let B be an arbitrary subset of E. If {An} is a sequence of S satisfying
B £ IXti 4/f.then note that B £ IXli ^ n 4„ and so
OO 00 PC
v*(B) < £ w(£ n A„) = £ m*(£ nA„)<J2 mG4„).
«=1 /i=l n=l
This implies v*(B) < /x*(£). On the other hand, if {An} is a sequence of S
satisfying B c (JJJli E ^ 4„, then we have
00 OO
Thus, jjl*(B) < v*(B) also holds, and so v*(B) = /z*(£) for each subset 5
of E.
(b) Let F be a subset of E. Assume first that F is v-measurable. If A e S,
then note that
fi(A) = /i*G4 H E) + ii*(A O (X \ £))
= v*04n£) + /^(An(X\£))
= v*((A H £) n F)+v*((i4 D £) H (£ \ F))+m*(A n (X \£))
> /x*(4 H F) + /z*([A n (£ \ F)] U [A n (X \ £)])
= /z*(4nF) + /x*(,4n(X \F)),
which shows that F is //.-measurable.
For the converse, assume that F is \i-measurable. If A is an arbitrary subset
of £, then note that
v\A) = m*(A) = /i*(4 nF) + /x'(An(X\ F))
= /z*04nF) + /x*(An(£ \ F))
= v*(yinF) + y*(An(£ \ F)),
which means that F is also v-measurable.
Problem 15.8. Show that a subset £ of a measure space (X, 5, fi)is measurable
if and only if for each € > 0 r/zere exists a measurable set A€ and two subsets B€
and C6 satisfying
E = (A€ UB€)\C€, /x*(5€)<^, and fi*(C€) <e.

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE 127
Solution. Let E be a subset of a measure space (X, S, fi). If E is a measurable
set and s > 0 is given, then let AE — E and BE = CE = 0, and note that these
sets satisfy the desired properties.
For the converse, assume that for each e > 0 there exist a measurable set AE
and subsets BE and CE satisfying
E = (AEUBE)\CE, /x*(£e)<£, and ii*{CE) < e. (•)
Replacing CE by (AE U Be) H CE1 we can assume that Ce is a subset of AE U BE.
From (•), we see that
EUCE = AEUBE. (*•)
Now, by Theorem 15.11, there exists a measurable set DE such that BE c De
and M*(De) = ijl*(Be). Using (••), we get
£UCfU(D£\ £,) = A£UfiEU (D£ \ BE) = AEU DE.
Clearly, AE U DE is a measurable set and
li\CE U (D£ \ BE)) < n*(CE) 4- /x*(D£) < 2f.
In other words, the preceding show that for each £ > 0 there exist a measurable
set FE and a subset Ge such that
E U Ge = FE and fi*(GE) < s.
Now, for each n pick a measurable set Fn and a subset Gn with fJL*(Gn) < £
and £ U G„ = Fn. Clearly, the set F — fXLi F„ is measurable. Also, the set
G = n^Li G" *s a nu^ set—^^ hence G \ £ is also measurable. In view of
OO co
£UG = Q(£UGJ = f>F,l = F,
«=i /t=i
we see that £ U G is a measurable set. Finally, the measurability of E follows
immediately from the identity
£ = (£UC)\(G\£) = F\(G\£).
Problem 15.9. Let (X, 5, /i.) be a measure space, and let A be a subset of X.
Show that if there exists a measurable subset E of X such that A C E,ja*(E) < oo,
and /x*(£) = M*(A) -f £t*(£ \ A), then A is measurable.

128
Chapter 3: THE THEORY OF MEASURE
Solution. By Problem 15.7, we know that the outer measure generated by the
measure space (£,5£,/i*) coincides with \i* and the a-algebra of all measurable
sets of the measure space (£, <S£, aO is [A e AM: A c £}.
Now, to complete the proof, assume E e AM and that a subset A of £ satisfies
/z*(A) + /**(£ \A) = /x*(£). If //,*(£) < oo holds, then it follows from
Theorem 15.8 that A is a measurable set for (£, <Se, /z*). Thus, by the preceding
discussion, A e AM.
Problem 15.10. Lef A^a subset qfR with X*(A) > 0. Show that there exists
a nonmeasurable subset BofJR such that B c A.
Solution. If A is nonmeasurable, then there is nothing to prove. So, assume that
A is measurable. Since some [n, n + 1] n A must have nonzero measure (why?),
by translating appropriately (and using Problem 15.5), we can also assume that
A c [0,1].
As in Example 15.13 define an equivalence relation 21 on A by saying that
x ~ y whenever x—y is a rational number. By the Axiom of Choice, there exists
a subset B of A containing precisely one member from each equivalence class.
Let {ri,r2,...} be an enumeration of the rationals of [—1, 1] and let Bn =rn + B.
Then:
a. The sequence [Bn] is pairwise disjoint;
b. X*(£„) = k*(B) holds (by Problem 15.5) for each /?; and
c ^cU~,Bwc[-l,2].
Now, note that if B is a measurable set, then each Bn is likewise a measurable
set (see Problem 15.5 again). Thus, from (c), it follows that
00 00
o<m < r(u*ij = £>*(*/)
/=! / = 1
n
= lim YV(B/) = Urn «X"(B) < 3,
which is impossible. Therefore, B is a nonmeasurable subset of A.
Problem 15.11. Give an example of a disjoint sequence {£„} of subsets of some
measure space (X, 5, /x) such that
oo oo
fjL*({J'En)<Y,ii*(En).
n—\ n=l

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE 129
Solution. Let £„ be the disjoint sequence of nonmeasurable sets described in
Example 15.13, where £„ = /•„ + £. Since £ is a nonmeasurable set, X*(E) > 0
holds, and so k*(En) = k*(E) > 0. In particular, J]^i ^*(^n) = oo. On the
other hand, (J~ , £„ c [-1, 2] implies r(|J~ , En) < 3 < oo.
Problem 15.12. Le/ (X, 5, fi) be a measure space, and let [An] be a sequence
of subsets ofX such that An C An+\ holds for all n. If A — U^li ^«»tnen snow
that iS(An) | ^(A).
Solution. Choose some E £ A with A C £ and /x*(A) = /i*(£). (This is
possible by Theorem 15.11.) By the same theorem, for each n there exists some
£„ e A with A,, c En c £ and /x*(/\„) == /i*(£„). Now, for each /2 put
£„ = H^/i el G A> and then let F = U,~ i ^/i ^ A. Then, we have:
a. A,, C Fn and /x*(i4„) = /x*(F„) for each ai; and
b. £„ t Z7 and /x*(i4) = n*(F).
By Theorem 15.4, it follows that
H*(An) = /x*(F„) t M*(F) = /x*(A).
Problem 15.13. For subsets of a measure space (X, <S, /x) let us define the
following almost everywhere (a.e.) relations:
a. A C £*.<?. // /x*(/\ \ 5) = 0;
b. A = B a.e. if fi*(AAB) = 0;
c. A„ | A a.e. // i4„ C /\„+i a.e. for all n and A = UULi ^n #•£• C^
meaning of' A„ I A a.e. is similar.)
Generalize Theorem 15.4 by establishing the following properties for a sequence
{£„) of measurable sets:
i. //£„ f E a.e., then /x*(£„) t M*(£).
ii. //£„ I E a.e. and /x*(£J < oo for some k, then /x*(£„) | fJL*(E).
Is (0 fn/e without assuming measurability for the sets En?
f
Solution, (i) Assume that {£„} is a sequence of measurable sets such that £„ f
£ a.e. holds. Let
OO 00
A = [([J £„) Afi] U [Q(fi„ \ £B+1)]
and note that /x*(A) = 0. Now, define F = £ u A and Fn = EnU A for each /?.

130
Chapter 3: THE THEORY OF MEASURE
Clearly, /**(£) = /x*(£) and n*(En) = ii*(F„) for each n (see Problem 14.2)
and Fn f F.
Now, apply Theorem 15.4(1) to get
/**(£„) = ii\Fn) t M*(F) = /**(£).
(ii) Assume that {Efl} is a sequence of measurable sets such that En I E
a.e. and /x*(£j0 < oo holds for some k. Define B = [(f^lj £„)A£] U
[LG1i(£"+i \ £»)]« Clearly, n*(B) = 0. Now, apply Theorem 15.4(2) to
EnUEUB I EUB.
Statement (i) is also true without assuming measurability for the En. This
follows from the arguments of (i) previously and Problem 15.12.
Problem 15.14. Give an example of a sequence [En} of measurable sets of some
measure space (X, <S, /x) such that En+\ c En holds for all n and
00
lim n*(E„)>ti'(f]E„).
n—\
Solution. Consider R with the Lebesgue measure, and let En = (/?, oo) for each
n. Then, En I 0 holds, while X*(En) = oo for each n.
Problem 15.15. For a sequence [An] of subsets of a set X define
OO OO OC 00
liminf An = I) C\ At and lim sup/4„ = O I) A/.
M?vt>f /e/ (X, tS, /x) be a measure space and let {£„} be the sequence of measurable
sets. Show the following:
a. /z*(liminf£„) < liminf/z*(£„).
b. /f M*(L£ti ^«) < °°'tnen M*(limsup£„) > lim sup/!*(£„).
Solution, (a) Note that f)™n Ef t liminf En and f|~n £/ Q En holds for each
n. By Theorem 15.4(1), we get
/x*(liminf£„) = lim /Wp|£,J < liminf/x*(£„).
/=«
(b) Use similar arguments and Theorem 15.4(2).

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE 131
Problem 15.16. Give an example of a sequence [An} of subsets of some measure
space (X, S, /x) such that An+\ C An holds for each n, /x*(/40 < oo, and
fllim ^(An)>^(p]An).
Solution. Let An = [J°ln £/, where [En] is the sequence of Example 15.13.
Note that An I 0 holds. Indeed, if x e A„, then x e £* for some k > n. Since
Ei H Ej = 0 whenever / ^ 7, it follows that x $ Al+\ so that An I 0 holds.
Now, observe that k*(A„) > k*(En) = k*(E) > 0 holds for all n.
Problem 15.17. Let (X, S\, n\) and (X.Si, p-i) be two measure spaces. Show
that fi\ and \in generate the same outer measure on X if and only if \i\ = /x* on
S\ and [ii = /x* on Si both hold.
Solution. If /xi and jjli generate the same outer measure, then clearly /xj = /x*
on S\ and 1x2 = M* on ^2 both hold.
For the converse, assume that /xi = /x$ on <Si and /X2 = /x* on 52 both hold.
Let A c X. If /xJ(A) = 00, then /x*(A) < /x^A) holds. If /xJ(A) < 00, then
given e > 0 there exists a sequence {i4„} of 52 such that A C (JJJLi ^ and
E~iM2(^)<^(A) + £.Thus,
00 00
Mf04) < £ Mi(A„) = J2^An) < M5(A) + *
holds for all e > 0, and so /x*(A) < /x^A).
Similarly, /x*(A) > /x^A) holds, and therefore /x*(A) = /x^CA) holds for all
A CX.
Problem 15.18. Let (X, <S, /x) be a measure space. A measurable set A is called
an atom ///x*(A) > 0 and for every measurable subset E of A we have either
/x*(£) = 0 or /x*(/4 \ E) = 0. //(X, 5, /x) d<?es not have any atoms, then it is
called a nonatomic measure space.
a. Find the atoms of:
i. the counting measure, and
ii. the Dirac measure based at a point a.
b. Show that the real line with the Lebesgue measure is a nonatomic measure
space.
Solution, a. (i) The atoms of the counting measure on a set are precisely the
one-point sets.

132
Chapter 3: THE THEORY OF MEASURE
(ii) The atoms of the Dirac measure based at a point a are precisely the sets
containing the point a.
b. Let A c R be measurable with X*(A) > 0. Pick some integer n such
that X*([n, n + 1] D A) = 8 > 0. Subdivide [w, w + 1] into a finite number of
subintervals all of the same length less than 8. For one of them, say /, we must
have
r([/i,Hi]nAn/) >o.
Now, note that the set E = [n, n -f 1] H A C\ I c. A is measurable and satisfies
0 < X*(£) < 8 < k*(A). This shows that A is not an atom, and hence IR.
with the Lebesgue measure is nonatomic. (For more about this problem, see
Problem 18.19.)
Problem 15.19. This exercise presents an example of a measure that has
infinitely many extensions to a measure on the a-algebra generated by S. Fix a
proper nonempty subset A of a set X (i.e., A ^ X) and consider the collection of
subsets S = {0, A}.
a. Show that S is a semiring.
b. Show that the set function /x: <S-> [0, oo] defined by p,(0) = 0 and fi{A) — 1
is a measure.
c. Describe the Caratheodory extension /x* of fi.
d. Determine the a-algebra of measurable sets AM.
e. Show that fi has uncountably many extensions to a measure on the o-
algebra generated by S. Why doesn't this contradict Theorem 15.10?
Solution. The validity of (a) and (b) should be obvious.
(c) The Caratheodory extension of \x is given by
(0 if B = 0;
pf(B) = j 1 if B # 0 and B c A;
I oo if B £ A .
(d) The a-algebra generated by S is
A={0,A,Ac,X}.
(e) If a is any non-negative extended real number, then the set function
v: A -> [0, oo], defined by
v(0) = O, v(i4)=l, KAc) = a, and v(X)=l+a,

Section 16: MEASURABLE FUNCTIONS
133
is a measure which is an extension of \x to all of A. This shows that there are
uncountably many extensions of /x to the cr-algebra generated by S.
The latter conclusion does not contradict Theorem 15.10 because /x is not a
a-finite measure.
16. MEASURABLE FUNCTIONS
Problem 16.1. Let (X, S, fu.) be a measure space. For a function f:X-+7R
show that the following statements are equivalent:
a. / is a measurable function.
b. /"' ((—oo, a)) is measurable for each oeR.
c /-1((tf, oo)) is measurable for each a eR.
Solution. (a)=Kb) Note that 7_I((—oo, a)) is a measurable set simply because
the interval (—oo, a) is an open set.
(b)=Kc) Observe that the identity
oo
ri((-oo,fl]) = p|/-,((-cx3,a + i))
/J=l
implies that f~l((-ooya]) is a measurable set for each a e 1R. Consequently,
the set f~] ((a, oo)) = X \ f~l ((—oo, a]) is also measurable for each a e R.
(c)=Ka) Clearly, /_1((—oo, a]) = X \ /_1((a, oo)) is measurable for
each a e R. Thus, by condition (5) of Theorem 16.2, the function / is
measurable.
Problem 16.2. Let {Xy 5, /x) be a measure space, and let A be a dense subset
of JR. Show that a function f:X -> ]R is measurable if and only if the set
{x e X: f(x)>a] is measurable for each a e A.
Solution. Only the "if" part needs proof. Let a e R. Since A is dense in R,
there exists a sequence [afl] of A with an < a for each n and an f a. Now,
note that the identity
00
/-,([fl,00)) = p|/-,([fl/|fOO))
shows that the set f~x ([a, oo)) is measurable. Therefore, by Theorem 16.2, the
function / is measurable.

134
Chapter 3: THE THEORY OF MEASURE
Problem 16.3. Give an example of a nonmeasurable function f such that \f\is
a measurable function and f~l ({a}) is a measurable set for each a e R.
Solution. Take a non-Lebesgue measurable subset E of [0, 1] and consider the
function /: [0, 1] —> 1R defined by
f ( \ _ | *' tf x e E>
/W~ {-a-, if x €[0,1] \ E.
It is straightforward to verify that the function / satisfies the desired properties.
Problem 16.4. Show that iff: IR -» Rw continuous a.e., then f is a Lebesgue
measurable function.
Solution. Let /: R —> R be a function that is continuous almost everywhere.
Put E = {x e R: / is continuous at a*} and note that X*(R \ E) = 0. Hence,
R \ E and E are both measurable sets.
Now, let O be an arbitrary open subset of R. Clearly, the set f~\0)C)(R \ E)
(as a null set) is measurable. Since / restricted to E is continuous, f~l(0)C\E
is an open set in £, and consequently there exists an open subset V of R such
that f~x(P) fl E = V O E. In particular, note that f~\0) O E is a measurable
set. Therefore,
r\o) = [r\o) he]u [r\o) n (r \ e)]
is likewise measurable, so that / is a measurable function.
Problem 16.5. Let f:)R^]Rbea differentiate function. Show that f is
Lebesgue measurable.
Solution. For each n define
ft,Cx) = n[f(X + I) - /C*)] = /(^-/U>.
n
and note that each gn is measurable (since it is continuous). In view of £„(*) —►
f'{x) for each x e R, it follows from Theorem 16.6 that /' is a measurable
function.
Problem 16.6. Let (X, <S, fx) be a measure space and let f:X-+1Rbea
measurable function. Show that:
a. \f\p is a measurable function for all p > 0, and
b. if f(x) 7^ Ofor each x e X, then \/f is a measurable function.

Section 16: MEASURABLE FUNCTIONS
135
Solution. Let /: X —> R be a measurable function.
(a) Assume p > 0. By Theorem 16.5, |/| is measurable. The conclusion
now follows from the identities [x e X: \f\p(x)>a}=-Xifa<0, and
{jc € X: \f\p(x) >a} = {x e X: \f(x)\ >'a$] if a > 0.
(b) Assume that f(x) ^ 0 holds for each x e X. Note that
{jc e X: i(.r) >0) = (^X
{* e X: j(x) > a] = [x e X
[x e X: j(x) > a] = {x e X
f(x) > 0),
f{x) < £} if a > 0, and
/Ct) <±}U{x eX: /(jc) > 0} if a < 0.
The preceding identities guarantee that j is measurable.
Problem 16.7. Let {/„} be a sequence of real-valued measurable functions on a
measure space (X, <S, fi). Then show that the sets
a. A = {x eX
b. B = [x e X
c. C = {x eX
are all measurable.
fn(x) -> co},
/«(*)-» -oo}, and
\im fn(x) exists in R}
Solution, (a) For each m and k let Am,k — [x e X: fn{x) > k for all n > m}.
From Amj = fXfLm{x € X: fn(x) > k], we see that Amtk e AM for each m, k.
Now, note that A = |X, IX=i >W-
(b) Put £„a = [x e X: f„(x) < -k for each n > m\ and note that £ =
HkL\ Um=l fim.*'
(c) Let 7 = X\(/IU5) and consider the measure space (Y, Sy, /x*). Also,
consider all functions restricted to Y. In view of Problem 15.7, all functions are
measurable with respect to this space. By Theorem 16.6, both functions liminf/„
and limsup/, are measurable. The conclusion now follows from Theorem 16.4(c)
by observing that
C = {jc e X: \imfn(x) exists in R}
= (i e F: limsup/„(.r) = liminf fn(x)}.
Problem 16.8. Let (X, <S, /x) be a measure space. Assume that f'.X-tlRisa
measurable function and g: R -» R is a continuous function. Show that go f is
a measurable function.
f 5
Solution. Consider the functions X —> R —> R with / measurable and g
continuous, and let O be an open subset of R. Since g is continuous, we know

136
Chapter 3: THE THEORY OF MEASURE
that g 1 (O) is an open set, and the conclusion follows from the identity
{gof)-\o) = ri{g-\o)).
Problem 16.9. Let T bea nonempty family of continuous real-valued functions
definedonlR. Assume thatthere exists a function g:R -> R such that fix) < g(x)
for each x G R and all f g T. Show that the supremum function /?:R -* Rr
defined by h(x) = sup{/(x): f e J7}, is (Lebesgue) measurable.
Solution. We shall show that h~] ((a, oo)) is an open set for each a G R (and
hence, a Lebesgue measurable set).
To see this, let a G R and fix xo G h~l {(a, oo)), i.e., h(xo) > a. So, there exists
some / G T such that /(ao) > a. Since / is a continuous function, there exists
some neighborhood V of a*o such that fix) > a for each x e V. This implies
h(x) > fix) > a for each x e V, and so V c /z_1 ((a, oo)). This shows that ao
is an interior point of h~} ((<z, oo)) and consequently, h~l((a, oo)) is an open set.
Note: A real-valued function /: X -► R defined on a topological space X is
said to be lower semicontinuous if f~x ((a, oo)) is an open set for each a e R.
The preceding arguments show that we have proven the following result: The
poinrwise supremum of a family of lower semicontinuous functions is likewise
lower semicontinuous.
Problem 16.10. Show that iff: X -+ UHisa measurable function, then either f
is constant almost everywhere or else (exclusively) there exists a constant c such
that
H*({x € X: fix) > c}) > 0 and //*({* e X: fix) < c}) > 0.
Solution. Let f:X -> R be a measurable function which is not a constant almost
everywhere. Assume first that fix) > 0 holds for each x e X and let
c0 = sup{c € R: fi*i[x e X: fix) < c) = 0}.
Clearly, 0 < cq < oo and /x*({a g X: fix) < cq}) = 0. Since / is not constant
almost everywhere, there exists some c > cq such that fi*({x G X: fix) >
c}) > 0. Now, if k satisfies cq < k < c, then by the definition of cq we have
/x*({a- g X: fix) < c)) > m*({a g X: fix) < k}) > 0, and the desired
conclusion is established in this case.
In the general case, either f+ or f~ is not equal to a constant almost everywhere.
We consider the case where /+ is not equal to a constant almost everywhere (the
other case can be treated in a similar fashion). By the preceding case, there exists
c > 0 with fi*([x G X: f+ix) > c})> 0 and (i*({x G X: f+ix) < c)) > 0.

Section 17: SIMPLE AND STEP FUNCTIONS
137
To finish the proof, notice that
{a- e X: f+(x) > c] = {x e X: f{x) > c]
and {a- e X: f+(x) < c] = {x e X: f(x) < c}.
17. SIMPLE AND STEP FUNCTIONS
Problem 17.1. For subsets A and B of a set X, establish the following
statements:
1. X0 = 0 and Xx = 1.
2. /ICfi if and only if xa <Xb-
3. X/*nz? = Xa • Xs = Xa *Xb-
4. X/HJB = Xa + Xb ~ Xahb = Xa v xa-
5. Xa \ fl = Xa - Xinfl.
6. If A = IJ^li ^// r//?^ M«) /5 o paii-wise disjoint sequence of subsets of
X,then xa = E,^1i Xa„.
7. X/t x a = Xa ' Xb {Here the set B can be considered to be a subset of some
other set Y.)
Solution. The proofs of the statements are straightforward. To indicate how one
can prove them, we shall establish the validity of statements (3) and (7).
(3) We have
f 1, if x e AH B
XahbM = jQ ifx^AnB
II, if a- e A and a- e B
0, if x £ A
0, if a' £ B
= XaM ' XbM
= Xa-XbW
= min{X/iU), XbM).
(7) Note that
. _ 1, if (A,y)e A x B
XAxfi(.v,y)- f0f tf CV|y)^AxB
II, if a- E A and y e £
0, if .v £ A
0, if y £ B
XaW'XbW.

138
Chapter 3: THE THEORY OF MEASURE
Problem 17.2. Let <p be a step function and \jr a simple function such that
0 < V < <P a-e- Show that iff is a step function.
Solution. Let E = {a: 0 < t/t(a) < 0(a)}, and observe that fi*(X \ E) = 0.
If F = [x e X: 0(a) > 0}, then the measurable set A = (X \ E) U F satisfies
fi*(A) < oo, and ij/(x) = 0 for each x e X \ A.
Problem 17.3. Show that if(X, <S, /x) is a finite measure space, then eveiy simple
function is a step function.
Solution. If 0 is a simple function and E = [x e X: 0(a) ^ 0}, then note that
/z*(£) < fi*(X) < oo holds.
Problem 17.4. Give an alternate proof of the linearity of the integral
(Theorem 17.2) based on Problem 12.14.
Solution. The linearity follows immediately from the following property.
• 7/0 is a step function and 0 = 5Zy=i£yXfl, is an arbitrary representation
ofcj), then
m
/(0) = £>y/x*(By).
7 = 1
We shall establish the preceding property below.
To this end, let 0 = X!/=i^*X/i, De tne standard representation of 0. Assume
first that the Bj are pairwise disjoint. Since neither the function 0 nor the sum
IZyLibj^iBj) changes by deleting the terms with bj = 0, we can assume that
bj t£ 0 for each j. In such a case, we have (J"=1 Ai = U7=i ^J- Moreover, note
that aifi*(Ai n Bj) = 6y/x*(A/ 0 Bj) for all / and j. Indeed, if Ai 0 Bj = 0 the
equality is obvious and if a € Aj D Bj, then a, = bj = 0(a). Therefore,
n m n
/(0) = £>/i*(A,) = ££fl//iU H By)
/=i /=y i=i
m // m
= £ &-M*(A/ H By) = X>y/i*(By).
y=i /=i y=i
Now, consider the general case. By Problem 12.14, there exist pairwise disjoint
measurable sets C\,..., C\ such that each C, is included in some By and By =
\J[Cf: C, c By}. For each / and y let SJ = 1 if C, c By and <$/ = 0 if C, £ By.

Section 17: SIMPLE AND STEP FUNCTIONS
139
Clearly, XBj = £?=,«/Xc- and fi(Bj) = E^S/V (C,-). Therefore,
m m k k m
♦ -E^-E^E'/** =EEm/*c,
7=1 7=1 /=i /=i y=i
So, by the preceding case, we have
k m m A /?>
'<*> = J2[EbJsi y^ = E*y[EWc'>] = E vw
/=! 7 = 1 7 = 1 1 = 1 7 = 1
Problem 17.5. Show that |/(0)| < /(|0|) holds for every step function (p.
Solution. From —10| < 0 < |0| and the monotonicity of the integral (Theorem
17.3), it follows that
-/(|0|) = /(-|0|)</(0)</(|0|),
and so |/(0)| < /(|0|) holds.
Problem 17.6. Let 0 be a step function such that /(|0|) = 0. Show that 0=0
a.e. holds.
Solution. Let 0 = £"=l a<'X/*, be tne standard representation of 0. Then, note
that |0| = Yl*i=\ \ai\XAi is a representation of |0|, and therefore
n
O = /(|0|) = E>/lAi*(*/)-
1 = 1
Since |a,| > 0 holds for each 1 < / < n, it follows that fi*(Ai) = 0 for each
1 < / < /z, and so 0 = 0 a.e. holds.
Problem 17.7. Let 0 be a step function. Let A = [x e X: 0(a) ^ 0} and
M = max{|0Gr)|: .t e X}. Show that |/(0)| < Mil*{A).
Solution. Apply the monotonicity of the integral (Theorem 17.3) to the inequality
-Mxa <0 < Mxa-
Problem 17.8. Let {0,,} be a sequence of step functions. Show that if<p is a step
function and 0„ | 0 a.e. holds, then /(0„) | /(0) also holds.

140
Chapter 3: THE THEORY OF MEASURE
Solution. If </>„ I 0 a.e. holds, then 0„ — 0 I 0 a.e. likewise holds. Thus, by
the order continuity of the integral (Theorem 17.4), 7(0,,) — 7(0) = 7(0„ — 0) I 0
so that 7(0„) | 7(0).
Problem 17.9. Ler {0„} be a sequence of step functions and 0 a simple function
such that 0 < 0;, t 0 a-e. holds. Show that if lim 7(0„) < oo, then 0 is a step
function.
Solution. Assume 0(a') > 0 for each x and let 0 = ]C/=i tf/X/i, be the standard
representation of 0. Now, let / be fixed. Then, for each n the function \frn =
0n A fl/Xi4(- is a step function, t/t„ < 0„ holds, and V« t„ 0 a tf/X/i, = #/X/i, a.e.
By Theorem 17.6, we see that
0 < aiii*(Ai) = lim 7(i/rn) < lim 7(0„) < oo,
n-*oo n-*oo
and so p*(Aj) < oo holds for each I < i < k. That is, 0 is a step function.
Problem 17.10. Let (X, 5, p) be a measure space, and let f:X —► Rka
function. Show that f is a measurable function if and only if there exists a sequence
{0«} of simple functions such that lim0„(x) = f(x) holds for all x e X.
Solution. Assume / to be measurable. Then both /+ and /" are measurable
functions. By Theorem 17.7 there exist two sequences of simple functions [sn]
and [tn] with 0 < sn(x) f f+(x) and 0 < tn(x) f /"(a*) for each x e X. Now,
note that 0„ = sn — tn satisfies 0„(a*) —> f(x) for all a\
For the converse, note that (by Theorem 16.6) the pointwise limit of a sequence
of measurable functions is always a measurable function.
Problem 17.11. Let (X, <S, p.) be a o-finite measure space, and let f: X -* 1R
be a measurable function such that f(x) > Ofor all x e X. Show that there exists
a sequence {0„} of step functions such that 0 < <pn f f(x) holds for all x e X.
Solution. By Theorem 17.7 there exists a sequence [\l/n] of simple functions
satisfying 0 < \lr„(x) t /(•*) for all x e X. Now, pick a sequence {£„} of
measurable sets with p*(En) < oo for each n, and En f X. Let 0„ = i/o, a xe„-
Then, {0„} is a sequence of step functions satisfying 0 < (pn(x) t /(*) for each
x eX.
Problem 17.12. Give a proof of the order continuity of the integral, i.e., 0„ 4- 0
a.e. implies I((pn) I 0, based on Egorov's Theorem 16.7.

Section 17: SIMPLE AND STEP FUNCTIONS
141
Solution. Assume that (0„} is a sequence of step functions of some measure
space (X, «S, fi) satisfying <j>n J, 0 a.e. Without loss of generality, we can suppose
that (pn(x) I 0 for each x e X. Let E - [x e X: 0i(x) > 0} and note that
/x*(£) < co. Also, let M = max{0i(A*): a* e J().
Now, let € > 0. By Egorov's Theorem 16.7 there exists a measurable set F c E
such that fjL*{F) < e and {0,,} converges uniformly to zero on E \ F. So, there
exists some k such that 0 < 4>n(x) < e for all x e E \ F and all n > k. Thus, for
n > k, we have
0 < (pn < 6Xe\f + Mxf < tXE + Mxf,
and consequently, by the monotonicity of the integral
0 < Ufa) < *M*(£) + Mfi'(F) < [fM*{E) + M]€
for all n > k. This shows that /(</>„) | 0.
Problem 17.13. Let {X,S, fi) be a measure space, and let f:X -> [0, co) be
a fiinction. Show that f is measurable if and only if there exist non-negative
constants C\, c2,... and measurable sets E\, Ei,... such that
oo
/(A') = X>X£,,CO
holds for each x e X.
Solution. Consider a measure space (X, <S, /x) and a non-negative real-valued
function /: X -> [0, oo). Assume first that there exist non-negative constants
Ci,C2,... and measurable sets £i,£2,... such that f(x) = Y1T=\ cnXE„M
holds for each x e X. If we let </>„ = YH=\ ci'X£,-» tnen </>/i is a measurable
function (in fact, it is a simple function) and (j)n{x) —> f(x) holds for each
x e X. Now, by Theorem 16.6(1), the function / is necessarily a measurable
function.
For the converse, assume that / is a measurable function. By Theorem 17.7
there exists a sequence of simple functions {<pn} such that 0 < (j)n{x) f f(x) for
each a e X. If we let <p0 = 0, then f(x) = f^L\[<t>"W ~ Ai-ito] holds for
each a* 6 X. For each n write 0„ — 0„_j = 5^/!L1 cJ-'xe,? with c" > 0 for each i
and n. Thus,
oo k„

142
Chapter 3: THE THEORY OF MEASURE
and by rearranging the terms of the preceding series in a single series, our
conclusion follows.
Problem 17.14. Let (X, <S, ji) be a finite measure space satisfying p.*(X) = 1,
and let E\, £2,..., £jo be ten measurable sets such that /i*(£/) = \ holds for
each i. Show that four of these sets have an intersection of positive measure. Is
the conclusion true for nine measurable sets instead often?
Solution. Consider the step function 0 = £/=i Xe,-« Clearly, the function 0
assumes only integer values and
(p(x) = the cardinality of the set {/ e {1,..., 10}: a' e E,}.
If 000 < 3 = 3xxM for almost all a*, then
10 10
3 < f = !>*(£/) = £'(*£,) = /(0) < /(3xx) = 3
/=i /=i
a contradiction. Hence, the measurable set
A = {a* € X: 0(a) > 4}
must have positive measure.
Next, let A], An,..., Aj, denote the collection of all (nonempty) intersections
of the sets £/ taken four at a time; clearly, k < (^ =210. Now, an easy argument
guarantees that A C (Ji=i ^ anc^ frorn tms it easily follows that at least one of
the Aj must have positive measure.
For nine sets the conclusion is false. For a counterexample take X = [0, 1],
\x — X, E] = Ei = £3 = (0, j), £4 = £5 = £6 = Q, |), and £7 = £8 = £9 =
(f.i)-
Problem 17.15. ///: X —> [0, 1 ] w a measurable function, then show that either
f = xa #•£• /or some measurable set A or else (exclusively) there exists a constant
0 < c < ^ such that
fi*([x € X: c < /(a) < 1 - c}) > 0.
Solution. For each n let A„ = {x e X: ± < f(x) < 1-^}. lfp*(An) > Ofor
some /?, then the constant c = ^ satisfies /x*({a g X: c < /(a*) < 1 — c}) > 0.

Section 17: SIMPLE AND STEP FUNCTIONS
143
Now, assume that p,*(An) — 0 for each n. Then from
Afl t {xeX: 0 < f(x) < l},
we see that ii*({x e X: 0 < /(jc) < 1}) = 0. This easily implies that / = xa
a.e. for the measurable set A = /_1({1}).
Problem 17.16. Let (X, <S, /x) be a measure space, and let 0: X -> 1R fee a
simple function having the standard representation (p = 5Z?=ifl/X/\,-- ^/0 ^ Ofl.e.,
f/zew f/ze jw/w ]C?=iflf//x*(^/) wa/:e.s jeAz.se as a« extended real number (it may be
infinite). Call this extended real number the Lebesgue integral of<p, and write
I (</>) = EILiaiVO*/).
a. 7/0 a^za1 0 are simple functions such that 0 > 0 a.e., then 0- > 0 a.e., then
show that 7(0 + 0) = 7(0) + 7(0).
b. 7/0 a«(i 0 are simple functions such that 0 < 0 < 0 a.e., then show that
/(0) < i(if).
c. S/zovv f/zaf if (0,,) artd {0,,} <?>'£ two sequences of simple functions and
f:X -> R* jz/c/z r/zar 0 < </>„ f / a?- and 0 < ^, | / a.e., r/7^/7
lim 7(0„) = lim 7(0-,,) /jo/as (w/r7z the limits possibly being infinite).
d. Assume that {0,,} zs a sequence of simple functions such that 0 < 0„ t Xa
a.e. /zo/as. Show that lim7(0„) = /z*(/4).
e. G/ve a/7 example of a sequence {0„} of simple functions on some measure
space such that 0„ I 0 (eveijwhere) and lim 7(0„) 7^ 0.
Solution. Clearly, a simple function 0 is a step function if and only if 7(0) < 00.
(a) Note that 0 -f 0 is a step function if and only if both 0 and 0 are step
functions. In this case, the equality 7(0 + 00 = 7(0) 4- 7(0") follows from
Theorem 17.2. On the other hand, if 0 4- 0" is not a step function, then either 0 or
0- fails to be a step function and hence, in this case, 7 (0 -f 00 = 7 (0) -h 7 (0) = 00
holds.
(b) If 7(0) = 00, then 7(0) < 1(f) holds trivially. On the other hand, if
7(0-) < 00, then 0 is a step function. It follows (from Problem 17.2) that 0 is a
step function, and the desired inequality follows from Theorem 17.3.
(c) If both (0„} and {0,,} are sequences of step functions, then the conclusion
follows from Theorem 17.5. Thus, we only need to consider the case when {0,,}
is a sequence of step functions and 7(0>) = 00 holds for some k.
In view of 0„ a 0;. f „ / a 0* = 0* a.e., it follows from Problem 17.9 that
lim^oo 7(0„ A 0;.) = 00. From 0„ a 0> < 0„, we obtain that lim7(0„) = 00.
Hence, lim 7(0„) = lim 7(0„) = 00 holds in this case.
(d) We can suppose 0 < 0„Ct) t XaM holds for each x. If for each n we let
An — [x e X: (f)n(x) > 0}, then each An is measurable and An t A holds. Since

144
Chapter 3: THE THEORY OF MEASURE
Xa„ f Xa, part (c) coupled with Theorem 15.4 gives
lim /«>„) = Urn I(XAn) = iim fim(An) = n\A).
n—*-oo n—>oo n—►oo
(e) Consider R with the Lebesgue measure, and let <j)n = X(«,oo) •
Problem 17.17. Let (X, Y,, fi) be a measure space with E being a a-algebra.
Let us say that a function f:X -* R is E -measurable if f~l(A) e E for each
open subset A of R. Also, let My denote the collection of all ^-measurable
functions. Establish the following:
a. My. is a function space and an algebra of functions.
b. My is closed under sequential pointwise limits.
c. If fi is a-finite and f:X -> IRis a measurable function, then there exists
a ^-measurable function g: X -* R such that f — g a.e.
Solution, (a) In order to show that My. is closed under addition and
multiplication, we need the following properties among E-measurable functions / and g:
The sets
1. {xeX: f(x)>g(x)},
2. {xeX: /(*)>*(*)}, and
3. \x e X: f(x) = g(x)}
all belong to L. To see (1), let r\, r2,... be an enumeration of the rational numbers
of R, and note that
{x € X: f(x) > g(x)} = Q[ {x e X: f(x) > rn] n {x e X: g(x) < /■„}],
which belongs to E, since it is a countable union of sets from the <r-algebra E. For
(2), note that {x e X: f{x) > g(x)} = [x e X: g(x) > /(a)}c, which belongs to
E by (1). Finally, for (3), observe that
{x e X: f(x) = gix)} = [x e X: fix) > gix)} n {x e X: gix) > fix)),
which belongs to E by (2).
To complete the proof of part (a), we shall establish that for E-measurable
functions / and g, the following statements hold:
i. / + g is a E-measurable function,
ii. fg is a E-measurable function.

Section 17: SIMPLE AND STEP FUNCTIONS
145
iii. |/|, /+, and / are E-measurable functions.
iv- / v S an(^ / A 8 are S-measurable functions.
The proofs of these claims are given below.
(i) Note first that if c is a constant number, then c — g is a E-measurable function.
[Reason: If a e IR, then [x e X: c - g(x) > a] - [x 6 X: g(x) < c - a) e £.]
Now, if a e IR, then the set
(/-r-Sr'^oo)) = {-v 6 X: f{x) + g(x) > a] = {a G X: /(*) > «-*(*)}
belongs to £ by the preceding observation and (2). This implies (how?) that f + g
is a E-measurable function.
(ii) Note first that f2 is a E- measurable function. To see this, let a e IR.
Then {a- e X: f2(x) < a] = 0 if a < 0 and {a e X: /2(a) < a) =
/-1([—<s/a, V?]) if a > 0. This implies that /2 is a E-measurable function.
Also, if c is a constant, then c/ is measurable. [Reason: If A — [x e X: cf(x) >
a], then A = {a e X: /(a*) > a/c} fore > 0 and A = {a 6 X: /(*) < a/c} for
c < 0.] The result now follows from the preceding observations combined with
(i) and the relation
fs = {[(f+s)2-f2-g2].
(iii) The E-measurability of \f\ follows from the relation
{xeX: |/(.v)| <a}=0 if a < 0,
and
[x e X: |/(.v)| <a} = [x s X: /(*) < a) n {x e X: /(.v) > -a) if a > 0.
For the E-measurability of /+ and /" use the identities
f+ = \{\f\ + f) and /- = i(|/|-/).
(iv) The identities
/vs = A(/ + * + |/-g|) and /Aj = I(/ + S-|/-g|)
show that / V g and / A g are E-measurable functions.

146
Chapter 3: THE THEORY OF MEASURE
(b) Assume that {/„} is a sequence of E-measurable functions such that
fn(x) -> f(x) holds for each x e X. Observe that the equality
00 oo
/-■((«.">)-uru-i((«+*.°°))
«=1 i=n
and the E-measurability of each // show that /~l((tf, oo)) belongs to E. This
implies that / is a E-measurable function.
(c) We can assume f(x) > 0 for each x e X (otherwise, we apply the arguments
below to f+ and /" separately). Assume first that / = xa for some A e E.
Since (i is cr-finite, it follows from Theorem 15.11 that there exists a /z-null set C
such that B = A U C e E. So, if g = xb» then g is E-measurable and / = g
/z-a.e. It follows that if 0 is a /z-simple function, then there exists a E-simple
function if/ such that x// = 0 /x-a.e.
Now, by Theorem 17.7, there exists a sequence {0„} of simple functions such that
0/iQO t /(*) f°r eacn x £ X. Replacing each <j)n by a E-simple function i/rw (as
above) we have yjrn(x) t f(x) for /x-almost all jc. So, there exists a /z-measurable
set E such that ^(jc) t fW for each x £ E. Now, use Theorem 15.11 to select
a set F e E with E c F and /x*(F) = 0. Clearly, iMjOxf'M t /(*)Xf«(*) =
g(A') for each a* € X. By part (b), g is E-measurable and satisfies g = / /x-a.e.
18. THE LEBESGUE MEASURE
Problem 18.1. Lef / = f]"=i A ^ <"* internal oflR". Show that I is Lebesgue
measurable and that \(I) = Yl"=] | /, |, where \ l\ \ denotes the length of the interval
//.
Solution. The verification of the formula can be done by cases as in Problem 15.3.
To show this, we establish the formula for two cases, and leave the rest for the
reader.
The first case is when // = [a,, bj], where — oo < at < bt < oo holds for each
1 < / < n. Then, n;.'=1l>/, &,- + i) ±k n^,// = /. Thus, from Theorem 15.4, it
follows that
X(I) = Hm x(fl[ahb, + i))= Hm fty, -«,- + {)
/=1 1 = 1
n n
= n<*'-*<>=n I7*'•
1=1 1=1
The second case is when / = [a, oo) x [^2,^2] x • • • x [tf/n 6„]. Then, note
that [a,a + k] x [ai.bi] x • • • x [a„, b„] t* /• Taking into account the preceding

Section 18: THE LEBESGUE MEASURE
147
case, it follows from Theorem 15.4 that
A.(/) = lim k([a, a + k] x [a2, b2] x
k -+<x>
= lim k-(b2-a2)---(bfl - an)
k—oo
Problem 18.2. Let O be an open subset of JR. Show that there exists an at-most
countable collection {Ia: a e A] of pairwise disjoint open intervals such that
O = UaeA '«• AlSO, Show that X(G) = ZaeA IU
Solution. Let O be an open subset of JR. By part (g) of Problem 6.11, we know
that there exists an at-most countable collection {Ia: a e A] of pairwise disjoint
open intervals such that O = Uae/i ^*-
Now, using the fact that the length of each Ja coincides with its Lebesgue
measure (Problem 15.3), we see that
HO) = *(U /„) = £>(/„) = J2 l7«l-
ct€A ct€A aG/4
Problem 18.3. Show that the Borel sets of R" are precisely the members of the
a-algebra generated by the compact sets.
Solution. Let C denote the a-algebra generated by the compact sets. Since every
compact set is closed (which is the complement of an open set), it follows that
C c B. On the other hand, if C is a closed set and C„ — {x e C: d{0, x) < n}y
then {Cn} is a sequence of compact sets satisfying C„ fC. This implies that C
contains all the closed sets (and hence, all the open sets). Thus, B C.C also holds,
and so B = C.
Problem 18.4. Show that a subset E of Rn is Lebesgue measurable if and only
if for each € > 0 there exists a closed subset F of R" such that F C E and
\{E\F)<c
Solution. Assume that E is Lebesgue measurable and let e > 0. Since Ec
is also Lebesgue measurable, there exists an open set V such that Ec C V and
X(V \ Ec) = X(E fl V) < e. Then, the closed set C = Vc satisfies C C E
and A(E \ C) = X{E C\V) < s. For the converse, either reverse the preceding
arguments and use Theorem 18.2, or else use Problem 14.8.
Problem 18.5. Show that if a subset E of[0, 1] satisfies X(E) = 1, then E is
dense in [0, 1].
x [a„,bn])
CO
CO
/=!

148
Chapter 3: THE THEORY OF MEASURE
Solution. Let / be a (nonempty) subinterval of [0,1]. If / H E = 0, then
we have X(E) + X(I) = A.(£ U /) < A([0, 1]) = 1, and hence, in this case,
X(E) < 1 — A.(/) < 1 holds, which is a contradiction. Thus, / C\E ^ 0 holds for
each subinterval / of [0, 1], and so the set E is dense in [0, 1].
Problem 18.6. IfE c IR" satisfies X(E) = 0, then show that E° = 0.
Solution. If V is a nonempty open set with Vc£, then note that 0 < X(V) <
X(E) holds. Therefore, the open set E° must be empty.
Problem 18.7. Show that ifE is a Lebesgue measurable subset of JR", then there
exist an Fa-set A and a Gs-set B such that A c E C B and X(B \A) = 0.
Solution. By Problem 18.4, for each k there exists a closed set Ck with Ck C E
and X(E \ Ck) < |. Similarly, by Theorem 18.2, for every k there exists an open
set Vk with E c V* and X(Vk \ E) < i. Put A = (J~, C* (an £a-set) and
£ = f|~ , Vk (a G«-set). Clearly, A C £ C £ holds, and in view of
MB \ A) < X(Vk \ Ck) = X((l/A \ £) U (£ \ Q))
< A(VA \ £) + *(£ \C*)<£
for each k, we see that X(B \ A) = 0.
Problem 18.8. Le/ {£„} be a sequence of nonempty (Lebesgue) measurable
subsets of10, I] satisfying limA(£„) = 1.
a. Show that for each 0 < € < 1 fAere ejcwte a subsequence [Ekn] of {£„}
such that X(f)™:=lEkn)>6.
b. 5/20 w f/jar n!u=n £* = 0 w possible for each n = 1, 2,
Solution. Let {£„} be a sequence of nonempty Lebesgue measurable subsets of
[0, 1] satisfying limA(£„) = 1.
(a) Fix 0 < e < 1. From limX(En) = 1, we see that there exists a subsequence
{Ekn} of {£„} satisfying X(Ekn) > 1 — ^. Now, consider the measurable sets
£ = f|~ , Ekn and £ = [0, 1] \ £. Then, we have
X(F) = X([0, 1] \ £) = A(Q([0, 1] \ Ekn))
= f>([0, 1] \ £*„) = f^[l-X(EL)] < £ ^ = 1 -e.
n=l n=l n=l
Hence, X(E) = 1 - X(£) > 1 - (1 - e) = e.

Section 18: THE LEBESGUE MEASURE 149
(b) Let A'l = [0, 1] \ [^-, £], 1 < * < n\ n > 2. Clearly, \{Ank) = 1 - i holds
for each 1 < k < n and P|"=1 ^l ~ $ holds for each n > 2. Let En denote the
sequence
A2 A1 43 43 A3 An An A" An+*
Now, note that A.(£/i) —>• 1 and f]^Ln Ek—0 holds for each n > 1.
Problem 18.9. /tew77£ //?af a function f: I —► R defined on a subinterval oflR
satisfies a Lipschitz condition. That is, assume that there exists a constant C > 0
suc/i r/?ar |/(a*) — /(y)| < C|* — y\ holds for all x, y e /. Show that f carries
(Lebesgue) null sets to null sets.
In particular, if a function f: I -+ R defined on a subinterval of R has a
continuous derivative, then show that f carries null sets to null sets.
Solution. Assume that a function /: / —> R satisfies the condition of the
problem. Clearly, / is a (uniformly) continuous function. In particular, note that
if J is a subinterval of /, then /(/) is also a subinterval of R (see part (g) of
Problem 6.11), and our condition implies (how?) that the length of /(/) is less
than or equal to C times the length of 7, i.e., X* (/(/)) < Ck*(J) holds.
Now, let A be a null subset of / and let e > 0. Pick a sequence {[#„, bn)\ of
half-open intervals such that
oo oo oo
A C [J[an, b„) and £] A.*([a„, bnj) = ]T(6„ - an) < e.
/i=l n=\ n=\
Hence, f(A) c f{\J?=l[a„, b„) n /) = U~ i /([«». *») n /), and so by the
preceding
r(/(/i)) <r(Q/([fl„.6a)n/))
00 oo
Since £ > 0 is arbitrary, we see that X* (f(A)) = 0 holds, as desired.
For the second part notice that if [a, b] is a closed subinterval of /, then there
exists some constant M > 0 satisfying |/'(f)l < M for all / € [a, 6]. Now, if
.v, v 6 [a, fr], then there exists (by the Mean Value Theorem) some z between x and
.y satisfying f(x)-f(y) = /'(z)U-y). This implies |/U)-/(v)| < M\x-y\
for all a*, y e [<?, fr]. So, by the first part, / carries null sets of [a, b] to null sets.

150
Chapter 3: THE THEORY OF MEASURE
Now, fix a sequence [[an,bn]} of closed subintervals of / such that / =
{J™=i[an, bn] and let A be a null subset of /. Then A fl [an, bn] is a null
subset of [an, bn], and so f(A fl [afn bn]) is a null subset of R. Now, notice that the
identity
CO 00
f(A) = /((J a n [*„, iB]j = (J /(A n fa.- W)
guarantees that /(>4) is a null subset of R.
Problem 18.10. Show that the Lebesgue measure of a triangle in R2 equals its
area. Also, determine the Lebesgue measure of a disk in R2.
Solution. Start by observing that every line segment has Lebesgue measure zero
(why?). Thus, the Lebesgue measure of a triangle is the same with or without
some of its edges. Also, every triangle is Lebesgue measurable (since without its
edges-it is an open set). Since X is translation invariant, we can assume that all
triangles have one of their vertices at zero. Let T be such a triangle, and let A(T)
denote its area. Following the graphs in Figure 3.1 (from left to right) we see that:
2X(T) = X(T) + X(-D = X(T) + X(T2) = X(T) + X(T})
= k(Tx U T) = X(P) = X(Q) = A(P) = 2A(T).
r1 = (G+/?) + r2
? = r1ur2
\
<h
1 w * '
^2 = c -
X
i
y
i
Q
FIGURE 3.1. The Lebesgue Measure of a Triangle

Section 18: THE LEBESGUE MEASURE 151
FIGURE 3.2. The Computation of the Lebesgue Measure of a Disk
That is, X(T) = A{T). In particular, this implies that the Lebesgue measure of
any polygon equals its area.
Now,let D be a closed disk of radius /*; see Figure 3.2. To compute its Lebesgue
measure, we use the Eudoxus-Archimedes Method of Exhaustion. For each n, let
Pn and Qn be the inscribed and circumscribed regular /i-polygons, respectively.
Clearly, Pn c D c Qn holds. Now, note that
A(/>„) = *r2
and so, by letting n —> co, we see that
A(D) = nr2 = /4(D).
Problem 18.11. If p. is a translation invariant Borel measure on R", then show
that there exists some c > 0 such that p*{A) = cA.*(A)/or a// sMfesef A o/R".
Solution. By Theorem 18.8,/x = cA holds on B for some constant c > 0. Now,
by Theorem 14.10, (cA)* = cA* holds, and consequently /z*(A) = (cA.)*(A) =
cA*(A) for each subset A of R".
Problem 18.12. Show that an arbitrary collection of pah-wise disjoint
measurable subsets o/R, each of which has positive measure, is at-most countable.
Solution. Let C be a collection of pairwise disjoint measurable subsets of R
such that A(C) > 0 holds for each C eC. For each n let
Cn = {C eC: A(CH [-nyn])> ±},
»n(l)"
cos(^) < A(D) < X(G„) = nr2
Hf)

152
Chapter 3: THE THEORY OF MEASURE
and note that C = \J%LX Cn. Now if C\,..., C* e Cn, then we have
Ln < £ KQ n [-/i, n]) = a(((JC/) H h/i, /i]) < X([-/if «]) = 2/7,
and so /: < 2/?2 holds. This shows that each Cn is a finite set, and consequently
C is at-most countable.
Problem 18.13. Let G be a proper additive subgroup of JR.". IfG is a
measurable set, then show that X(G) = 0.
Solution. If X(G) > 0, then, by Theorem 18.13, the element zero is an interior
point of G — G. Since G is an additive group, G — G = G holds, and from this
it follows that G = R", which is a contradiction.
Problem 18.14. Let /: R -> R fo? additive (i.e., f(x + v) = f(x) + f(y)for
all a, y e R) andLebesgue measurable. Show that f is continuous—and hence,
of the form f(x) — ex.
Solution. Assume / ^ 0 and let s > 0. Since / is an additive function,
nf'x ([0, e]) = f~] ([0, ne]) holds (why?). Thus, if k(fl ([0, e])) = 0, then
Mr1 ([«ne, 0])) = X(/"1 ([0, ne])) = n^/"1 ([0, £])) = 0
holds for each n, and so X(/_1 ([—/?£, ne])) = 0 for all n. From
Z"1 ([-*£,*£]) fR,
it follows that X(R) = 0, which is impossible. Thus, X(/_1([0, e])) > 0. Since
/ is also measurable, there exists (by Theorem 18.13) some 8 > 0 with
(-a,8) c r1 (to, *]) - /-1 (to, e]) = r1 ([-e, e]).
That is, —6 < a* < 5 implies — £ < f(x) < e so that / is continuous at zero.
Now apply Lemma 18.7.
Problem 18.15. Show that an arbitrary union of proper intei-vals of R is a
Lebesgue measurable set.
Solution. Let {Ia: a e A] be a family of "proper" intervals (an interval is
proper whenever its endpoints a and b satisfy a < b) and let E = \JaeAIa-
Write E = Ujce£ C*» where Cx "denotes the component of x in E. Since each
a* belongs to a proper subinterval of £, we see that each Cx is a proper interval;
see part (g) Problem 6.11. Since the distinct components Cx are pairwise disjoint,

Section 18: THE LEBESGUE MEASURE
153
we see that there are at-most countably many Cx and so E is the union of at-
most countably many intervals. Now, use the fact that each interval is a Lebesgue
measurable set to infer that E itself is a Lebesgue measurable set.
Problem 18.16. Let C be a closed nowhere dense subset of IR" such that
X(C) > 0. Show that the characteristic function xc cannot be continuous on
the complement of any Lebesgue null set of R". Also, show that xc w/// be
continuous on the complement of a properly chosen open set whose Lebesgue measure
can be made arbitrarily small.
Solution. Let A c R" be a Lebesgue null set. Since X(C) > 0 and X(A) = 0,
it follows that Ac n C ^ 0. Fix some a e Ac n C. We claim that xc' ^c —► R
is not continuous at x = a.
Indeed, if xc: ^c —> R is continuous at a, then there exists some open ball
B{a, r) with xc(-0 = 1 for all x e B(a, r) n Ac\ i.e., B(a, r) n Ac c C holds.
Since A.04) = 0, it follows that B(a, r) 0 Ac is dense in B{a, ;*), and therefore,
B(ay r) C C (since B{a, r) H Ac C. C and C is closed), contradicting the fact
that C is nowhere dense.
Now, let e > 0. By Theorem 18.2, there exists an open set V with C C V and
X(V \ C) < e. Note that the set O = V \ C = V D Cc is open, and A.(0) < e.
We claim that xc' Oc —> IR is continuous.
To see this, let a £ O = V fl Cc. We have two cases.
1) a e C. Since CCV, there exists some open ball B(a, r) with #(<2, r) c V.
Now, note that
B(a% /') O Oc = £(a, r) fl [Vc U C] = £(a, r)HC CC.
Thus, if jc G B(a,r) fl <9C, then Xc00 = 1- This shows that the function
Xc- Oc —> R is continuous at * = a.
2) a e Cc. Choose an open ball B(ay r) such that B(a, r) c Cc. Then,
B(a, r)C\Oc = fi(a, r) fl [Vc U C] = £(a, r) H Vc C Vc C Cc.
Thus, x e B(a,r) C\ Oc implies Xc00 = 0, which shows that in this case
Xc'-Oc —> R is continuous at x = a.
Problem 18.17. Lef /: R" -» R Z?e a continuous function. Show that the graph
G = {(.v,,...,*„, /(*,,..., *„)): Ui,..., x„) 6 R"}
off has (n 4- \)-dimensional Lebesgue measure zero.

154
Chapter 3: THE THEORY OF MEASURE
Solution. Denote by Xn+\ and Xn the {n + l)-dimensional and n-dimensional
Lebesgue measures, respectively. Fix some k and let A = [—k, k] x • • • x [—/:, k].
Now, let e > 0. By the uniform continuity of / on A, there exists some 8 > 0
such that jc, y e A and |jc/ — yt\ < 8 for 1 < / < n imply |/(a*) — f(y)\ < £• Fix
a partition P of [—k, k] with mesh \P\ < 8, and let Q = /> x • • • x P. Then,
Q subdivides A into a finite number of distinct closed cells, say A\,..., Ap.
(Note that the open cells corresponding to A\,..., Ap are pairwise disjoint). For
each 1 < / < p fix some a, e Aj, and let // = [/fe) — £, /(tf/) + £]. Then,
G* Q Uf=i(^/ x h) holds, and so
p p
K+i(Gk) < Ys^n+dAt X /,-) = J^kniAi) . 2£ = (2*)n • 2s
1 = 1 /=!
holds for all £ > 0. This shows that Xn+i(Gjt) = 0 for each k. To complete the
proof, now apply Theorem 15.4 to G* t G.
Problem 18.18. Let X bea Hausdoifftopological space, and let fibea regular
Bore I measure on X. Show the following:
a. If A is an arbitrary subset ofX% then
li*{A) = inf{/x(0): O open and A C O}.
b. If A is a measurable subset ofX with fi*(A) < oo, then
fJL*(A) = sup{fi(K): K compact and K C A}.
c. Iffi is a-finite and A is a measurable subset ofX, then
/jl*(A) = sup{/x(/0: K compact and K C A}.
Solution, (a) Since every a-set is a Borel set, Problem 15.2 shows that
fi*(A) = inf{fi(B): B is a Borel set satisfying A c. B).
Now, use property (2) of Definition 18.4.
(b) Let A be a measurable set with fi*(A) < oo and let £ > 0. Pick an open set
V with A C.V and /z*(V) < /z*(A)H-£. Similarly, choose an open set W such
that V\A£WQV and
fi*{W) < il*(Y \ A) 4- £ = m*(V) - M*(A) + e < 2s.

Section 18: THE LEBESGUE MEASURE
155
Next, pick a compact set C such that C C V and fi*(V) < fi*(C) 4- e. Set
K = C fl VKC, and note that /£ is a compact subset of A. Moreover,
0 < ii*{A) - vl*(K) = n*(A \ K) < fx*(V \ K)
= /z*((V \C)Uiy)< [>* (V) - M*(C)] + ijl*(W) < 3s
holds, and the desired conclusion follows,
(c) Straightforward using (b).
Problem 18.19. If A is a (Lebesgue) measurable subset o/R of positive measure
and 0 < 5 < X(A), then show that there exists a measurable subset B of A
satisfying X(B) = <5.
Solution. We shall present two solutions. The first one will employ the Axiom
of Choice (via Zorn's Lemma); the second one will establish the validity of the
conclusion without using the Axiom of Choice and without assuming that A is a
measurable set.
(a) Consider a measurable subset A of R and some 8 > 0 satisfying 0 < 8 <
k(A). Since k(A n [—w, n]) f X(A) holds, replacing A by some A D [—/2, n],
we can assume that \(A) < oo also holds.
Next, we shall denote by A the set of all collections C of pairwise disjoint
measurable subsets of A such that:
a) \{C) > 0 holds for each C e C (and so C is at most countable); and
b) The Lebesgue measurable set UceC^ satisfies X([JCeCC) <S.
From Problem 15.18, it is easy to see that A^0. Under the inclusion relation
c the set A is a partially ordered set. We claim that the partially ordered set
(A £) satisfies the hypothesis of Zorn's Lemma. To see this, let {C/: i e 1} be
a chain of A (i.e., for each pair /, j e I either C/ C C; or C} C C, holds
true). Our claim, will be established, if we can show that C — (J/6/ C, € A Note
first that if B, C e C, then £, C gQ must hold for at least one / e /, and
so j5 n C = 0. In particular, it follows that C is at most countable. Now, if
B\, ..., Bk g C, then Bj,..., B^ e C/ also must hold for some / (why?), and
so A.(Ur=1 Br) < ^((Jflec ^) — ^- Since C is at most countable, it follows that
Now, by Zorn's Lemma, the collection A has a maximal element, say C. If
5 = UceC £*'tnen we claim tnat tne measurable set 5 satisfies X(B) = 5 (and
this will complete the proof). To see the latter, assume by way of contradiction
that X(B) < 8. Then, we have 0 < rj = 8 - X(B) < X(A) - X(B) = X(A \ B)
holds, and so by Problem 15.18 there exists a measurable subset D of A \ B

156
Chapter 3: THE THEORY OF MEASURE
satisfying 0 < X(D) < r\ (clearly, D $ C). In view of B D D = 0 and
X(B UD) = k(B) + k(D) < k(B) + <5 - X(B) = <5,
we see that C\ = C U {£>} € A However, this contradicts the maximality property
of C, and so X(B) = 5 must hold, as desired.
(b) For this solution the set A is an arbitrary subset of IR satisfying X(A) > 0.
As in the preceding, we can assume that A c [—k, k] holds for some k. Now,
consider the function /: [—fc, k] —► IR defined by
/(0 = «AnhM]), re[-U].
Clearly, f(—k) = 0 and /(&) = A(i4). We claim that / is a continuous function.
Indeed, if — k < s < t < k, then
f(t) = k(A n [-*, /]) < A.(i4 n [-*, j]) + x(i4 n c?, /]) < /(j) + / - s.
Therefore, \f(s) - f(t)\ < \t - s\ holds for all sj e [-£, k] and so / is a
continuous function.
Finally, by the Intermediate Value Theorem, there exists some — k < x < k such
that the subset B = A D [—k, x] of A (which is measurable if A is measurable)
satisfies f(x) = X(B) = 8.
Problem 18.20. Let Ebea Lebesgue measurable subset of R of finite Lebesgue
measure. Show that the function f&\ R —> R, defined by
fE(x) = k(EA{x + E)),
is uniformly continuous.
Solution. The solution goes by steps.
(1) Assume first that E = (a, b) is a bounded open subinterval of R. In this case,
an easy calculation shows that
|2|*|, if \x\<b-a
jekx) - |2(ft-fl)t if \x\>b-a.
This guarantees that /*£ is uniformly continuous in this case.
(2) Assume that E and F are two* Lebesgue measurable subsets of R of finite
measure such that fs and ff are both uniformly continuous. Put G = E U F. We
shall show that fc is also uniformly continuous.

Section 19: CONVERGENCE IN MEASURE
157
To see this, notice first that
IxgOO - Xx+c(y)\ < \xeW - x.x+E(y)\ + \xfW - x.v+fO0|
implies
k(GA(x + G)) < X(EA(x + £)) + A.(FA(jc + F)).
Hence,
\fcW-fciy)\ = |MGA(* + G))-MGA(y + G))|
< A.([GAU + G)]A[G A(y + G)])
= X((jc + G)A(y + G)) = X(GA(y -x + G))
< k(EA(y - jc + £)) + X(£A(y - jc + f))
= fE(y-x) + fF{y-x).
Since /*£ and /> are uniformly continuous, it follows that fc is likewise uniformly
continuous. (Actually, the continuity of ff and ff at zero is what is needed
here.)
(3) By induction, we can show that if E = (J"=1 £/ with each £,- Lebesgue
measurable having finite measure and £/ n £y = 0 if/ ^ 7, then ff is uniformly
continuous.
(4) Now, let 6 > 0. Pick a finite collection of pairwise disjoint bounded open
intervals I\,..., ln such that the set G = (JJL, ^ satisfies X(EAG) < €. Then,
as previously, we have
|/fto " /e()0| = \UEA(x + £)) - X(£A(y + £))|
< X(£A(y-jc + £))
< k(E&G) + k{GA(y-x + G)) + k({y-x + G)b{y-x + E))
< 2€ + X(GA(^-jc + G))
= 2€ + /c(3'--r).
This easily implies that /£ must be a uniformly continuous function.
19. CONVERGENCE IN MEASURE
Problem 19.1. Let {/„} be a sequence of measurable functions and let f: X ->
R. Assume //7af lim £<,*({* € X: |/,,(*) - /(jc)| > e}) = 0 holds for every e > 0.
5/?<9w //zaf / /s fl measurable function.

158
Chapter 3: THE THEORY OF MEASURE
Solution. Pick a sequence {kn} of strictly increasing positive integers such that
H*([x 6 X: \fk(x) - f(x)\ > i})< 2-" for all k > kn. Set
E„ = {xeX: !/*.(*)-/(*)!>£}
for each n and let E = f|~=i IC £" • Then,
OO 00
H*(E) < fi*(\J E„) < £>*(£„) < 2'-m
holds for all m, so that /z*(£) = 0. Also, if x £ £, then there exists some /w
such that x £ \J%Lm En, and so l/^Qc) — f(x)\ < £ holds for each n > m.
Therefore, lim/*n(jc) = f(x) for each x £ £, and so fan —> f a.e. holds. The
latter (by Theorem 16.6) easily implies that / is a measurable function.
Problem 19.2. Assume that (/„) C M satisfies fn f and fn -^ /. Show that
fn t / a.e. holds.
Solution. By Theorem 19.4, there exists a subsequence {/^} of the sequence
{/„} with fkn —> f a.e. Since /„ |, it easily follows that /„ t / a.e. holds.
Problem 19.3. If[fn] Q M satisfies fn -^* / and fn > 0 a.e. for each n, then
show that / > 0 a.e. holds.
Solution. Since, by Theorem 19.4, some subsequence of {/„} converges almost
everywhere to /, we must have / > 0 a.e.
Problem 19.4. Let {/„} c M and [gn] C M satisfy fn-^+ f, gn -^ g, ond
fn = gn a.e. for each n. Show that f = g a.e. holds.
Solution. Since /„ —> f implies fon —> f for each subsequence {/^J
of {/„}, by passing to two subsequences (if necessary), we can choose a strictly
increasing sequence [kn] of positive integers such that ftn —> f a.e. and
8kn —► g a-e- This easily implies / = g a.e.
Problem 19.5. Let (X, <S, ix) be a finite measure space. Assume that two sequ-
encps {fin} and [gn} of M satisfy /„ -^ f' andgn^ g. Show that fng„ -A- fg.
Is this statement true if /x*(X) = oo?

Section 19: CONVERGENCE IN MEASURE
159
Solution. By Theorem 19.4, the only possible limit of {/„£„} is fg.
Consequently, if fngn —> fg does not hold, then there exist e > 0 and 8 > 0 and
some subsequence of {/„#„} (which we shall denote by {fngn} again) such that
Ijl*{{x e X: \fa(x)g„{x) - f{x)g{x)\ >s})>8 (*)
holds for all n. In view of fn —> f and g„ —>• g, Theorem 19.4 shows that
for some subsequence {fk„gk„} of {/„#„} we must have fk„gk„ —► fg a.e.
Now, note that (by Theorem 19.5) fk„gk„ —> fg holds, contrary to (•). Thus,
fngn -^ fg holds.
If n*(X) = co, then the conclusion is no longer true. An example: Take X =
(0, co) with the Lebesgue measure. Consider the functions fn{x) = .Vx4 + ^ and
f(x) = x\ Then, /„ -U /, while f}x A /2.
Problem 19.6. Show that a sequence of measurable functions {fn} on a finite
measure space converges to f in measure if and only if every subsequence of{fn}
has in turn a subsequence which converges to f a.e.
Solution. The conclusion follows immediately from Theorems 19.4 and 19.5.
Problem 19.7. Define a sequence [fn] of M to be /x-Cauchy whenever for
each € > 0 and 8 > 0 there exists some k {depending on e and 8) such that
ti*([x e X: \fn{x) - fm(x)\ > *}) < 8 holds for all n% m > k.
Show that a sequence {/„} of M is a ji-Cauchy sequence if and only if there
exists a measurable function f such that fn -^> /.
Solution. If /„ —y /, then the inclusion
[x: \fnM-fm{x)\>2e] c [x: }fn(x)-f{x)\ > e} U [x: \fm{x)-fW\>s]
easily implies that {fn} is a /x-Cauchy sequence.
For the converse, assume that {/„} is a /x-Cauchy sequence. It suffices to show
that {/„} has a subsequence that converges in measure (why?). To this end, start
by selecting a subsequence {#„} of {/„) satisfying
^{{x: \gn(x)-gm{x)\>2-"})<2-n

160
Chapter 3: THE THEORY OF MEASURE
for all m > n. Let En = {jc: \gn+\(x) - gn(x)\ > 2""}. Also, let
oo
Fn = {jEk = [x: |&+,M - gk(x)\ > 2~k holds for some k>n).
Clearly,/z*(F„) < YlT=n /**(£*) 5 21"" holds for all h, and hence the measurable
set F = Pl^li ^ satisfies /x*(F) = 0. Now, note for each fixed x £ F there
exists some positive integer kx such that x $ Fn holds for all n > kx. Thus, for
n>kx, we have
00
\g*+P(x) ~ 8nW\ < £>+,(*) - g,(x)\ < 21-".
i=n
Therefore, {gn(x)} is a Cauchy sequence of real numbers for each x £ F. Thus,
there exists a function g € M such that £„(*) -» #(a) holds for each x £ F.
Now, if ;z > k and jc $ F„, then
oo
\gn+\W - ft,+/,0r)| < ]T |ft+i - ftW| < 2~n
implies that |g„+i(A) — £(a)| < 2"n < 2~A for all n > k. Thus,
{*€*: |^,(a-)-^)|>24(cF„
holds for all n > k. Finally, to see that gn —> g holds, note that for n > k, we
have
{xeX: \gn(x)-g(x)\>2l-k}
C {a- g X: |^(a) - gn+dx)\ > 2"a}u{a g X: \gn+](x) - g(jr)| > 2~k)
c EnUFn = Fn.
20. ABSTRACT MEASURABILITY
Problem 20.1. Le/ H be a nonempty collection of subsets of a set X. Show
that 1Z is a ring if and only ifR, is -closed under symmetric differences and finite
intersections.

Section 20: ABSTRACT MEASURABBLITY
161
Solution. Assume first that the nonempty collection 1Z is a ring. That is, assume
that A, B e 71 imply A U B e 71 and A \ B € 71. Then the identities
A&B =(A\B)U(B\A) and AHB = A\(A\B)
easily imply that 71 is closed under symmetric differences and finite intersections.
For the converse assume that 71 is closed under symmetric differences and finite
intersections. Then, the identities
A\B = AA(AC\B) and A U B = (A&B)&(A n B)
guarantee that 71 is a ring.
Problem 20.2. // 71 is a ring of subsets of a set X, then show that the collection
A={A<ZX: Either A or Ac belongs to 71}
is an algebra of sets.
Solution. From the definition of A it easily follows that if A e A, then Ac e Ay
i.e., that A is closed under complementation.
Now, assume that A, B e A. If A, B e 7£, then since 71 (as being a ring) is
closed under finite unions, we have A\JB eTZ and so A U B e A. If Ac, Bc eTl,
then Ac \ (Ac \ Bc) £ 71, and so
AUB =(ACC\ Bc)c = [Ac \ (Ac \ Bc)]c e A.
Now, assume that A e 71 and Bc e 71. Then, Bc \ A = Bc n Ac e K, and
consequently (from the definition of .4), AUB = (Ac Pi Z?c)c e A. The preceding
show that A is an algebra.
Problem 20.3. In the implication scheme of Figure 3.3 show that no other
implication is true by verifying the following regarding an uncountable set X.
^* rinu
cr-nug
a-algebra *^ "** ring >» semiring

162
Chapter 3: THE THEORY OF MEASURE
a. The collection of all singleton subsets ofX together with the empty set is a
semiring but not a ring.
b. The collection of all finite subsets ofX is a ring but is neither an algebra
nor a a-ring.
c. The collection of all subsets ofX that are either finite or have finite
complement is an algebra but is neither a a-algebra nor a a-ring.
d. The collection of all at-most countable subsets ofX is a cr-ring but not an
algebra.
e. The collection of all subsets ofX that are either at-most countable or have
at-most a countable complement is a a-algebra (which is, in fact, the a-
algebra generated by the singletons).
Solution, (a) If A and B are singletons, then A n B and A \ B are either empty or
singletons. This shows that the collection of all singletons together with the empty
set is a semiring. However, it should be obvious that finite unions of singletons
need not be a singleton, and so the collection of all singletons is not an algebra.
(b) Let 7£ denote the collection of all finite subsets of (the infinite) set X. If
A, B eTl, then A U B and A \ B are finite sets and so A U B and A \ B belong to
71. This shows that 71 is a ring. Since the complement of a finite set is infinite, it
follows that 71 is not closed under complementation, and so 71 is not an algebra.
To see that 71 is not a cr-ring, let A = [a\, #2,...} be a countable subset of X,
and for each n let An = {a,,} e 71. Then, IJ^li An = A £ 71, and this shows that
7Z is not a cr-ring.
(c) If 71 is the ring of all finite subsets, then by part (b) the collection
A= {AQX: Either A or Ac belongs to 71}
is an algebra of sets. To see that A is not a a-ring (and hence neither a a -algebra),
let A = [a\, ai,...} be a countable subset of X such that X \ A is an infinite
set. Clearly, A £ A. On the other hand, if An = {an}, then An € A and
U^li An = A £ A. This shows that A is not a a-ring.
(d) Let C denote the collection of all at-most countable subsets of X. Clearly,
A, B e C imply A \ B e C. Also, C is closed under countable unions (recall that
an at-most countable union of sets each of which is at-most countable is at-most
countable; see Theorem 2.6). Therefore, C is a cr-ring. However, when X is an
uncountable set, C is not closed under complementation, and hence it cannot be
an algebra.
(e) This is Problem 12.7.
Problem 20.4. Show that a Dynkin system is a a -algebra if and only if it is
closed under finite intersections.

Section 20: ABSTRACT MEASURABILITY
163
Solution. Let V be a Dynkin system that is closed under finite intersections.
Since V is also closed under complementation, it is easy to see (by using the
identity A U B = (Ac n Bc)c) that V is in fact an algebra. So, if A = (J~ , An
with [An] c P, then by letting Bn = \Jnk=]Ak € £>, and noting that Bn t A, we
see that /\ e V. In other words, V is a cr-algebra.
Problem 20.5. Give an example of a Dynkin system which is not an algebra.
Solution. Consider the set X = (1,2,3,4), and let
P={CS,{1,2},{3,4},{1,3},{2,4},X}.
Then V is a Dynkin system (why?), which (since {1, 2} U {1, 3} = {1, 2, 3} does
not belong to V) fails to be an algebra.
Problem 20.6. A monotone class of sets is a family M of subsets of a set X
such that if a sequence [An] of M satisfies An t A or An I A, then A e M.
Establish the following properties regarding monotone classes:
a. We have the following implications:
a-algebra => Dynkin system => monotone class
Give examples to show that no other implication in the preceding scheme
is true.
b. An algebra is a monotone class if and only if it is a a-algebra.
c. The a -algebra a {A) generated by an algebra A is the smallest monotone
class containing A.
Solution, (a) The implication scheme follows immediately from the definitions
of the three classes of sets involved. An example of a Dynkin system which is not
an algebra was exhibited in the preceding problem. Now, if X — {1,2}, then the
collection M = [X, {1}} is a monotone class but not a Dynkin system.
(b) Let A be an algebra of sets. If A is a cr-algebra, then it is clearly a monotone
class. For the converse assume that the algebra A is a monotone class.
Assume [An] c .4 and put A = (JUli ^»- Let Bn = UHUi ^* G ^ ^ note
that Bn t A. Since A is a monotone class, it follows that A £ A and so A is a
cr-algebra.
(c) Let A be an algebra of sets and let M be the smallest monotone class that
contains A, i.e., M is the intersection of the collection of all monotone classes
that include A. Clearly, Ac. M C. a (A).

164
Chapter 3: THE THEORY OF MEASURE
Let C = {B e M: B\A e M for each A e A}. An easy verification shows
that C is a monotone class that includes A, and so M = C. Now, let
V={B eM: M\B eM for each M e M).
Again, V is a monotone class which (in view of M = C) satisfies A C V.
Thus, £> = M. This shows that M. is, in fact, a Dynkin system. By Dynkin's
Lemma 20.8, a{A) c M, and so M = a(A).
Problem 20.7. S/icw r/zaf //" X and Y are two separable metric spaces, then
Bxxy=Bx®By.
Solution. Assume that X and Y are two arbitrary topological spaces. For each
subset A of X, let
VA = {BQY: AxB eBXxY}.
From the identities A x (B \C) = (A x B)\(A x C), we see that if #, C € £*,
then 5 \C € 5V From A x (f]T=\ Bn) = fXliO4 x **)•il follows that XU is
closed under countable intersections. Observing that 0 e E/\,we infer that D^ is
a a-ring. Clearly, E^ is a a-algebra if and only if y e E/\.
Next, note that for any open subset O of X, V e So for every open subset V
of y. Since Y is itself open, if O is open, then Ep is a a-algebra of subsets of Y
that includes the open subsets of Y. Thus, By C Ec? for each open subset 0 of X.
Now, let
■A= {A CX: By c X^}.
As we have just noticed, V e A holds for each open subset V of X. Since
(as easily checked) E^ = E^c for each A e A, we see that .4 is closed under
complementation. Moreover, if [An] C .A, then for any Borel subset 5 of y, we
have/4„x£ e Bx*y for each/?. Thus,inviewofp|^,(Anx5) = (D^li j4„)xZ?,
we obtain B e En«=li4n. In other words, A is closed under countable intersections,
and so A is a a-algebra including the open subsets of X. This implies Bx Q A.
We have just established the following: If A is a Borel subset of X and B is a
Borel subset of Y, then >4 x B e Bx*y• Therefore,
BX®BY CBM.
For the reverse inclusion, assume that X and Y are two separable metric spaces.
Then every open subset of X x Y is an at-most countable union of sets of the form

Section 20: ABSTRACT MEASURABELITY
165
V xU, where V is an open subset of X and U an open subset of Y. Consequently,
BxxY £ &x ® #k, from which it follows that #x ® &y = #*xr-
Problem 20.8. Show that the composition function of two measurable functions
is measurable.
Solution. Assume that (X, Ej)—> (Y, E2) -^-> (Z, E3) are measurable
functions. IM € E3,then<r104)<= E2, and so (g o /)"l(A) = f~]{g-](A)) e E,.
This shows that g o / is measurable.
Problem 20.9. If(X, E) is a measurable space, then show that
a. the collection of all real-valued measurable functions defined on X is a
function space and an algebra of functions, and
b. any real-valued function on X which is the pointwise limit of a sequence of
(E, B)-measurable real-valued functions is itself (H, B)-measurable.
Solution. Repeat the solution of Problem 17.17.
Problem 20.10. Let (X, Y,)bea measurable space. A E-simple function is any
measurable function 0: X -» 1R having a finite range, i.e, if(j) has finite range and
its standard representation <j) = Yl'i=\aiXai satisfies At e S/or each i.
Show that a function f: X —> [0, 00) is measurable if and only if there exists a
sequence {(pn} of Ti-simple functions such that (pn(x) f f {x) holds for each x e X.
Solution. The proof is identical to the proof of Theorem 17.7. Here it is.
For each n let Aln = [x € X: (i-l)2-w </(*) <i2"n]fori = \,2,...,n2n,
and note that A'n fi AJn = 0 if / ^ j. Since / is measurable, all the Aln belong to
E.
Now, for each n define 0„ = X^=1 2~n(i — l)X/v, and note that {</>„} is a
sequence of E-simple functions. Also, an easy verification shows that 0 < <pn(x) <
0/i-fiU) < /(a) holds for all x and all n. Moreover, if x is fixed, then 0 <
/(a) — (pn(x) < 2~" holds for all sufficiently large n. This implies 0„(a) t fW
for all a eX.
Problem 20.11. Use Corollaiy 20.10 to show that if a measure p is o -finite,
then jjl* is the one and only extension of p to a measure on a(S).
Solution. Let v: a(S) -> [0, 00] be a measure satisfying v(A) — p(A) for each
A e S. We shall establish that v(A) = p*(A) for each A e a(S).

166
Chapter 3: TOE THEORY OF MEASURE
Fix E e S with /x(£) < oo and let
SE = {E n A: A e S] = {B e S: B o E}.
Clearly, Se is a semiring of subsets of E and \jl restricted to £ is a measure.
Moreover, we know (see Problem 15.7) that the outer measure generated by the
measure space (£, <S£, /x) is simply the restriction of /x* to V(E). In addition, we
claim that if o(Se) denotes the a-algebra generated by Se in P(£), then
a(SE) ={ADE: Ae a(S)} = {B e cr(5): B c E}. (•)
To see this, note first that since [B e cr(S): B c £} is a a-algebra containing Se,
we have
tf(SE)c {B €ff(5): B c£}.
On the other hand, the collection
A={A CX: AHE ea(SE)}
is a cx-algebra of subsets of X satisfying S c. A. Hence, a(S) c A. In particular,
if B c £ satisfies £ € a(<S), then £e.Aandso£ = £n£€ a(5£).
Consequently, {5 G <j(5): K£)C ct(Se), and the validity of (•) follows.
Next, note that since Se is closed under finite intersections, jjl*(E) = /x(£) =
v(£) < oo, and v(£) = /x(£) = /x*(£) for all £ e S£, it follows from
Corollary 20.10 that v(£) = /x*(£) for all £ e a(S) with £ c £.
Now, let {£„} be a pairwise disjoint sequence of S satisfying X = (J£li ^ anc*
/z(£„) < oo for each n. If A e cr(S), then by the preceding discussion we have
v(A fl X„) = /x*(A H X„) for each /z, and so
OO 00
v(A) = v(A nx) = vflj Anxn) = Y^ V^A n *«>
oo oo
= /z*04 H X) =/x*(A),
and we are finished. (For more about the extension of/x, see Problem 15.19.)
Problem 20.12. Show that the uniform limit of a sequence of measurable
functions from a measurable space into a metric space is measurable.

Section 20: ABSTRACT MEASURABILITY
167
Solution. Let {/„} be a sequence of measurable functions from a measurable
space (X, Z) into a metric space (Y, d). Suppose / is the uniform limit of {/„}.
That is, assume that for each e > 0 there exists some n0 such that d(fn(x), f(x)) <
e for all x G X and all n > riQ. By passing to a subsequence, we can assume that
d(f,,(x), /(*)) < I holds for each n and all x e X.
Since the family of closed sets generates the Borel sets of Y, in order to establish
the measurability of /, it suffices to prove that f~\C) G E for each closed set C.
To this end, let C be a closed subset of Y.
Let Vn = [y eY: d{y,C) < £}. We claim that
CO
rl(C) = f)r\v„). (**>
To see this, assume a* g f~\C), i.e., let f(x) e C. From d{fn{x),C) <
d(fn(x), fW) < £» we get fnM € Vn or a G f~x(Vn) for each n. Conversely,
if fnW € Vn for each n, then d(fn(x),C) < j- for each /7, and so if we pick some
c„ G C with d(ffl(x)% c„) < yr then we have
d{f(x), C) < d(f{x), cn) < d{f(x\ fn(x)) + d(fn(xl c„) < i 4- ± = 2
for each /i. This implies J(/(a), C) = 0. Since C is a closed set, it follows that
/(*)€ C.or* €/-'((:).
Next, use the measurability of each /„ and the fact that each V„ is open to obtain
that f~](V„) G E for each a?. Now, invoke (••) to conclude that /_1(C) G S.
Problem 20.13. Let /, g:X -> R fee fwo functions and let B denote the a-
algebra of all Borel sets oflR. Show that there exists a Borel measurable function
h: R -» R satisfying f = h o g //flwrf onfy if f~\B) c g-^B) fo/cfc.
Solution. Assume f = hog holds for some Borel measurable function /z: R ->
R. Fix 5 G S and note that h~\B) e B. Therefore, f~l(B) = g-](h~\B)) e
g-l(B), and so f\B) C g-^B) holds.
For the converse, assume f~l(B) C g_1(B). The existence of the Borel
measurable function h will be established by steps.
Step I: Assume f = Xa f°r some A e /~l(B).
Since f~x(B) c g-1(B) is true, there exists some B G B such that A = g~l(B).
Let /* = Xfi, and note that h o g = /.
Step //: Let / = 5Z?=ia/Xi4, witn tne ^» pairwise disjoint and Ai G /-1(B)
for each /. For each / choose some Bt g g_1(B) such that At = g~l(5/). If
we consider the Borel step function h = ]JC"=ia/Xfl(-» men i* *s easv t0 see tnat
hog = f.

168
Chapter 3: THE THEORY OF MEASURE
Step III: The general case.
The preceding problem applied with £ = /_1(S) guarantees the existence of
a sequence {0„} of f~l(B)-simp\t functions satisfying (pn(x) t /(*) for each
x e X. Now, by Step II, for each n there exists a Borel measurable function
hn: R -* R such that hn o g = 4>n. Next, let
B — \x e R: lim h„(x) = h(x) exists}.
It follows (as in Problem 16.7) that B e B and hn(x)xB(x) -» h(x) for each
A' € R. If we let h(x) = 0 for a £ 5, then h: R —► R is Borel measurable and
satisfies h o g = /.
Problem 20.14. Lef (X, E) be a measurable space, Y,Z\, and Z-i be separable
metric spaces and ty a topological space. Now, assume also that the functions
fi■: X x Y -> Z/, (/ = 1, 2), are Caratheodory functions and g: Z\ x Z-i -» ^ (s
Borel measurable. Show that the function h: X x Y -> ty, defined by
Kx,y) = g(Mx,y)*f2(x,y)).
is jointly measurable.
Solution. By Theorem 20.15, each /,: X x F —► Z, is jointly measurable. This
implies that the function F:X x Y —> Z\ xZ-i, defined by
F{x,y) = (Mx,y),f2{x%y))
is measurable (why?). Since g\Z\ x Z2 ->• ^ is 0Bz,xZ;, Z3vj/)-measurable and
(by Problem 20.7) Sz, ® #z2 = jBz,xz2, it follows that // = ^ o F is likewise
measurable.
Problem 20.15. Let (X,Y,)bea measurable space and (Yy d) a separable metric
space. Show that a function f\X-+Yis measurable if and only if for each fixed
y eY the function x \-+ d(y, f(x)),from X into R, is measurable.
Solution. Let /:(X, E) —► Y be a function from a measurable space to a
separable metric space. For each y e Y define the function gy:X -» R by
gy(x) = d(y, f(x)). Note that for each r > 0 and each y G 7, we have
/-'(SOV)) = {a' € X: f(x).e B(y,r)} = {x e X: d{y, f(x)) < r)
= {a- g X: £3,(a-) < r} = ^((-oo.r))-

Section 20: ABSTRACT MEASURABILITY
169
Assume that each gy is measurable. Then, by the preceding identity, we have
f~] (B(y, /')) = gj] ((-co, /•)) e £ for each y e Y and all r > 0. Since Y is a
separable metric space, every open set can be written as an at-most countable union
of open balls, and so f~l(0) e £ holds foreach open set O. By Theorem 20.6,
/ is a measurable function.
For the converse, suppose that / is a measurable function and let y e Y. From
^((-oo.r)) = f-](B(y,r)) if r > 0 and gJl({-oo,r)) = 0 if r < 0, we
easily infer that gv' is a measurable function for each y eY.
Problem 20.16. Let (X, S, fi) be a a-finite measure space, where S is a o-
algebra. If f:X -» R is a A ^-measurable function, then show that there exists a
S-measurable function g: X -> R such that f = g a.e.
Solution. We can assume f(x) > 0 for each x e X (otherwise, we apply the
arguments below to /+ and /" separately). If / = xa for some A e AM, then
an easy argument (using Theorem 15.11) shows that there exists a null set C such
that B = AUC e S. So, if g = xb, theng is5-measurableand / = g a.e. holds.
It follows that if / is a AM-simple function, then there exists a 5-simple function
g such that / = g a.e.
Now, we consider the general case. By Problem 20.10 there exists a sequence
{0„} of AM-simple functions satisfying <pn(x) t /U) for each .r e X. Foreachflfix
a 5-simple function yj/n such that ijfn = $„ a.e. By Theorem 15.11, foreach/? there
exists a null set An e 5 with V0i(*) = </>„(*) for all* £ An. Put A = (J^li An e ^>
and note that A is a null set. Moreover, we have V^X/^CO t fXAcM for each x.
If g = /x/ic»tnen (°y Problem 20.9(b)) g is a 5-measurable function satisfying
f = 8 a.e.

CHAPTER 4
THE LEBESGUE INTEGRAL
21. UPPER FUNCTIONS
Problem 21.1. Let L be the collection of all step functions 0 such that there exist
a finite number of members A \,..., An ofS all of finite measure and real numbers
a\,..., an such that 0 = Yl'i=\aiXA<- Show that L is a function space. Is L an
algebra of functions?
Solution. Let 0 = £"=1 aiXA, and V = Y1J=\ bjXBr where the Aj and B}
belong to S and they all have finite measure. By Problem 12.14, there exist
pairwise disjoint sets C\,..., Q of S such that each A-t and each B} can be
written as a union from the C/. We can assume that IJr=i Cr = [U/=i ^'] u
[Uy=i Bj\ ft ls easy t0 see tnat 0 ^^ ^ can be written in the form 0 =
]Cr=i crXcr and ^ = Hr=i ^Xcr- Now, everything follows from the
equalities:
1. at + PJf = T.kr=l{acr+pdr)xc,\
2. 0 v f = Er=i(cr v dr)xcr and 0 a f = £*=1(cr a rfr)xcr; and
3. 0^ = Er=ICr^rXCr.
Problem 21.2. Consider the function /:R -* R ate/wed fry /(jc) = 0 //* £
(0, 1], a/id /(*) = v^ '/* € (^T> nl^r 5C,we *• Sfow ^^ / w <3/z upper
function and that —f is not an upper function.
Solution. Put Ak = (^, £] and note that \(Ak) = ^ 1}. Thus, if we let
0/, = £V*xa4.
then {0,,} is a sequence of step functions satisfying 0„(a") \ f(x) for each x. On
171

172
Chapter 4: THE LEBESGUE INTEGRAL
the other hand, the relations
r» /; n oo
/<
A-=l *=1 /.=1
guarantee that / is an upper function.
Since — / is not bounded from below, there is no step function 0 satisfying
(/> < —f. This implies that — / cannot be an upper function.
Problem 21.3. Compute ff dX for the upper function f of the preceding
exercise.
00
Solution. We have // dk = ^Ortfcs?-
Problem 21.4. Verify that every continuous function f:[a,b] —>■ 1R is an upper
function—with respect to the Lebesgue measure on [a,b].
Solution. For each n let Pn = {ao, X\, . . . , A'2" } be the partition that divides
[a,b] into 2" subintervals all of the same length (b — a)2~n\ that is, a, =
a + i(b — a)2~n. Let /w/ = min{/(A): a* e [a*,_i, a*,] }, and then define
0" = X]™/X[.v,-,,v().
1=1
Clearly, each 0„ is a step function. Using the uniform continuity of /, it is not
difficult to see that $„(a) f /(*) holds for all x e [a, b). On the other hand, if
/(a) < M holds for each a, then f(pn dX < M(b — a) holds for all /2, implying
that / is an upper function.
Problem 21.5. Let A be a measurable set, and let f be an upper function. If
Xa < f ae-> then show that fM*(A) < oo.
Solution. Choose a sequence {0„} of step functions with 0„ t / a«e. Then,
0/» a xa t / A Xa = Xa a-e., and so, by Theorem 17.6,
fi*(A)= lim (pn ^XAdfi < lim <pndfjL= fdfi
n-+oo J n-»°° J J
< OO.

Section 21: UPPER FUNCTIONS
173
Problem 21.6. Let f be an upper function, and let A be a measurable set of
finite measure such that a < f(x) < b holds for each x e A. Then, show that
a. fxA is an upper function, and
b. an*{A)<ffXAdfi<bii*{A). •
Solution, (a) Pick a sequence {<j)n} of step functions with 0„ t / a.e. For each
n define the step function \j/n — ((pnXA) A bXA- Then,
/ ir„ dp< bxA d\x = bp*(A) < oo.
Since ij/n \ fxA a.e. holds, fxA is an upper function.
(b) Apply the monotone property of the integral (Theorem 21.5) to the inequality
aXA < fXA <oxa-
Problem 21.7. Let (X, S, p) be a finite measure space, and let f be a positive
measurable function. Show that f is an upper function if and only if there exists a
real number M such that J 0 dp < M holds for every step function <p with 0 < /
a.e. Also, show that if this is the case, then
J f dp = sup! / (pdp: 0 is a step function with 0 < / a.e. J.
Solution. If / is an upper function, then by Theorem 21.5 every step function
0 with 0 < / a.e. satisfies fcfrdp, < ff dp < oo.
Conversely, by Theorem 17.7 there exists a sequence {0,,} of simple functions
with 0„ t / a-e- Since (X, 5, p) is a finite measure space, we know that each
0„ is a step function. If /0„ dp < M holds for all n, then this readily implies
that / is an upper function.
The last formula is immediate from Theorem 21.5.
Problem 21.8. Show that every monotone function f:[a,b] -» R is an upper
function—with respect to the Lebesgue measure on [ayb].
Solution. We assume that /: [a, b] -> 1R is an increasing function. The
"decreasing case" can be proven in a similar fashion and is left for the reader.
By Problem 9.8, we know that the set D of all discontinuities of / is an at-most
countable set. In particular, X(D) = 0. Now, for each n, let Pn be the partition that
subdivides [a, b] into 2n equal subintervals. That is, let Pn = {ag, a",..., a^h

174
Chapter 4: THE LEBESGUE INTEGRAL
where a? =a + ^-i for i = 0, 1,..., 2". Next, for each 1 < / < 2" let
< = inf{/(A): ag[<_p<]},
and put <pn = z2i=i m" X[a?_va?)- Clearly, each </>„ is a step function and, in view
of the monotonicity of /, we have
<t>nW<(f>n+\(x)<fW
for all x e [a, b). Put E = D U P{ U P2 U F3 U • • • and note that X(E) = 0. We
shall establish that (f>n(x) f /(a) for each a e [a, b] \ E.
To this end, fix some t € [a, b] \ E and let € > 0. Since / is continuous at f,
there exists some 6 > 0 such that
x e [a,b] and t-8 < x <t +8 imply f(t)-e<f(x)<f(t) + €. (•)
Next, pick some k such that ~r < 5 for all n > k, and then choose the subinterval
[jc/_i, jc,] of F;. such that / g (a*/_i , a*/). From (•), it easily follows that
MO = inf{/(*): a € [a,-!, a,]} > /(f) - 6.
Therefore, /(/) - e < <pk(t) < 0„(f) < /(/) holds for all n > k, and this shows
that <pn(t) f /(/), as claimed.
Finally, note that the monotonicity of / guarantees that m" < f(b) holds for
all 1 < / < 2\ This implies
/2" 2"
/=] i = l
for each «, and this establishes that / is an upper function.
22. INTEGRABLE FUNCTIONS
Problem 22.1. Show by a counterexample that the integrable functions do not
form an algebra.
Solution. Consider the function •/: R -> R defined by /(a) = 0 if a £ (0, 1],
and /(a) = +Jn if a € (^j, jt] for some «. From Problem 21.2, we know that /
is an integrable function.' Now, note that f2 is not an integrable function.

Section 22: INTEGRABLE FUNCTIONS
175
Problem 22.2. Let Xbea nonempty set, and let 8 be the Dirac measure on X with
respect to the point a {see Example 13.4). Show that every function f: X -> R is
integrable and that f f d8 = f{a).
Solution. Note that / = f{a)x{a) a.e. holds. Consequently, the function / is
integrable and JfdS = f(a)S({a}) = f{a).
Problem 22.3. Let \jl be the counting measure on IN {see Example 13.3). Show
that a function /: JN -* R is integrable if and only z/X^ii \f(n)\ < °°- Also,
show that in this case f f d\i = Y1T=\ fW-
Solution. Let /: IN —► R. Since every function is measurable, / is integrable
if and only if both /+ and /" are integrable. So, we can assume that f{k) > 0
holds for each k.
If (pn = J2'k=\ f(k)X[k)i then {</>„} is a sequence of step functions such that
<M&) tn /W for each k, and
//I oo
0ndM = £/(*) t« £/(*)•
k=\ k=]
This shows that / is integrable if and only if Y1T=\ /W < °°» anc* *n tn^s case
//rfM = E21i/(*) holds.
Problem 22.4. Show that a measurable function f is integrable if and only if\f\
is integrable. Give an example of a nonintegrable function whose absolute value
is integrable.
Solution. Apply Theorems 22.2 and 22.6. For a counterexample: Let E
be a non-Lebesgue measurable subset of [0, 1] and consider the function
/ = XE ~ X[0J]\£.
Problem 22.5. Let f be an integrable function, and let {En) be a sequence of
disjoint measurable subsets ofX. If E = (J^li &»> then show that
ffdp, = jrf f^-
Solution. LetF„ = U?=i £/.Clearly, \fXFn\< |/|foreach/?and/x^ —> fXE

176
Chapter 4: THE LEBESGUE INTEGRAL
a.e. Thus, by the Lebesgue Dominated Convergence Theorem, we have
I fdn= / fXE d\i = JLim^ / fxFn d\i
=iimii://^^)=E//^.
Problem 22.6. Let f be an integrable function. Show that for each e > 0
there exists some 8 > 0 (depending on e) such that \fEfdii\ < € holds for all
measurable sets with /x*(£) < 8.
Solution. Consider an integrable function / and let e > 0. From 0 < |/| A n t
|/| and the Lebesgue Dominated Convergence Theorem we get / |/| A n dfM t
/ |/| dfi. So, there exists some «o such that /(|/| — |/| a no)dn < | for all
n > A20. Now, put 8 = 2^- and note that if a measurable set E satisfies fi*(E) < 8,
then
\f fdfi\ = [\f\dn= f{\f\-\f\An0)dn+ f \f\Anodfi
lJE ' JE JE JE
< /,(I/I~I/|a/20)^/x + f n0dn
< § + AZ0/X*(£)< § + §=£,
as desired.
Problem 22.7. 5/^vv that for every integrable function f the set
{a-€X:/(a-)#0}
az/7 be written as a countable union of measurable sets of finite measure—referred
to as a a-finite set.
Solution. Each En = [x e X: \f(x)\ > £} is a measurable set and, by
Theorem 22.5, ix*(En) < oo holds. Now, observe that
[xeX: fW?0] = \jE„.
Problem 22.8. Let /: IR -» R be integrable with respect to the Lebesgue
measure. Show that the function g: [0, oo) -» R defined by
g(t) = sup\f \f(x + y) - /(*) I dX(jc): |y| < r|
/or / >0 is continuous at t = 0.

Section 22: INTEGRABLE FUNCTIONS
177
Solution. Let /: R —> R be an upper function and let {0,,} be a sequence of
step functions with 0„ t / a.e. Fix some yy and note that cpn(x -f y) t f(x + y)
holds for almost all x. Since //*(* + 30 = XA-yOO and M^) = MA — 3O, it
follows that f(x + y) as a function of x is integrable and ff(x + y)^X(.r) =
// dX. Thus, if / is integrable, then f{x+y) is integrable with respect to x for
each fixed y and //(* + y)dX(x) = ffdX holds. In particular, for each fixed
y we have / |/(.v + y) - f(x)\ dX(x) <f\f(x + 3OI dX(x) + / \f(x)\ dX(x) =
2f\f\dX <oo.
Now, for each integrable function /: R —> R and each / > 0, define
gf(t) = supjJl/U + 3O - /U)| dX(x): \y\ < t j > 0.
Then, we have
£/+/,(') < g/(0 + £/,(0 and gaf(t) = |of|g/(/).
These relations show that the set
L = {/: /is integrable and g/ is continuous at zero }
is a vector space. Moreover, L has the following approximation property:
• // / is an integrable function such that for each e > 0 //zere exists some
h e L with f\f - h\dX < e, then f e L.
Indeed, if / is such a function and s > 0 is given, then choose h € L with
f\f — h\dX < s. Pick some 8 > 0 with gi,(t) < £ whenever 0 < / < <5, and
note that for \y\ < t we have
J\f(x + y)-fW\dk{x)
< j\ftx + y) - h(x + y)\ dX(x) + j\h(x + y) - A(*)| rfX(.r)
+ f\h- f\dX <3s.
Thus, #/(0 < 3e holds for all 0 < t < 8 so that / e L.
Now, assume that / = X[a,b)- If 0 < / < b — a, then for | v| < f we have
f\ftx + y)~ fW\ dX{x) = J\x[a-y.b-y)(x) - X[a.b)W\ dXW
= X([a - y, 6 - y)A[fl, b)) = 2|v| < 2f,

178
Chapter 4: THE LEBESGUE INTEGRAL
and so gf(t) < 2t holds for all 0 < / < b — a, i.e., / e L. By the approximation
property, we have xa € L for every a-set A of finite measure, and hence, by the
same property, xa £ L for every A e A with \(A) < oo (see Problem 15.2). It
follows that L contains the step functions. Since for every integrable function /
and each e > 0 there exists a step function 0 with f\f — <p\ dX < e, we infer
that L consists of all the integrable functions.
Note. We basically verified here that the collection L satisfies properties (1), (2),
and (3) of Theorem 22.12. This guarantees that L coincides with the vector space
of all integrable functions.
Problem 22.9. Let g be an integrable function and let {/„} be a sequence of
integrable functions such that \fn\ < g a.e. holds for alln. Show that if fn -^ /,
then f is an integrable function and iim J\fn — f\dfi = 0 holds.
Solution. By Theorem 19.4, the sequence {/„} has a subsequence that converges
to / a.e., and so |/| < g a.e. Thus, by Theorem 22.6, the function / is integrable.
Assume that for some e > 0 there exists a subsequence {gn} of {/„} such
that J\gn — f\dfji > e. By Theorem 19.4, there exists a subsequence {hn} of
[gn} with hn —> f a.e. Now, note that the Lebesgue Dominated Convergence
Theorem implies 0 < e < f\hn — f\dfi —> 0, which is impossible. Hence,
lim/|/„-/|rf/z = 0.
Problem 22.10. Establish the following generalization of Theorem 22.9: If
{fn} is a sequence of integrable functions such that Y1T=\ / l/«l ^M < °°>tnen
Y^L\ fn defines an integrable function and
/'(E/«)^=f;/,/»^.
J n=\ n=lJ
Solution. By Theorem 22.9, the series g = X!^=i \fn\ defines an integrable
function and |]C«=i fn\ < g a.e. holds for each k. Since X^li fn is
convergent for almost all points, it follows from the Lebesgue Dominated Convergence
Theorem that Y1T=\ fn defines an integrable function and that
J n=\ «=1J
Problem 22.11. Let f be a positive (a.e.) measurable function, and let
^^eX^'-1 </(*)< 2''})
for each integer i. Show that f is integrable if and only if X^-oo^'*/ < oo.

Section 22: INTEGRABLE FUNCTIONS
179
Solution. Let E, = [x e X: T'] < f{x) < 2'' }, / = 0, ±1, ±2,... . For
each n let 0„ = YH=-n ^ X£, • Then, there exists some function g with (pn t g
a.e. Clearly, g is a measurable function and 0 < / < g a.e. holds.
Assume that / is integrable. Then, each <pn is a step function, and in view of
4>n < 2/ (why?), it follows that
E2lei = lim (pndfi <2 f dfj.
< 00.
On the other hand, if X^.^ 2'e,- < oo, then each </>„ is a step function, and
so g is integrable. Since 0 < / < g, Theorem 22.6 shows that / is also
integrable.
Problem 22.12. Let {/„} be a sequence of integrable functions satisfying 0 <
fn+\ < fn a-e- for each n. Then, show that fn I 0 a.e. holds if and only if
ffn dfl I 0.
Solution. Assume ffndfi 4- 0. Let /„ | / a.e.; clearly, / > 0 a.e. It follows
that // dfx = 0, and thus (by Theorem 22.7) / = 0 a.e.
Problem 22.13. Let f be an integrable function such that f{x) > 0 holds for
almost all x. If A is a measurable set such that fA f dfi — 0, then show that
H*{A) = 0.
Solution. Let A be a measurable set satisfying fAf d\x = 0. Next, consider
the set B = [x € A: f(x) < 0}, and note that, by our hypothesis, JjL*(B) — 0.
Also, for each n put
An = {xeA: /(*)>£}.
Then, A = (USLi ^/») u ^» ^^ fXA„ < fXA a.e. holds for each n. Thus,
0<^U)< f fdfi< f /rf/x = 0.
JAn Ja
which shows that n*(An) = 0 for each n. This easily implies fi*(A) = 0.
Problem 22.14. Let (X, <S, /z) be a finite measure space and let f:X-*1Rbe
an integrable function satisfying f{x) > 0 for almost all x. If 0 < s < /x*(X),

180
Chapter 4: THE LEBESGUE INTEGRAL
then show that
inf! / fdfi: E e AM and fi*(E) > e\ > 0.
Solution. We can assume that f(x) > 0 holds for each x e X. If for some
0 < e < fi*(X) we have
inff / fdfi: E e AM and fi*(E) > cj = 0,
then there exists a sequence {En) of AM satisfying /x*(£„) > 6 and /£ f dfi< -^
for each /2. Put T7,, = U!u=n ^* an<^ note mat:
• etfc/z F„ is measurable;
• ^n+i ^ ^« holds for each n; and
• /x*(F„) > fi*(En) > £ holds for each n.
If F = n^Li ^n> men ^ is a measurable set and (by Theorem 15.4(2)) we have
fi*(F) = lim /x*(F„) > e > 0. (*)
From /xf„ < £^„ /X£<> we infer mat
/OO /» oo
and so / f XFn I 0- The latter implies fxF„ I 0 a.e. (see Problem 22.12), and
since fxF„ I /Xf, we infer that fxF = 0 a.e. In view of f(x) > 0 for each a',
the latter (in view of Problem 22.13) implies /x*(F) = 0, contrary to (•), and the
desired conclusion follows.
Problem 22.15. Let f be a positive integrable function. Define v: A -> [0, oo)
by v(A) = fAf dfi for each A € A. Show that
a. (X, A, v) is a measure space.
b. If Av denotes the a-algebra of all v-measurable subsets of X, then show
that A C Au. Give an example for which A^AV.
c. // fji*({x e X: f(x) = 0}) = 0, then show that A = Av.
d. Ifg is an integrable function with respect to the measure space (X, A, v),

Section 22: INTEGRABLE FUNCTIONS
181
then show that fg is integrable with respect to the measure space (X, S, p),
and that
/ gdv ='/ gf dp.
Solution, (a) This part follows immediately from Problem 22.5.
(b) The measure v has initial domain A. Hence, by Theorem 15.3, A C Aw
holds.
Consider R with the Lebesgue measure, and let / = Xu,2). Since, in this
case, u([0, 1]) =0, it follows that every subset of [0, 1] is a v-null set (and
hence v-measurable). On the other hand, not every subset of [0, 1] is Lebesgue
measurable. Thus, A ^ Au holds in this case.
(c) First, observe that v is a finite measure. Now, let A € Av with v*(A) = 0.
By Theorem 15.11, there exists some B e A such that A C B and v(B) = 0.
The relation fBf dp = v(B) = 0 combined with Problem 22.13, shows that
p*{A) = 0. Thus, A e A. Now, if V e Au, then pick some W e A with
V c W and v(W) = v*{V) (Theorem 15.11). Note that v*(W \ V) = 0, and so,
by the preceding discussion, W \ V e A. Finally, V = W \{W\V) e A holds,
which shows that A = Av.
(d) We shall assume g(x) > 0 and f(x) > 0 for all x. Pick a sequence {</>„} of
v-step functions such that 0 < 0„ t g v-a.e. Let
G = {xe X: f{x) > 0).
Clearly, G e A. Since Gc = [x e X: f{x) = 0), it follows that v(Gc) =
fGQf dfi = 0. Since / is strictly positive on G, the arguments of part (c) show
that whenever A c G, we have:
1) If A e Aw, then A 6 A, and
2) By Problem 22.13, u*(A) = 0 if and only if fji*(A) = 0 (and in this case
A g A).
In particular, it follows that 0,,/ t /g M-a.e. holds. Now, if A e Av satisfies
v*(A) < oo, then
[XAdv = v(AnG)= f fdfi= fxAfdfji.
J Jadg J
This implies that if 0 is a v-step function, then (f>f is /x-integrable, and that
f<f>dv = J4>f dp. holds. Now, note that 0 < 0,,/ f fg /x-a.e. and f4>ndv =
f(f)nf dp,, show that /g is /x-integrable and that fg dv = fgf dp holds.

182
Chapter 4: THE LEBESGUE INTEGRAL
Problem 22.16. Let I be an interval 0/R, and let f: I -> R be an integrable
function with respect to the Lebesgue measure. For a pair of real numbers a and
b with a ^ 0, let J = {(x - b)/a: x e I}. Show that the function g:J->R
defined by g(x) = f(ax-\-b)forx e J is integrable and that fjfd\ = \a\ JjgdX
holds.
Solution. Assume first that / = xa for some measurable set A c /. Clearly,
^{A — b) C J. Thus, in view of the identity, XAicix + b) = x±{A-b)(x)> it follows
from Problem 15.5 that
JjgdX = ±X(A)=±JifdX.
Thus, the formula is true for the characteristic function of a measurable set. It
follows that it is also true for step functions.
Now, let / be an upper function. Choose a sequence [<pn] of step functions with
(p„ f / a.e. on /. If irn(x) = <Pn(a* + b) for x e /, then \f/n is a step function
on J and \/rn f g a.e. holds on J. (Note that if B C / satisfies \(B) = 0, then
by Problem 15.5, we have X(±(B - b)) = ^HB) = 0.) Therefore,
\a\ \ gdX = \a\ lim fifndk = lim f^„dk = //dX.
Thus, the formula holds true for every integrable function / on /.
Problem 22.17. Let (X, 5, /x) &e <z yzwte measure space. For every pair of
measurable functions f and g let
a. S/iovv that (M, d) w a metric space.
b. S/zovv that a sequence {/„} of measurable functions (i.e., {/„} C Af) w/w-
fies fn-^f if and only if lim </(/„, /) = 0.
c. Show that (M,d) is a complete metric space. That is, show that if a
sequence {/„} of measurable functions satisfies d(fn, fm) —> 0 as n,
m —► oo, //ze/z r/zere ex/ste 0 measurable function f such that \\md(fn, f)
= 0.
Solution, (a) We assume that functions equal /z-a.e. are considered identical.
Only the triangle inequality needs verification. To this end, let /, g, h € M. The

Section 22: INTEGRABLE FUNCTIONS
183
triangle inequality follows immediately from the inequality
\f(x)-g(x)\ ^ \f(x)-h(x)\ \h(x)-g(x)\
1 + l/U) - g{x)\ ~ 1 + |/(x) - k(x)\ 1 4- \h(x) - g(x)\''
For details see the solution of Problem 9.11.
(b) Start by observing that for x > 0 and £ > 0 we have:
Now, assume that lim^C/",,, /) = 0 holds. Then, the inequality
/z*({* e X: \fa(x)-f(x)\>e)) = n*({x e X: 1^^)|>1fe})
< ]-¥-d(f„,f)
easily implies that /„ —» /.
For the converse, assume /„ —> f. If \\md{f,u f) ^ 0, then there exists
some £ > 0 and some subsequence {g„} of {/„} with d(gn, f) > £ for all n.
By passing to a subsequence, we can assume that g„ —> f a.e. (Theorem 19.4).
In view of {+?~!}fi < 1 and the finiteness of the measure space, the Lebesgue
Dominated Convergence Theorem yields 0 < £ < lim<i(g„, /) = 0, which is
absurd. Hence, \imd(fn, /) = 0 holds.
(c) Assume d(fn, fm) —► 0 as n, m —> oo. The inequality
H*({x 6 X: \fn(x) - fm(x)\ > e}) < ^ • d(fni fm)
shows that {/„} is a /x-Cauchy sequence. Thus, by Problem 19.7, /„ —> f
holds for some /, and by (b) above, limd(fn, /) = 0 also holds.
Conversely, if lim<i(//M /) = 0 holds, then by part (b) above fn —> /, which
implies that {/„} is a /z-Cauchy sequence.
Problem 22.18. Let /: R -> R be a Lebesgue integrable function. For each
finite interval I let ft = ^ fjf dX and Et — [x e I: f(x) > //}. Show
that
j\f~fi\d\ = lj{f-fi)d\.

184
Chapter 4: THE LEBESGUE INTEGRAL
Solution. We follow the notation of the problem. Start by observing that
/ (/ - f,)dk + f (/ - /,)dX = /(/ - /,) dX
J Ei JI\E, JI
= jfdX- ff,dX
= ffdX- ffdX = 0.
Consequently,
[ {f-fi)dk= f (fi-f)dk.
JE, Jl\E,
Now, note that
f\f-f,\dX=[\f-fl\dX+f \f-f,\dX
JI J Ei Jl\Ei
= I {f-f,)dX+ f (f,-f)dX
JE, Jl\Ef
= I' (f-fi)dk+ [ (f- /,) dX = 2 f (f - //) dX.
JE, JE, JE,
Problem 22.19. Let f: [0, oo) —> R be a Lebesgue integrable function such
that fl)f(x)dX(x) = 0 for each t > 0. Show that f(x) = 0 holds for almost
allx.
Solution. Start by observing that
f fdk= f fdk- f fdk = 0
J[a,b) J[0,b) J[0,a)
holds for each interval [a, b). By Problem 22.5, we see that fAfdX = 0 holds
for each o-set A. From Problem 15.2 (and the Lebesgue Dominated Convergence
Theorem), we see that fAfdk = 0 holds for each Lebesgue measurable subset
A of R.

Section 22: INTEGRABLE FUNCTIONS
185
Now, let X = {x e R: f(x) > 0} and Y = {x e X: f(x) < 0). Clearly, X
and K are both Lebesgue measurable sets, and
I,"*-/,
fdX = 0.
Y
Now, invoke Problem 22.13 to obtain X(X) = A.(7) = 0. Therefore, f(x) = 0
holds for almost all a\
Problem 22.20. Lef (X, 5, /z) be a measure space and let /, f\, /2,... be /20«-
negative integrable functions such that fn —> f a.e. and lim//„ dfi = /'/' d\x.
If E is a measurable set, then show that
lim f„dfj.= / fdfM.
n-+°°JE Je
Solution. Assume that the integrable functions /, f\, fi,... are non-negative
satisfying the hypotheses of the problem and let E be a measurable set. Then the
functions /xe» f\XE, fiXE, ... are non-negative and integrable (because 0 <
fXE < / and 0 < f„XE < /,,) and f„XE —> fXE holds. Using Fatou's
Lemma, we see that
/ fd(i= / liminf/,,x£d/z<liminf / f„XEdfJi = \immf I fnd\x. (•)
Similarly, we have
/ f dfi < lim inf / /,, dfji. (•-*)
Jec Jec
Therefore,
// d[i— \ f d\i + I f dfi < lim inf / fn d\x 4- lim inf / fn d\i
Je Jec Je Jec
< lim inf ( / f„ d[A+ I fn dfi J
= lim inf / /„
E
d\x

186
Chapter 4: THE LEBESGUE INTEGRAL
where the second inequality holds by virtue of Problem 4.7(b). It follows that
I f dn+ I f dfM = lim inf / /„ d/i + lim inf / /„ dfi,
Je Jec Je Jec
and from (•) and (••), we see that
lim inf I fndfj,= I f dfi.
Now, let [gn] be a subsequence of {/„}. Then,
gn —> f a.e. and JKm^ gndfi = / f dfi.
By the preceding conclusion, we infer that
lim inf gndfi = / f dfi,
and so there exists a subsequence {g^} of the sequence {gn} such that
^mfE8kadfi = fEfdV>-
Thus, we have demonstrated that every subsequence of the bounded sequence
of real numbers [fEfndfi} has a convergent subsequence to fEf dfi. This means
that
lim I f„dfji= f dfi
"-00 Je Je
holds; see Problem 4.2.
Problem 22.21. If a Lebesgue integrable function /: [0, 1] -> R satisfies
J^x2nf{x) dX{x) = 0 for each n = 0,1, 2,..., then show that / = 0 a.e.
Solution. Let an integrable function /: [0, 1] —► R satisfy
/ x2nf(x)dk(x) = 0 for n = 0, 1, 2,....
Jo
Since the algebra of functions generated by {1,jc2} is uniformly dense in C[0, 1]
(see Problem 11.5), it follows that f*g(x)f(x)dk(x) = 0 holds for all g in

Section 22: INTEGRABLE FUNCTIONS
187
C[0, 1]. Consider the two measurable sets
E = {x € [0, 1]: f(x) > 0} and F = [x € [0, 1]: f(x) < 0}.
We have to show that X(E) = X(F) = 0. We shall establish that X(E) = 0 holds
and leave the identical arguments for F to the reader.
Pick a sequence {Kn} of compact sets and a sequence [On] of open sets
of [0, 1] satisfying Kn C E c On for each n, Kn t, On I, and X(£) =
YimX(Kn) = liml((9n). (Here we use the regulantyoftheLebesgue measure.) For
each n there exists (by Theorem 10.8) a continuous function gn:[Q, 1] —> [0, 1]
satisfying gn(x) = 1 for each * e /£„ and f(x) = 0 for each „r £ <9„. Clearly,
lg«/l < l/l and g„/ —► fxE a.e. By the Lebesgue Dominated Convergence
Theorem, we get
lim f gn(x)f(x)dX(x) = [ f(x)XEMdX(x)= [ f dX.
n-+°° Jo Jo Je
Taking into account that f0gn(x)f(x)dX(x) = 0 holds for all n, we infer that
fEf dX = 0. Now, invoke Problem 22.13 to infer that X{E) = 0, as claimed.
Problem 22.22. For each n consider the partition
{0, 2"\ 2 • 2~\ 3 • 2~\ ..., (2" - 1) • 2"\ 1}
o/r/ie interval [0, 1] and define the function rn: [0, 1] -> F. fry/'„(1) = —1 tfwi
/•„(*) = (-1)*"1 /or (k - \)2~n < x < kT~n {k = 1, 2,..., 2").
a. Draw r/ze graphs ofr\ and ri.
b. 5/zow that if f: [0, 1] -» R /j a Lebesgue integrable function, then
lim / /•nU)/(^)dAU) = 0.
,'->0° Jo
Solution, (a) The graphs of r\ and r2 are shown in Figure 4.1.
(b) By Theorem 22.12, it suffices (how?) to establish the claim for the case
/ = X[a,b)> where [a, b) is a subinterval of [0, 1]. Clearly, fQ rn(x)x[a,b) dX{x) —
f rn(x) dx. Therefore, it suffices to show that lim fa rn{x) dx = 0 holds for each
0°< a < b < 1.
To this end, fix 0 < a < b < 1 and let s > 0. Fix n0 such that 2"rt° <
min{£, ^p}. Pick n > no and consider the partition {0, ^,|,..., ^p-, l}; for

188
Chapter 4: THE LEBESGUE INTEGRAL
i
y
i •
i,
i
y = rx (x)
1
i 1 x
y = r2 (x)
> i *
FIGURE 4.1. The Graphs of n and r2
simplicity, let jc/ = ^r and note that the points a and b are related to the a*, as
shown in Figure 4.2.
Since for any three consecutive points a*/-i, a:/, a;+i we have j^'_ /•„(*)</* =
0, we see that fgr„(x)dx = fgLr„(x)dx + fcr„(x)dx, where c = Am_i or
c = Am; see Figure 4.2. Hence,
|J rn(x)d*| < jXL\rn(x)\dx + j \rn{x)\dx
= (a* - a) + (/? - c) < e + 2e = 3e
for each /7 > /20. This means that lim fa r„(x) dx = 0, as desired.
Problem 22.23. Let {en} be a sequence of real numbers such that 0 < en < 1
for each n. Also, let us say that a sequence [An] ofLebesgue measurable subsets
of[0, 1] is consistent with the sequence {e„} // X(An) = €n for each n. Establish
the following properties of [€n} :
a. The sequence [en} converges to zero if and only if there exists a consistent
sequence [An] of measurable subsets of [0, 1] such that Y1T=]Xa„M < co
for almost all x.
b. The series Yln*L\€" converges in JR if and only if for each consistent
sequence {An} of measurable subsets of [0, 1] wehaveY^L\XA,M') < oo for
almost all x.
Solution, (a) If en —> 0, then let A„ = (0, en) (n = 1,2,...), and note that
X(An) = en holds for each n and that Y1T=\ Xa„M < oo for each x e [0, 1].
• • •
xk-\a xk
■*-•-
*Ar + 1
xm — 1 xn\b xm-\-\
FIGURE 4.2.

Section 22: INTEGRABLE FUNCTIONS
189
For the converse, assume that there exists a consistent sequence of measurable
subsets (An] of [0, 1] satisfying Y1T=\ Xa„M < °° for almost all x. For each
n let Bn = |Ja1w Ak and note that Bn 15 = f|^i **• If M#) > ° holds>
then note that J2T=\ Xa„M = °° for each x e B (why?), which contradicts our
hypothesis. Thus, k(B) = 0. From the continuity of the measure (Theorem 15.4),
we see that k(B„) I 0. In view of An C 5„, we have 0 < en = k(A„) < k(Bn)
for each n, and so lim en = 0.
(b) Assume X!^=i e« < °° and that [An] is a consistent sequence of measurable
subsets of [0, 1]. Then,
OO p CO CO
Y] / XAndk = Y] A.(i4n) = V] £n < CO,
and so, by the series version of Levi's Theorem 22.9, we have Y1T=\ Xa„M < °°
for almost all x.
For the converse, assume that for every consistent sequence {An} of measurable
subsets of [0, 1], we have Y1T=\ Xa„M < oo for almost all x. Suppose by
way of contradiction that Y2T=\ e" = °°- Using an inductive argument (how?),
we see that there is a sequence [kn] of strictly increasing natural numbers such
that Y^i=k +\£i > * holds for each n. Next, for each n we can choose (how?)
subintervals >U„+i, /Un+2,..., A;n+1 of [0, 1] such that k(Ai) = e,- for kn + 1 <
/ < kn+\ and U/l^+i^' = [0, 1]. Now, note that the sequence of measurable sets
{An} is consistent with {£„} and J2T=\ Xa„M = oo holds for each x e [0, 1],
contrary to our hypothesis. So, Y1T=\ £n < °° must hold.
Problem 22.24. Ler (X, <S, /x) fee a finite measure space and let f:X->Kbea
measurable function.
a. Show that if fn is integrable for each n and that lim ffn d\x exists in R,
then |/C*)| < 1 holds for almost all x.
b. // /" is integrable for each n, then show that f fn dfi = c (a constant)
for n = 1,2,... if and only if f = xa for some measurable subset A
ofX.
Solution. Keep in mind that fn denotes the function fn: X —> R defined by
f"W = IfWY for each x e X.
(a) Assume that fn is Lebesgue integrable for each n and that \\mffndii exists
in R. Assume by way of contradiction that the measurable set
E = {xeX:\f{x)\> 1}
satisfies fJi*{E) > 0. From the identity E = (JHLi £*» where
EA = {*eX:|/(.t)|>l + i},
we see that there exists some <5 > 1 such that the measurable set

190
Chapter 4: THE LEBESGUE INTEGRAL
F = [x e X: \f(x)\ > 8} satisfies fi*(F) > 0. Now, note f2n > 82nXF holds for
each /?, and so from
82"ix*(F) = J82nXFdvL<ff2ndfjL,
we infer that lim ff2n dfi = oo, contradicting the existence in IR of the limit
\imffndp,. Hence, |/(jc)| < 1 must hold for almost all x.
(b) Assume ff" dfi = c holds for each n = 1,2,.... By part (a), we know
that |/(a)| < 1 holds for almost all A' e X. Now, define the sets A = [x e
X: f(x) = 1), B = {a- € X: f(x) = -l},and C = {jc € X: \f(x)\ < 1}. Then,
for each n we have
Jfndfi = j fndfi + f fndfi + f fndn
= [ldti+ f(-\)ndiJL+ f fndn
J A JB JC
= il*(A) + (-1)V(*) + j fndfi = c.
Since fn{x) —> 0 holds foreach a € C, it follows from the Lebesgue Dominated
Convergence Theorem that lim fcfn d\i = 0. Hence,
lim [/x*(A) + (-l)V0B)]=c.
Since lim(—1)" does not exist, we infer that fi*(B) = 0, and therefore, c =
H*(A) = iA*(A) + fcfn dfji foreach n. In particular, we have fc f2 dfi = 0, and
so /(a) = 0 must hold for almost all x eC. The latter implies that f = Xa a.e.
holds.
23. THE RBEMANN INTEGRAL AS A LEBESGUE INTEGRAL
Problem 23.1. Lef /: [a, fe] -> IR fee Riemann integrable. Show that f is
Riemann integrable on every closed subintemal of [a, fe]. Also, show that
J f(x)dx=: ^ f(x)dx + j f(X)dx
holds for every three points c, d, and e of [a, fe].

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 191
Solution. Let /: [a, b] —► R be a Riemann integrable function and let [u, v]
be a closed subinterval of [a, b]. If /: [u, v] —> R is discontinuous at some
point x e [u, v], then /: [a, b] —> R is also discontinuous at the point x—note
that, in this case, there exists a sequence {xn} of [u, v] (and hence of [a, b])
such that [f(xn)} does not converge to f(x). Thus, the set D of all points of
discontinuity of /:[«, v] —> R is a subset of the set of all points of discontinuity
of /: [a, b] —> R. Since /: [a, b] —> R is Riemann integrable, we know that
\(D) = 0, and so (by Theorem 23.7) the function /: [u, v] —> R is Riemann
integrable.
Now, assume that a < c < e < d < b. Since /: [c, e] —> R is Riemann
integrable (and hence Lebesgue integrable), there exists a sequence of step functions
{0„} over [c, e] (i.e., <pn{x) = 0 holds for x £ [c, e]) with <pn(x) f fix) for
almost all x e [cye]. Similarly, there exists a sequence of step functions {^n} over
[e, d] such that i/n(x) f fix) holds for almost all x e [e, d]. Then, (0„ -f- ^n}
is a sequence of step functions over [c, d] satisfying (pn(x) -f i/n(x) t fix) for
almost all x e [c, d]. Therefore,
f fix)dx = f fdk= lim [ (</>„ +1rn)d\
Jc J[c.d} n-+°° J[c,d)
= lim / 0„ dX + lim / \//n dX
n-+°° J[c,d) n^°° J[c.d]
= [ fdX+ [ fdX
J[c,e) J[e,d]
= j6fix)dx + j fix)dx.
Now, the equality fc fix)dx = f* fix)dx + fe fix)dx for arbitrary elements
c, d, and e of [a,b] can be obtained by considering all possible cases. We prove
it for one such case and leave the rest for the reader. Assume that a < e < c <
d < b. Then, by the preceding case, we have
f fix)dx= [Cf(x)dx+ f fix)dx = - fef{x)dx+ f f{x)dx,
Je Je Jc Jc Jc
from which it follows that f* fix)dx 4- fe fix)dx = fc fix)dx.
Problem 23.2. Let f:[ayb]-+1Rbe Riemann integrable. Then, show that
fix)dx= lim > /(a + ).
i
Ja

192
Chapter 4: THE LEBESGUE INTEGRAL
Solution. The conclusion follows from Theorem 23.5 by observing that
!
1 = 1
where the partition Pn = {a*o, ai, ..., x„] and T = {/i,...,**} satisfy A; =
a + J'^r (0 < / < w) and ry = jc, (1 < / < n).
Problem 23.3. Let {fn)bea sequence of Riemann integrable functions on [a, b]
such that {/„} converges uniformly to a function f. Show that f is Riemann
integrable and that
lim [ f„{x)dx = f f(x)dx.
Solution. Choose some k such that I/* (a) — /„(a)| < 1 holds for all n > k
and all x e [a, b]. Thus, \fk(x) - /(a)| < 1 holds for all x e [a, b]. Since fk
is bounded, it is easy to see that there exists spme M > 0 such that |/(a)| < M
holds for all x e [a,b].
If Dn £ [#» b] denotes the set of discontinuities of /„, then (by Theorem 23.7)
D = U^li Dn satisfies \(D) = 0. Since each fn is continuous on [a, b] \ D, it
follows from Theorem 9.2 that / is continuous on [a, b] \ D, i.e., / is continuous
almost everywhere. By Theorem 23.7, / is Riemann integrable.
For the last part, let € > 0. Pick some k such that \fn(x) — f(x)\ < e for all
n > k and all x e [a, b]. So, for n > k we have
\f fnWdx-J f(x)dx\<J \ftt{x)-f(x)\dx<€{b-a),
and this shows that lim,,-^ fa fn(x)dx = fa f(x)dx.
Problem 23.4. For each n, let fn(x): [0, 1] -» R be defined by fn(x) = ^
for all x € [0, 1]. Then, show that lim f0 fn(x)dx = ^.
Solution. Integrating by parts, we get
fo fnWdX = £ [+jf ^^ = i +fQ j^dx. (*)
Since 0 < ^—tj < 1 holds for all a* € [0, 1] and lim ~~y = 0 for each x in

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 193
[0, 1), the Lebesgue Dominated Convergence Theorem yields lim/0 *"s dx=0.
Thus, from (•), we see that lim/0 fn(x)dx = ^.
Problem 23.5. Let f:[a,b] -> R be an increasing function. Show that f is
Riemann integrable.
Solution. Let Pn = {x0, x\,..., xn) be the partition that subdivides [a, b] into
n subintervals of equal length ^. Since / is increasing, note that m,- = /U/_i)
and Mi = /(a*/) hold for each 1 < / < n. Next, observe that
0 < /*(/) - /*(/) < S*(f9 Pn) - W> Pn)
n
i = l
holds for each n. Thus, /*(/) — /*(/) = 0 and so / is Riemann integrable.
An alternate proof goes as follows: According to Problem 9.8 the set of
discontinuities of / is at-most countable—and hence, it has Lebesgue measure zero. Now,
Theorem 23.7 guarantees that / is Riemann integrable. (See also Problem 21.8.)
Problem 23.6 (The Fundamental Theorem of Calculus). Iff:[a,b]-+R is
a Riemann integrable fiinction, then define its area function A: [a, b] -> R by
A(x) = fj f(t)dt for each x e [a, b]. Show that
a. A is a uniformly continuous function.
b. /// is continuous at some point c of[a,b], then A is differentiable at c and
A'{c) = f(c) holds.
c. Give an example of a Riemann integrable function f whose area function
A is differentiable and satisfies A! ^ f.
Solution, (a) Choose some M > 0 with |/U)| < M for each x in [a, b]. The
uniform continuity of A follows from the inequalities
\A(x) - A(y)\ = \Jyf(t)dt | < \j}\m\ dt | < M\x - y\.
(b) Let / be continuous at some point c e [a,b] and let e > 0. Choose some
8 > 0 such that |/(a*)—/(c)| < e holds whenever x € [a, b] satisfies \x—c\ < 8.
Then, for a- € [fl, b] with 0 < |a" — c\ < 8 we have \f{t) — f(c)\ < € for all t in

194
Chapter 4: THE LEBESGUE INTEGRAL
the interval with endpoints x and c, and so
A(x) - A(c)
X — c
-/(c)
x - c
\x-c\
f(t)dt-f{c)(x-c)
[
f\fO) - /w]
dr
< F^re|Jc-c| = e.
This shows that i4'(c) exists and that A'(c) = /(c) holds,
(c) We consider the function /: R —> R of Problem 9.7 defined by /(*) = 0 if
x is irrational and f(x) = £ if x = ~ with h > 0 and with the integers m and h
without having any common factors other than ±1. It was proven in Problem 9.7
that / is continuous at every irrational and discontinuous at every rational. This
implies that / restricted on an arbitrary closed interval [c, d] is continuous almost
everywhere and / = 0 a.e. From Theorems 23.6 and 23.7, we infer that / is
Riemann integrable over [c, d] and fc /(jc) dx = f f dk = 0.
In particular, if [a, b] is any closed interval, then A(x) = f* /(/) dt = 0 for each
x e [a, b]. Thus, A'(x) = 0 for each x € [a, 6], and consequently A'{x) # f(x)
at each rational number x in [a, b].
Problem 23.7 (Arzela). Let [fn}bea sequence ofRiemann integrable functions
on [a, b] such that lim fn(x) = f{x) holds for each x e [a, b] and f is Riemann
integrable. Also, assume that there exists a constant M such that |/„(a-)| < M
holds for all x e [a, b] and all n. Show that
lim f fn{x)dx= f f(x)dx.
"-°° J a J a
Solution. Using Theorem 23.6 and the Lebesgue Dominated Convergence
Theorem, we see that
/ f(x)dx= f fd\= lim f fndk= lim [ fn(x)dx.
J a J[a,b) n^°°J[atb} n^°° J a
Problem 23.8. Determine the lower and upper Riemann integrals for the
function /: [0, 1] —> R definedby f(x) = 0 ifx is a rational number and f(x) = 1
// x is an irrational number.

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 195
Solution. Let P be a partition of [0, 1]. Since each interval contains rational and
irrational numbers, we have m,- = 0 and M, = 1 for all i. Thus, S*{f, P) = 1
and S*(/, P) = 0 for all partitions P. Therefore, /*(/) = 1 and /*(/) = 0.
Problem 23.9. Let C be the Cantor set {see Example 6.15). Show that xc is
Riemann integrable over [0, 1], and that f0 xc dx = 0.
Solution. Note that xc is continuous at every point of [0, 1] \C and
discontinuous at every point of C. Since X{C) = 0, it follows from Theorem 23.7 that
Xc is Riemann integrable over [0, 1]. Since xc = 0 a.e. holds, we see that
f xc(x)dx= [ XcdX = 0.
JO J [0.1]
Problem 23.10. Let 0 < € < 1, and consider the e-Cantor set C€ of [0, 1].
Show that xc( is not Riemann integrable over [0, 1]. Also, determine /*(xc<) and
/*(Xcf).
Solution. Consider the e-Cantor set for some 0 < s < 1. Since CE is nowhere
dense in [0, 1], it is easy to see that xcc is discontinuous at every point of Ce and
continuous at every point of [0, 1] \CE. Since X{CE) = s > 0, it follows from
Theorem 23.7 that xc€ is not Riemann integrable.
Now, let P = {xo, x\,..., xn) be a partition of [0, 1], Since C£ is nowhere
dense, it follows that mi = 0 for each 1 < / < n. Thus, S*(xcf» P) = 0 for
each partition P, and so /*(xcc) = 0. Clearly, M, = 1 if |>/_i,;c/] C\CE ^ 0,
and Mi = 0 if [jcf-_i, jc,-] H C£ = 0. Since CE = (J/Li [*/-i»xi] H Ce, it follows
that
n n
e = HCE) < Y,^Xi-\>xA nCs)<J2 M^Xi " *-i> = 5*(^' P)'
i=i /=i
and so /*(xcf) > ^-
On the other hand, if 0 < 8 < 1 — e, then there exist pairwise disjoint open
subintervals {a\, b\),..., {an, bn) such that
n
fa, 6,-] c [0, 1] \ Ce (1 < / < /i) and - ^(6/ - a,-) > 1 - e - 8.
/=i
The endpoints of all these subintervals together with 0 and 1 form a partition P

196
Chapter 4: THE LEBESGUE INTEGRAL
of [0, 1] such that
e < /*(xcf) < S*(Xc,, P) < 1 - (1 - s - 8) = e + 8.
Since 0 < 8 < 1 — e is arbitrary, it easily follows that I*(xce) = £•
Problem 23,11. Give a proof of the Riemann integrability of a continuous
function based upon its uniform continuity (Theorem 7.7).
Solution. Let e > 0. Since (by Theorem 7.7) / is uniformly continuous, there
is some 8 > 0 such that x, y e [a, b] and \x — y\ < 8 imply |/(a) — f(y)\ < s.
Let P be a partition of [a, b] with mesh \P\ < 8. Then, M\—mx < e holds for
each 1 < / < n (why?), and so
n
0 < /•(/) - /*(/) < £(Af, r mMxi - a-/-i) < <?(& - a).
Since e > 0 is arbitrary, we see that /*(/) = /*(/) holds, and therefore / is
Riemann integrable.
Problem 23.12. Establish the following change of variable formula for the
Riemann integral of continuous functions: If [a,b] —^-> [c, d] —> R are
continuous functions with g continuously differentiable (i.e., g has a continuous
derivative), then
nb ng(b)
/ f(g(x))g'(x)dx= / f(u)du.
Ja Jf>(a)
Solution. We shall apply the Fundamental Theorem of Calculus in connection
with the Chain Rule. We consider the two functions F, G: [a, b] —> R defined by
F(x)= f(u)du and G(x) = f(g(x))g,(x)dx
Jgia) Ja
for all x € [a, b]. Next, we shall compute the derivatives of F and G separately.
For the derivative of F we use the Fundamental Theorem of Calculus and the
Chain Rule to get F'(x) = f(g(x))g'(x) for each x € [a, b]. For the derivative of
G, the Fundamental Theorem of Calculus yields G'(x) = f(g(x))g'(x) for each
a- € [a, b]. So, F'(x) = G'(x) for all x e [a, b].
The latter implies that there exists a constant c such that F(a) = G(x) + c for
all x e [a, b]. Letting x = a and taking into account that F(a) = G(a) = 0,

Section 23: THE RDEMANN INTEGRAL AS A LEBESGUE INTEGRAL 197
we get c = 0. Thus, F(x) = G(x) for all x £ [a, b]. Finally, letting x — b, we
obtain
rb n<>(b)
/ f{g{x))g\x)dx = f(u)du,
Jo Jg{a)
as desired.
Problem 23.13. Let f: [0, oo) -* R fee a continuous function such that
limr—co f(x) = 8. Show that lim,,.^ /0° f(nx)dx = <2<5 /or e<2c/z a > 0.
Solution. Fix a > 0 and then define the sequence of continuous functions {/„}
by /„(*) = /(/z.v). Clearly, lim,,^/„(*) = 5 holds for all* e (0,a]. We
claim that the sequence of functions {/„} is uniformly bounded on the interval
[0, a]. Indeed, since lim^oo f(x) — 8 holds, there exists a number M > 0 such
that \f(x)\ < \8\ -f 1 whenever x > M\ Also, since / is a continuous function,
it is bounded on the interval [0, A/]. Thus, there exists a constant C such that
|/(.v)| < C holds for all *, and hence \fn(x)\ = \f(nx)\ < C holds for all jc.
Now, an application of the Lebesgue Dominated Convergence yields
pa pa pa
lim / f(nx)dx = lim / f„(x)dx — I 8dx — a8.
n-*xJo n-+°°Jo Jo
Problem 23.14. Let f: [0, oo) -> R be a real-valued continuous function such
that f(x + 1) = f(x) for all x > 0. If g: [0, 1] ->• R is an arbitrary continuous
function, then show that
lim^j g{x)f(nx)dx = (J g(x)dx) • (J /(*)d.t).
Solution. Let the functions / and g satisfy the hypotheses of the problem.
Observe first that an easy inductive argument establishes that f(x + k) = f(x)
holds for all .v > 0 and all non-negative integers k.
The change of variable u = nx yields
f g(x)f(nx)dx = I [ngCj;)f(u)du
Jo Jo
= *£/" *(*)/<«)*«■

198
Chapter 4: THE LEBESGUE INTEGRAL
Letting t = u — / + 1, we get
f gCi)f(u)du= [ g(!±L±)f« + i-l)dt = f g(i±±L)f(f)dt.
Ji-\ Jo Jo
Consequently,
f g(x)f(nx)dx = / [E TiSC-^^nOdt = f hn(f)dt% (*)
where hn(t) = [X)?=i w^^^/W* Clearly, /zn is a continuous function
defined on [0, 1]. In addition, note that if \g(x)\ < K and \f(x)\ < AT hold for
each x e [0, 1], then
\h„(t)\ <K2 for all r 6 [0, 1],
i.e., the sequence {hn} is uniformly bounded on [0, 1]. Next, note that 0 < / < 1
implies — < ^^^ < L. Thus, if ml and M? denote the minimum and
maximum values of g, respectively, on the closed interval f^, ^], then
< < gi1^) < M?
holds for each 0 < t < 1. Next, put
*. = E>? and 5» = E;<-
1=1 1=1
and note that Rn and Sn are two Riemann sums—the smallest and largest ones,
respectively—for the function g corresponding to the partition {0, £, \,...,
~^, 1}. Hence, limn_oo Rn = lim^oo Sn = f0 g{x)dx. From
|Ai,(0 - Rn ■ /(0| = \[J2 ^^)l/(0 - Rtt • /(/)
/=i
= ([tU(!±1r1)]-«n)-\m\
< (Sn - R„)\f(t)\,

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 199
we see that lim,,-^ h„(t) = f(t)fQg(x)dx—in fact, the sequence [h„]
converges uniformly (why?).
Now, use (*) and the Lebesgue Dominated Convergence Theorem to obtain
lim f g(x)f(nx)dx = lim f hn(t)dt
Ti-oo JQ n-oo JQ
= f [lim h„(t)]dt
Jo "-°°
= f [mf gwdx)]dt
Problem 23.15. Let f: [0, 1] -* [0, co) be Riemann integrable on every closed
subinterval 0/(0, 1]. Show that f is Lebesgue integrable over [0, 1] if and only
if \im€io f f(x)dx exists in R. Also, show that if this is the case, then we have
ffdX = \imnof€lf(x)dx.
Solution. Assume that / is Lebesgue integrable. Let {en} be an arbitrary
sequence of (0, 1] with en I 0. For each n, we consider the upper function
gn = fx[En,\]- Then, gn f / a.e. holds and so, by Theorem 21.6, we have
[fdk = lim [gndX= lim [ f{x)dx.
This shows that lim^o fe f(x) dx exists and that
lim f f(x)dx= [fdk.
no JE J
Conversely, assume that the limit exists. Let en = £ and consider the sequence
of upper functions [gn] as previously (i.e., let gn = /Xfo.i])- Then, gn t / a.e.
and
lim I gndk= lim / f(x)dx = lim / f{x)dx < oo.
n^oo J „_«, J^ Ei0 J£
By Theorem 21.6, / is an upper function, and hence, Lebesgue integrable.

200
Chapter 4: THE LEBESGUE INTEGRAL
Problem 23.16. As an application of the preceding problem, show that the
function f: [0, 1] -► 1R defined by f(x) = xP ifx e (0, 1] and /(0) = 0 is Lebesgue
integrable if and only if p > — 1. Also, show that if f is Lebesgue integrable,
then
f
fd\= l
1+P
Solution. If 0 < e < 1, then note that f£xp dx = Mpr for p # — 1 and
f£x~] dx = — lne. Thus, lime±o fe xp dx exists if and only if p > —1, and,
in this case, the limit is -~y. The conclusion now follows immediately from the
preceding problem.
Problem 23.17. Let f: [0, 1] -> R be a function and define g: [0, 1] -» R by
g(x) = e/(x).
a. Show that if f is measurable (or Borel measurable), then so is g.
b. /// is Lebesgue integrable, is then g necessarily Lebesgue integrable?
c. Give an example of an essentially unboundedfunction f which is continuous
on (0, 1] such that fn is Lebesgue integrable for each n = 1, 2,.... (A
function f is "essentially unbounded" if for each positive real number
M > 0 the set [x e [0, 1]: \f(x)\ > M) has positive measure.)
Solution, (a) Let h(x) = ex and note that g = h o /. The conclusion follows
from the identity (Ao/)"1(V) = /"^(g"1^)) and the fact that h is a continuous
function.
(b) The measurable function g need not be necessarily Lebesgue integrable.
Here is an example. Consider the function /: [0, 1] —► R defined by f(x) = 4^;
at x = 0 we let /(0) = 0. If 0 < e < 1, then the change of variable t = y/x
yields
f f(x)dx = f ^= = 2 f dt = 2(1 - Ji).
Therefore, from Problem 23.16, we see that / is Lebesgue integrable and / / dk =
2. On the other hand, for each 0 < € < 1 the change of variable u = 4= yields
AM
f g(x)dk(x)= f e^ dx = 2 f* $du >2 f* eu du = 2(e'* - l).
This implies \ime±ofe ^ dx = oo, and so by Problem 23.15 the function g is
not Lebesgue integrable over [0, 1].

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 201
(c) The function /: [0, 1] -» R defined by /(a) = In* for 0 < x < 1 and
/(0) = 0 satisfies /J fn(x)dX(x) = (-l)"/z! for each « = 1,2
Problem 23.18. Lef /: [0, 1] -> R be Lebesgue integrable (with respect to the
Lebesgue measure). Assume that f is dijferentiable at x = 0 and /(0) = 0. Show
that the function g: [0, 1] -> R defined by g(x) = x~*f(x)forx e (0, 1] <2/?d
g(0) = 0 is Lebesgue integrable.
Solution. Start by observing that (by Problem 23.16) the function h(x) = a"
for A" e (0, 1] is Lebesgue integrable over [0, 1], Since /(0) = 0 and /'(0)
exists, there exist 0 < <5 < 1 and M > 0 such that |/(.r)| < Mx for all
0 < a < 5. Since for 8 < x < 1 we have a"5 < <5~2, we can assume that
M > S~i. Now, note that for 0 < a < 1, we have
[ |/(a)| if <§ < a < 1
< M(/7 + |/|)(A).
Since h -f |/| is integrable and (obviously) g is measurable, Theorem 22.6
guarantees that g is also Lebesgue integrable.
Problem 23.19. Let f:[a,b]x [c\ d] —»» R be a continuous function. Show that
the Riemann integral of f can be computed with two iterated integrations. That
is, show that
If
f(x, y)dxdy = f [f /(*. y)dy ] dx = f [j f(x, y)dx ] dy.
Generalize this to a continuous function of n variables.
Solution. Note first that the functions
f(x,y)dy and y\—► / f(x,y)dx
are both continuous—and so both iterated integrals are well defined. Indeed, since
the function / is uniformly continuous, given e > 0 there exists some 8 > 0
such that |aj — a2| < 8 and \y\ — yi\ < 8 imply \f(x\% y\) — f{xit yi)\ < s.

202
Chapter 4: THE LEBESGUE INTEGRAL
Thus, if |a*i — ao| < <$ and \y\ — yi\ < 8 both hold, then
f Hxuy)dy-f
f(x\,y)dy- / f(x2,y)dy
< s(c — d)
and
/
b fib
f(x,yx)dx- I f(x,y2)dx
Ja
< e{b — a).
Let P = {a'o, aj ,..., A'„} be a partition of [a, b] and Q = [yo, y\,..., yn}
be a partition of [c, d]. Put /?,y = [a*;_i, a*;] x |jy_i, jy], and then define
niij = inf {/(a, y): (a, )>)e/?/y} and M/y = sup{/(A\ y): (a*, )>)€/?/y}.
From the inequality w;y < /(a*, )>) < M,y for (a, y) e /?,y, it follows that
mu(yj ~yj-\)< / /(*. y)dy < Mjjiyj - yj-\)
for all A/_i < a < a,, and so
rriijixi - Xi-\)(yj - yf-\) < / ( / /(a, y)dyj dx
< Mjj(Xi -A-/-i)()'y -^y-i).
Consequently, we have
n k
S*(f P x Q) = X^X^'A "" A'-l)(^ -^y-i)
i=l 7 = 1
n k /»\j /»y.
J a ^Jc
n k
i=i y=i
= 5*(/, P x Q).

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 203
Since P and Q are arbitrary and / is Riemann integrable, it follows that
f f f{x,y)dxdy = J (J f(x%y)dy)dx.
The other equality can be proven in a similar manner.
Problem 23.20. Assume that f:[a,b]-+1R and g: [a, b] -> R are two
continuous functions such that f(x) < g(x)for each x e [a, b]. Let
A = i(Xt y) e R2: x e [a, b] and f(x) <y< g{x)).
a. Show that A is a closed set—and hence, a measurable subset o/R2.
b. // h: A —> 1R is a continuous function, then show that h is Lebesgue
integrable over A and that
fhdX= f \f h(x,y)dy]d:
Solution, (a) Let {(*„, y,,)} be a sequence of A such that xn -> x and yn ->• y.
From the inequality f(xn) < yn < g(xn) and the continuity of / and g, it follows
that f(x) < y < g(x), i.e., (jc, y) e A. Thus, A is a closed set.
(b) Let c < inf{/Cr): x e [a, b]} and d > sup{g(„t): x e [a, b]}. Consequently,
^ £ [<3, b] x [c, d] = £. Extend h to £ by h(x, y) = 0 if (;c, y) £ ^4, and note
that the set of all discontinuities of h is a subset of
D = {(x, y) € R2: a <x <b and y = /(*) or y — g(x)).
By Problem 18.17, X(D) = 0, and so /z is Riemann integrable on E (and hence,
Lebesgue integrable). Now, by modifying the arguments of Problem 23.19, we
easily see that
J hdX =jl h(x,y)dxdy= f (f h{x,y)dy\dx
Problem 23.21. Let f: [a, b] -» R be a differentiate function—with one-sided
derivatives at the end points. If the derivative fr is uniformly bounded on [a,b],

204
Chapter 4: THE LEBESGUE INTEGRAL
then show that f is Lebesgue integrable and that
f'dk = f(b)-f(a).
L
Solution. Let /: IR —> IR be a differentiable function such that for some M > 0
we have |/'(a)| < M for all x e [a, b]. By letting /(a) = f(a) + f'(a)(x - a)
for x < a and f(x) = f(b) + f'(b)(x — b) for x > b, we can assume that / is
defined (and is differentiable) on IR.
Next, consider the sequence of differentiable functions {/„} defined by
fnW = n[f(x + I) - /(*)] = fiX + ^'f^\ x € Rf
and note that fn(x) —> f'(x) holds for each a* g IR. Also, by the Mean Value
Theorem, it is easy to see that |/n(x)| < M holds for each x. Consequently, by
the Lebesgue Dominated Convergence Theorem, /' is Lebesgue integrable over
[a, b] and
f f'dk= lim f f„(x)dx. (*)
Now, using the change of variable u = x -f- £, we see that
J /„(*)</* = /i[J f(x + i)^A - J /(a)</a-]
= *[/ °f(u)du- J f(x)dx]
= n[f " f(x)dx- j° nf(x)dx]
rb+
fT'-md* c-fwdx
fib) - f{a\
where the last limit is justified by Virtue of the Fundamental Theorem of
Calculus. A glance at (•) guarantees that /. „/'dk = f(b) — f(a), and we are
finished.

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 205
Problem 23.22. Let f,g:[a,b] —> R be two Lebesgue integrable functions
satisfying
[ f(t)dk(t)< f'g(t)dX(t)
J a J a
for all x e [a,b] If(p: [a, b] -» R is a non-negative decreasing function, then
show that the functions (pf and (j)g are both Lebesgue integrable over [a, b] and
that they satisfy
f 4>(t)f(t)dk(t)< f
J a J a
4>(t)f(t)dX(t)< / 4>(t)g(t)dX(t)
for all x e [a,b].
Solution. Since 0 is decreasing there exists some M > 0 satisfying |0(f)| < M
for each t e [a,b]. Since / and g are Lebesgue integrable, it follows from the
inequalities |0(/)/(OI < M\f{f)\ and \(p(t)g(t)\ < M\g(t)\ for each t e [a,b]
that (j)f and cpg are both Lebesgue integrable functions over [a, b].
To obtain the desired inequality, fix x e [a, b]. Assume first that 0 is a non-
negative decreasing function of the form 0 = Xw=i c/X|>,-_i.«/,•)» where {a = ao <
a\ < • • • < ai< = b] is a partition of [a, b]. Since / is decreasing, we know that
C| > c2 > • • • > a > 0. Clearly,
0 = (C\ - C2)X[aMi) + (C2 - C3)X[fltfl2) H 4- (Q-l - CjL)X[fl.<n_,) + ckX[a,b)
k
with yi > 0 for each /. Pick 1 < m < k such that am-\ < x < ami and note that
/ 4>{t)f(t)dx{t) = YJy' \ nodm+Ym no dm
J a ,-_j J a J a
m — \ pm n.x
< X> / 8(0dk(t) + Ym / g(t)d\(t)
/_! J a J a
= f <P(t)g(')dX(t).
J a
Now, we consider the general case. Fix x e [a, b). As in the solution of
Problem 21.8, we see that there exists a sequence {0„} of non-negative decreasing

206
Chapter 4: THE LEBESGUE INTEGRAL
step functions (as above) satisfying <pn(t) f 0(0 for almost all / € [a, b]. Since
IA,(0/(0I < M\f(t)l |0„(Og(OI < M\g(t)\, 0„(O/(O -> 0(0/(0, and since
0n(Og(O -* 0(Og(O for almost all t e [a, b], it follows from the inequality
^ 0n(0/(0^(0 < /" *„(0*(0^(0
J a J a
and the Lebesgue Dominated Convergence Theorem that
T 0(0/(0^(0= lim f 0,(0/(0^(0
< iim r <pn(t)g(t)dx(t)= r<Kt)g{t)dm.
"-°° J a J a
24. APPLICATIONS OF THE LEBESGUE INTEGRAL
Problem 24.1. Show that
/ xlne A Jjc = ——- • —-
h 22"/2! 2
holds for n = 0, 1, 2,
Solution. We shall establish the formula by induction on /?. For n = 0 the
formula is true by virtue of Theorem 24.6. If the formula is true for some n > 0,
then an integration by parts yields
rx*»+»e-*2 dx = _i rx^d(e-x2)
Jo Jo
Jo
for each r > 0. This implies
/•OO /T /»00
/ jc2(,,+,)c-jr2rfjc= lim / jt2(fl+,)*-*2djc = 2s±i / x2^-r2^
Jo r-^°°Jo Jo
— (2w+l)(2/i+2) (2/Q» -v/tF
~~ 22(/i+l) ' 22nn\ ' 2
__ f2(/i+D1! Vtt
~ 22<"+lKw+l)! ' 2 '

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL 207
Problem 24.2. Show that J™e~tx2 dx = \J± for each t > 0.
Solution. Let u = x^/t. Then, J^e~tx~ dx = -jj Jq *e~l(1 du holds for each
r > 0. Therefore,
J/»oo pr pry/7
f e'tx2dx= lim / e~tx2 dx = ± Vim / e~"2 du
o r-+°°Jo v'r->0°./o
— _l J* — ] r*
~ 4~i ' 2 ~2Vf
Problem 24.3. Show that f(x) = ~£ w Lebesgue integrable over [1, oo) and
thatffdk=l.
Solution. Since ^f > 0 holds for each x > 1, it suffices (in view of
Theorem 24.3) to show that /j°°^f dx exists.
If r > 1, then an integration by parts yields
fr^dx = - flnxd(\) = -llr[+ /'f djc = ! " f - T1-
Therefore,
//»oo pr
fdk= Hdx= lim / Hdx= iim(l-i-^) = 1.
J] x~ r-^°°J\ r" r-°° r r
Problem 24.4. Show that
lim [n(l + -)ne-2xdx = l.
n-*°°Jo v n/
Solution. Note that
poo pr
/ e~x dx = lim / e~x dx = lim (1 - e~r) = 1.
Therefore, the function e~x is Lebesgue integrable over [0, oo). Now, let
gn(x) = (l -f j;)"e~ZxX[o.n](x), and note that each gn is Lebesgue integrable
over [0, oo).

20S
Chapter 4: THE LEBESGUE INTEGRAL
From elementary calculus, we know that (l -f j;)" t e* f°r eacn x ^ 0» and
so £„(a) t e~x holds for each a* > 0. Thus,
lim
/l-*-00
/ (1 + ^) VA 4a = J|im / ft, dk= e~x dx = 1.
Problem 24.5. Let f: [0, oo) -> (0, oo) fee a continuous, decreasing, and
Lebesgue integrable function. Show that
1 f°° /(a* + t)
lim —-r / f(s)ds = 0 j/dwd only if lim —-—— = 0 for each t > 0.
A—oo f(x) Jx A—OO /(a-)
Solution. Assume that lim*.-+00 -77^ f™f(s)ds = 0 and let / > 0 be fixed.
Since / is decreasing, we see that /(a +t) < f(s) for all a < s < a: + f, and so
Hx+t)ds<] f(s)ds.
Consequently, we have
U ^ /(A) - / * /(X) - 7 * /(A) '
from which it follows that lim^oo ^f£r = 0. For the converse, assume
that limA_>oo ^f^r = 0 holds for each fixed r, and, for simplicity, let us write
F(a) = j^j J™f(s)ds for each x e [0, 00). Fix e > 0 and then choose
some 0 < <5 < 1 such that yz? < £. (Since lims_+o+ 737 = 0 such a 8 always
exists.) From lim^oo ^j^f- = 0, we infer that there exists some M > 0
such that ^0± < 8 holds for all x > M. That is, f(x + 8) < 8f(x) holds for
each x > M. Now, note that for x > M, we have
f^=w) r&ns)ds+m rmds
Jx Jx+8
/A+8 /»00
/x+8 /»oo
f(s)ds + -±-)Jx Sf(u)du
rx+8
7cT
x- J* f(s)ds + 8F(x).

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL 209
Consequently, if a* > M, then
(l-S)F(x)<jlr)jA f(s)ds<jlr)j'X f{x)ds = 8,
and so 0 < F{x) < y~ < e holds for all x > M. Thus, lirnr_*oo F(x) — 0.
Problem 24.6. Show that the improper Riemann integrals
<zos{xl)dx and / sin(A2)dA
nOO /»CO
/ cos(A*2)dA and / si
c/o Jo
{which are known as the Fresnel integrals) both exist. Also, show that cos(a2)
and sin(x2) are not Lebesgue integrable over [0, oo).
Solution. We shall work with /0°°sin(A2)<iA. Similar arguments will establish
the corresponding result for f£°cos{x2)dx.
Let 0 < s < t. The substitution u = a2 followed by an integration by parts
gives
|/W'2H = *|jf ^du\=^l-f^osud(u^)]\
^i[l + 7+£V*)]=}-
This inequality, combined with Theorem 24.1, guarantees the existence of the
improper Riemann integral /0°°sin(A2) dx. The inequality
Jy/kn—Tr Jkn—n Jkn—n
implies
/ |sin(*2)|dx = £ LhMli > * £ 7P
which shows that /0°°|sin(A*2)| dx does not exist in R—and hence, that sin(A2)
is not Lebesgue integrable over [0, oo).
Problem 24.7. Show that /0°° ^ dx = f.

210
Chapter 4: THE LEBESGUE INTEGRAL
Solution. Consider the function
1 if x = 0
/(*) =
A2
1
if x > 1 ,
and note that / is Lebesgue integrable over [0, oo). In view of the inequality
0 < —y^ < /(a), we see that the function ^^ is Lebesgue integrable over
[0, oo).
Now, note that for each r, e > 0, we have
r^dx = - rSin2A-tfQ)=--^r+ r*™*™* ^
__ sin2g sin2r • l sin a j .
r2r
JlE
Thus, by Theorem 24.8, we see that
nOO or nOQ
/ ™&dx= lim / anj±djt= / sai^. = *
Jo A" £§?./* ^ Jo A'
Problem 24.8. Let (X,S, n) be an arbitrary measure space, T a metric space,
and /:XxT->R a function. Assume that /(•, t) is a measurable function for
eacht € T and /(a, •) is a continuous function for each x e X. Assume also that
there exists an integrable function g such that for each t e T we have |/(x, /)| <
gWfor almost all x e X. Show that the function F:T -> R, defined by
F{t)= f f(X%t)dfJLMt
Jx
is a continuous function.
Solution. Let tn —> t in T. Define the function gn: X —► R by the formula
gn(x) = /(jc, tn). By our assumptions each gn is integrable, |g„| < g a.e., and
gnW —> f(x,t) holds for each x e X, Thus, by the Lebesgue Dominated
Convergence Theorem, we have
F(tn) = Jf(xtta)dfl(x) =.fgndfl —* ff(xj)dfJL(x) = F(/).
This shows that F is a continuous function.

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL 211
Problem 24.9. Show that
C°° e~x — e~xt
dx = In /
F
JO
/o
holds for each t > 0.
Solution. Consider the function f(x,t) = c *~c for .t > 0 and / > 0.
Observe that the value /(0, t) = / — 1 extends / continuously to the point (0, r),
/ > 0. Next, note that the function g(x, /), defined by
Q(x M= I 1/(^01 if0<*<l andr>0
^ % } \e'x + e~xt if x > 1 and / > 0 ,
is Lebesgue integrable for each t > 0. Moreover, |/(.r, t)\ < g{x, t) holds. This
implies that
poo
F{t)
rOO nOO
= / f(xj)dx= / Cl^ldx
Jo Jo
exists both as an improper Riemann integral and as a Lebesgue integral; see also
Theorem 24.3.
Next, note that §f(jc, t) = e~xt holds for all x > 0 and all t > 0. The
inequality 0 < e~xt < e~xa for all t > a > 0 and all x > 0, coupled with
Theorem 24.5, shows that
/»oo /»oo
F'(0= / |f(x,/)J.v= / e~«dx = \
Jo Jo
holds for all t > 0. Thus, F(t) = lnf + C. Since F(l) = 0, it follows that C = 0
and so
/•OO
F(r)= / ^^;c = lnf.
./o
Problem 24.10. For each t > 0, let F(t) = /0°° -^ d*.
a. S/zow that the integral exists as an improper Riemann integral and as a
Lebesgue integral.
b. Show that F has a second-order derivative and that F"(t) + F(t) = j
holds for each t > 0.

212
Chapter 4: THE LEBESGUE INTEGRAL
Solution, (a) The integrability of F follows from Theorem 24.3 and the
inequality
l+A-
^-
(b)If /(.v, 0 = -^, then
ff(*,0 = ^ and ^,,) = ^.
Since |ff(*,0| < e~" and |fj£(*,/)| < e~xl both hold, by applying
Theorem 24.5 twice we get
/•OO (-00
Consequently,
POO /»oo /»00
Problem 24.11. Show that the improper Riemann integral fQ2 \n(t cos x) dx
exists for each t > 0 and that it is also a Lebesgue integral. Also, show that
2L
jHln(fcos*)rf* = fln(0
holds for all t > 0.
Solution. Let f(x,t) = ln(fcosA') for 0 < jc < f, / > 0, and let g(x) =
(^ — A') 2 for 0 < x < |. An easy argument shows that the improper
Riemann integral (and hence, the Lebesgue integral) of g exists over [0, |).
Also, L'Hopital's Rule shows that lim^f-^^] =0. Thus, for each t > 0 there
exists some 0 < xq < £ such that |/(jc, /)| < g(x) holds for all xq < x < |.
Since /(a, /) is continuous for 0 < x < |, an easy application of Theorem 22.6
guarantees that
F(t)= I ~\n(tcosx)dx
-i:

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL
213
exists both as a Lebesgue and as an improper Riemann integral. Next, note that
Bl
Theorem 24.5, we have
Bl(x,t) = }, and that for 0 < a < t we have |§f(*,/)| < £. Thus, by
r'(0 = [2%(x,t)dx = %
Jo
for each / > 0, and therefore F(t) = § In t + C.
Since /0: \n(cosx)dx = /02 ln(sin.r)cf;c (why?), it follows that
2C =2F(1) = 2j2\n(cosx)dx
Jo
= / ln(cos x) d.v 4- / ln(sinA*)<i.Y
Jo Jo
= /TIn(^)
JO
-r
d-r
ln(sin2x)d-v- fin 2
ln(sin,x)cfjc — f In 2
C-§ In 2
Thus, C = — f In 2, and so
f'\n(tcosx)dx = f In/ - |ln2= f ln(4)
holds for each f > 0.
Problem 24.12. Show that for each t > 0 the improper Riemann integral
fo° va+v2) d* exists as a Lebesgue integral and that
Solution. For each f e R let F(t) = /0°° -^^ dx. From
I sinA7 I \xt\ |r|
Lv(1+jc2)I ~ |jc(1+jc2)| = 1+X2'

214
Chapter 4: THE LEBESGUE INTEGRAL
we see that F is indeed a well defined real-valued function on R and that the
integral defining F exists both as a Lebesgue integral and as an improper Riemann
integral. Moreover, the relations
cos*/
1+A2
1
<
1+A2
d r sin a/ i cos at
in connection with Theorem 24.5 guarantee that F is a differentiable function and
holds for each t e R.
Since ^[f^y] = -xff anc^ ^e natura^ dominating function in the
inequality | —Y+ffl — T+I3 ls not Lebesgue integrable over [0, oo), we cannot use
Theorem 24.5 to conclude that
Jo l+x2
As a matter of fact, the identity
xsinxt a2 sin at sinA7 sinxt
1+A2 JC(1+A2) A A'(l+A2)
(**)
shows that, on one hand, the function a h> x*™*2' is not Lebesgue integrable over
[0, oo) for each t > 0 and, on the other hand, that
f00 xsinxt , f°° sinxt , f°° sinxt n
/ 1 Tdx= dx = - — -dx = --F{t)
Jo 1+*2 Jo * Jo *(1+A2 2
(t)
for each t > 0.
We shall establish the validity of (*) for each t > 0 using another method. For
each n, let
""TjfS
Xt A
—zdx.
Clearly, Gn(t) -*■ /0 f^ff jA- - F'(t) for each r e JR. Now, from Theorem 24.5

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL
215
and (••), we see that
s'mxt
. N [n xsmxt , fn smxt , fn
G'(0 = -/ Tdx = -\ dx + /
Jo 1+*2 Jo • x Jo x(l+x*)
= -/ dx + r-d-t,
Jo x J0 x(l+x2)
dx
and consequently for each t > 0, we have
lim G'n(t) = - T ^ dx + r -^- dx = -«- + Fit) = g(t).
»-°° Jo x Jo jt(1+jt2)
We claim that for each a > 0 the sequence of derivatives {G'n} converges uniformly
to the function g(t) = — y -f- F(t) on the open interval (a, oo). To see this, fix
a > 0 and let e > 0. Choose some xo > 1 such that
f00 smx i I f00 sinxt i f00 d.r
/ dx \ < € and / — dx \ < I
Js x \ \JS x{\+x*) \-Js 1+.t2
< e
for all s > Xo. Now, if we fix some natural number k satisfying k > xq and
ka > jcq, then for each n > k and all r > a, we have
ic;,(„-s<„i=ir^-r
smx/
x(l+;t2)
rf:c
<2e.
This shows that {G'J converges uniformly to the function g(t) = — y + ^(r).
Finally, using Problem 9.29, we get F"{f) = [Hindoo Gn(t)] = -\ + F(r), or
F"(f) — F(/) = — \ for each / > 0. Solving the differential equation, we obtain
F(t) = | 4- Ci*"' + c2e' for each f > 0.
Since F and F' are continuous at zero (why?), it follows from ^(0) = 0 and
F'(0) = /0°° y^r = f and the preceding formula of F(t) that c, = -f and
c2 = 0. Hence, F(f) = § (l - e~') for each t > 0.
Problem 24.13. The Gamma function for t > 0 is defined by an integral as
follows:
no
/»0O
= / .t'-'<rv
Jo
dx.

216
Chapter 4: THE LEBESGUE INTEGRAL
a. Show that the integral
f xf-xe~xdx= Jim f x^e'-'dx
JO (Z.Q+ J€
exists as an improper Riemann integral (and hence, as a Lebesgue integral).
b. Show that T(^) = ^n.
c. Show that T(r -f 1) = tT(t) holds for all t > 0, and use this conclusion
to establish T(n + 1) = n\ for n = 1,2,....
d. Show that V is differentiate at every t > 0 and that
/»00
r'(/)= / x'-le-x\nxdx
Jo
holds.
e. Show that V has derivatives of all order and that
r(,,)(f)= I xl-xe-xQnx)ndx
Jo
holds for n = 1, 2,... and all t > 0.
Solution, (a) Since xl~xe~x < x!~l holds for 0 < x < 1, it follows from
Problem 23.16 that f0 x'~]e~x dx exists both as an improper Riemann integral
and as a Lebesgue integral.
Now, for each fixed t > 0 we have limA_(X>*,~,£~- = 0- Thus, there exists
some M > 0 (depending upon t) such that 0 < a*'"1^ < M holds for all
a* > 1. Hence, x*~xe~x < Me~^ holds for each x > 1. This shows that
f™xt~xe~x Ja- exists both as an improper Riemann integral and as a Lebesgue
integral for each t > 0.
The preceding show that J™xl~xe~x dx exists both as an improper Riemann
integral and as a Lebesgue integral.
(b) Substitute u = a'2 to get
/»oo /»oo
r(i)=/ x-L2e~xdx=2 e-"2du = yfr.
Jo Jo
(c) Integrating by parts, we get
/»oo /»oo
V(t + 1) = / x'e'x dx = - xl d(e~x)
Jo Jo
o /»oo poo
+ / tx'-le-*dx = t / x'~}e-xdx
Jo Jo
= tr(t).

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL
217
Consequently, we see that
r(/i + \) = n\r{l) = n\ [ e'xdx = nl.
vo
(d)and(e). Note that £;(x'-[e-x) = xl-]e~x(\nx)n holds for all t > 0 and all
x > 0.
Now, let a < t < b be fixed and consider the continuous function h{x,t) =
•l^(xl~le~x) = xl~le~x(\nx)n, a < t < /?, x > 0. We claim that there exists a
Lebesgue integrable function g:(0, oo) —> (0, oo) such that \h{x, t)\ < g(x)
holds for all x > 0 and all a < t < b. If this is the case, then Theorem 24.5
allows us to "differentiate under the integral sign" and since 0 < a < b are
arbitrary this shows that T must have derivatives of all orders and that the desired
formulas hold. So, we must construct a positive Lebesgue integrable function g
over (0, oo) such that \h{xj)\ < g(x) holds for each a < t < b and each
x > 0.
Note that for .v > 1, we have 0 < x'~l < xb. Using L'Hopital's rule, we see
that
lim xbe'i (lnx)tt = lim 4 • lim ^ =0-0 = 0,
and so there exists some M > 0 such that ^"^(ln.r)" < M for all x > 1.
Therefore,
\h(x,t)\ < |xf"16"Jf(lnx)'l| < Me~i
holds for all x > 1 and all a < t < b.
For the rest of our discussion, we shall need two facts from calculus.
lim xa (In x)"=0 and f xa~[ (in x)n dx = i^i.
Both can be proven by induction. For this limit use induction and L'Hopital's rule
by observing that
lim xa \nx = lim ^ = lim -^ = 0
v-v0+ t-v0+ {x > t-*0+ a
and
lim .va(ln.v)"+1 = lim I2sg£l = a±i lim xa(\nx)n.

218
Chapter 4: THE LEBESGUE INTEGRAL
For the integral, use induction and take into account that
/ xa~l]nxdx = !- f \nxd(xa)=*-xa\nxf -± / xa'1 dx = -\
Jv a Jo- a lo+ aU *
and
flxa-l(\nx)n+ldx = i f\\nx)n+ld(xa)
Jo+ Jo+
= -s±l f]xa-l(lnx)ndx.
Jo+
Since either (In*)" > 0 holds for all x e (0, 1] or (ina)" < 0 holds for all
x e (0, 1], it follows that the function 0(;c) = xa~lQnx)n,x € (0, 1], is Lebesgue
integrable over (0, 1]. Now, let
, v } xa-]\\nx\n if 0<x < 1
I Me 2 if x > 1
and note that g is Lebesgue integrable over (0, oo). To finish the proof, notice
that
|A(*,0|<*(*)
holds for all x > 0 and all a < t < b.
Problem 24.14. Let f: [0, 1 ] —► R be a Lebesgue integrable function and define
the function F: [0, l]-+WLby F(t) = /J f(x) sin(x t)dk(x).
a. Show that the integral defining F exists and that F is a uniformly continuous
function.
b. Show that F has derivatives of all orders and that
?0-n)(t\ _ / iy / v2w
= (-Dfl / ■
Jo
and
Fun)(t) = (-l)n / xznf(x)sm(xt)dk(x)
)o
F(2n-I)(r) = (-ir / x2n-{f(x)cos(xt)d\(x)
Jo
/o
for n = 1,2,... and each t e [0, 1]

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL
219
c. Show that F = 0 {i.e., F(t) = Ofor all t e [0, 1]) if and only if f = 0
a.e.
Solution, (a) Note that for each fixed t € [0, 1] the function x f-> sin(.vr)
is continuous and hence, measurable. The inequality |/Cr)sin(;cr)| < |/U)|
guarantees that x »-> /(.r)sin(jc/) is integrable for each t e [0, 1]. So, F is a
well-defined function.
For the uniform continuity of F note that
\F(t)-F(s)\ = \f f(x)sm(xt)dX(x)~ f f{x)sin(xs)dX(x)\
Uo Jo '
/(jc)[sin(jcr) - sin(jcs)] dX(x)\
< f \f(x)\\sm(xt) - sm(xs)\dX(x)
Jo
< f \f(x)\\xt-xs\dX(x)
Jo
= [j \f(x)\dX(x)\\t-s\
holds for all 5, r e [0, 1].
(b) Consider the function of two variables h(x, t) = f(x) sin(A'f). Then an easy
inductive argument shows that for each n = 1, 2,... we have
^^ = (-l)"*2VU)sin(*0 and ^£4 = {-\)» x2n~x f {x) zos{xt)
for each t € [0,1] and almost all*. This implies \^prL\ < \xnf(x)\ = g„(jc)for
all t e [0, 1] and almost all x. Since g„ is Lebesgue integrable for each n, it easily
follows from Theorem 24.5 that we can "differentiate under the integral sign" and
get the desired formulas.
(c) Assume F(t) = 0 for each / € [0, 1]. Then F{2n\t) = 0 for all /i, and so
from (b) we get /J x2n f{x) s'm(xt) dX(x) = 0 for each n and all t e [0, 1]. Letting
r = l, we get
l
x2n[f(x)smx]dX(x) = 0
for each /?. Now, invoke Problem 22.21 to conclude that f(x) sin* = 0 for almost
all x. Since sinx > 0 for each 0 < x < 1, we easily infer that f(x) = 0 for
almost all x.
HI
1

220
Chapter 4: THE LEBESGUE INTEGRAL
25. APPROXIMATING INTEGRABLE FUNCTIONS
Problem 25.1. Let /: R —► R be a Lebesgue integrable function. Show that
lim f
t->ooJ
f(x)cos(xt)dX(x)
= lim /
t-*ooJ
f(x)sin(xt)dX(x) = 0.
Solution. By Theorem 25.2, it suffices to establish the result for the special case
/ = X[a.b)- So, let / = X[a,6)» where —oo < a < b < oo. In this case, for each
t > 0 we have
/
f (x)cos(x t)dX(x)
=\C
cos(a7) dx
_ I sin(A7) \x-b
sin(6r)-sin(fl/)
/
and so lim,.^ ff(x) cos(a7) dX(x) = 0 holds. In a similar fashion, we can show
that linwoo ff(x)sin(xt)dX(x) = 0.
Problem 25.2. A function f:0-+lR (where O is a nonempty open subset of
R") is said to be a C°°-function // / has continuous partial derivatives of all
orders.
a. Consider the function p: R -* R defined by p(x) = exp[-^zy] '/ I* I < 1
and p(x) = 0 // |jc| > 1. Then show that p is a C°°-function such that
Suppp = [-l,l].
b. For € > 0 and a e R show that the function f(x) = p(^-) is also a
C™-function with Supp / = [a — e, a -f e].
Solution, (a) We shall establish that pin)(l) exists for each n.
Start by observing that, by UHopitaTs Rule, lim,—oc tke~^ = 0 holds for
k = 0, 1, 2,... . Notice that if for 0 < x < 1 we let t = j^, then we have the
inequality
xme *—'
|U2-1)4(jr-1)
from which it follows that
(l-x)4+l
= /*+ltf"ift
lim
ljY<x>-i)*c*-i) -^
= lim^+,e"s/ =0 for k,m =0,1,2,
(*)
Now, by a simple induction argument, we see that for — 1 < x < 1 the
derivative p(n)(x) is a finite sum of terms of the form TJrrnr. Using (•) and another

Section 25: APPROXIMATING INTEGRABLE FUNCTIONS
221
simple inductive argument, we can also see that p(n)(l) = 0 holds for n =
1,2
(b) Note that: f(x) ^ 0 if and only if -1 < ^ < 1 if and only if a - e < x <
a 4- 6. Therefore, Supp / = [a — st a 4- e].
Problem 25.3. Let [a, b] be an interval, € > 0 such that a 4- € < b — €, and
p as in the previous exercise. Define h: R -> R by h(x) = /a p(LjL)dt for all
x e R. 77ze/? s/jow r/?ar
a. Supp /i c [a — 6, fr + e],
b. h(x) = c (a constant function) for all x € [a + €,b — e],
c. A is a C™-function and h("\x) = f*Jfep(Lzi)dt holds for all x e R,and
d. r/ze C°°-function f = fc/c satisfies 0 < /(a) < I for all x e R, /(a) = 1
/<?/•a//a E [a + e, £ - e], cwd/ |x[fl.ft) — f\dk < 4e.
Solution. For simplicity, let g.v:R —► R be the function defined by gx(t) =
p(^)^ndsoh(x) = fahgx(t)dt.
(a) By part (b) of the preceding problem, we know that Supp gx = [a* — £, a 4- £].
Thus, if a < t < b and x & [a — £, fr 4- e], then gv(f) = ^(^t1) = 0 (since
|^| > 1). This implies that h(x) = 0 holds for all x $ [a - e, b + e], so that
Supp/z c [a — e, b 4- £].
(b) If a -f ^ < -v < b — £, then SuppgK = [a — e, a* 4- £] and so
h(x) = / &(*)<// = f 'p^dt = e / p(!*)rfM = c> 0.
J a J.x—E J— \
(c) Since every partial derivative j^pi'-y-) is continuous, it must be bounded on
[a, b] (and hence, on R). Now, the desired conclusion follows from Theorem 24.5.
(d) Since Suppgv = [x — e, x 4- e] and gx is a positive function, it follows that
pb rb rx+e
0 < h{x) = / fa(0dr = / P{^)dt < / p(*=*)<fr = c
Ja Ja Jx—e
holds for all a. Thus, / = h/c satisfies 0 < /(a) < 1 for all x.
Finally, observe that
\X(a,b) — f\ < X{a-E,a+E) + X(b-E,b+s)
holds, and so /|X[fl.w - f\dk = f\X{a.b) - f\dk < 4e.
The graph of / is shown in Figure 4.3.

222
Chapter 4: THE LEBESGUE INTEGRAL
y=m
a + e
b-e b b+z
FIGURE 4.3.
Problem 25.4. Let /:IR ->• 1R be an integrable function with respect to the
Lebesgue measure, and let e > 0. Show that there exists a C™-function g such
thatf\f-g\dk<€.
Solution. Let /:1R —> JR be a Lebesgue integrable function and let e > 0. By
Theorem 25.2, there is a step function 0 = X7=i QX[a.-A) (with c, ^ 0 for each
/) such that J\f — 4>\dX < e. Now, by the preceding problem, for each i there
exists a C°°-function g/ with compact support such that f\X[a,,bi)~~8i\dk < j^r
Now, note that the C°°-function g = YH=\ci8i ^as compact support and satisfies
f\f~&\dk < J\f-<l>\dX + f\<p-g\dk
< £+ / E C'*&*■*»> ~£C'*'
^ i=I i=\
<e + ^2\c,\ / |xt«,A)-ft|^
4A
1=1
1=1
= e + e = 2e.
Problem 25.5. The purpose of this problem is to establish the following
general result. If f:JR.n -> R is an integrable function (with respect to the

Section 25: APPROXIMATING INTEGRABLE FUNCTIONS 223
Lebesgue measure) and € > 0, then there exists a C°°-function g such that
f\f-g\dk<€.
a. Letai < btfori = 1,..., n, and let I — ]~l/=i(fli» M- Chooser > Osuch
that a{ + € < b\ — e for each i. Use Problem 25.3 to select for each i a
C00-function f:K. -* R such that 0 < f(x) < 1 for all x, f(x) = 1 if
x e [a + €,bi-e],andSuppfi C [a{ -e, 6/+e]. Now, define h:Rn -» R
fry /z(.ri,..., *„) = n?=i /»'(■*/)• Then show that h is a C°°-function on R"
artd //jor
Ax/ - *l^ <2[n(6, -fl; +2€)-fj(6/ -fl/)l.
' 1=1 1=1
b. Lef /: R" -> R be Lebesgue integrable, and let e > 0. TTzefl z^e pa/*/ (a)
re show that there exists a C°°-function g with compact support such that
J\f-g\dk<€.
Solution, (a) Clearly, h is a C°°-function. Let A = fl/Lito ~~ e^i + £)»
£ = ri/=ife "+" f- *' + e)' anc* C = YYi=\(ai ~ e» ^/ ~~ £)• N°w» the desired
conclusion follows from the inequality
|X/ - h\ < (xa - Xb) + (xa - Xc)-
(b) Let / be an integrable function and let s > 0. Pick a step function of the
form (p = ]Tj=1 CiXii (where each /, is a finite open interval of R") such that
f\f — <t>\d\ < s. From part (a) it follows that there exists a C°°-function g
with compact support such that f \(j> — g\dX < e. Consequently, f\f — g\dX
< 2e.
Problem 25.6. Let fibe a regular Borel measure on R", / a [x-integrable
function,ande > 0. Show that there exists a C°°-function g such that f\f—g\dfx < €.
Solution. Let / = n?=i[fl/* ^'1 De a finite closed interval. Given 8 > 0, pick
e > 0 such that the closed interval / = f]/Li [fl/ "~2ef bx +2e] satisfies fi{J \ I) =
fi(J) — jjl(I) < 8. (This is always possible since Yl'Ui lai ~ \^i +1\ Ik /•) As in
Problem 25.5, there exists a C°°-function h: R" —> R such that 0 < h(x) < 1
for all x e R\ h(x) = 1 for x e /, and Supp/z c 7. Therefore, if / = x/,
then
y*|x/ - *| rfM = J (A - x/) ^M < J{XJ ~ Xi) dp. = ii(J) ~ M(/) < 5.

224
Chapter 4: THE LEBESGUE INTEGRAL
Thus, the desired result holds true for the characteristic functions of the finite
closed intervals.
Now, let / = n?=ifo' */) be finite- Since n?=itflii bi - *] t* '. ^ follows
that the approximation result is also true for the characteristic functions of sets
of the form n/=i[fl/» ^')- Since these sets form a semiring and every open set
is a (j-set (for this semiring), it is not difficult to see that the result is true for
the characteristic functions of open sets of finite measure. The regularity of /x
guarantees the validity of the approximation result for characteristic functions of
/z-measurable sets of finite /x-measure. This in turn implies that the result holds
true for /z-step functions. Finally, since for each /x-integrable function / and each
6 > 0, there exists some /z-step function <p with f\f — (p\dp < e, it follows
that the C°°-functions with compact support satisfy the desired approximation
property.
Problem 25.7. Let f: [a, b] -» R be a Lebesgue integrable function, and let
€ > 0. Show that there exists a polynomial p such that f\f — p\dk < e, where
the integral is considered, of course, over [a, b].
Solution. Let f:[a,b] —> R be integrable (over [ayb]), and let e > 0.
By Theorem 25.3 there exists a continuous function g: [a, b] —► R such that
/ 1/ ~ g\ d\ < e. Now, by Corollary 11.6, there exists a polynomial p such that
\g{x) — p(a*)| < £ holds for all x e [a, b]. Thus,
J\f-p\dX<j\f-g\d\ + j\g-p\dX<e + e(b-a) = e(l+b-a),
and our conclusion follows.
26. PRODUCT MEASURES AND ITERATED INTEGRALS
Problem 26.1. Let (X, <S, p.) and (Y, E, v) be two measure spaces, and assume
that A x B e A^ ® Av.
a. Show that p*(A) • v*(B) < (p x v)*(A x B).
b. Show that if p\A) • v*{B) ^ 0, then (p x v)*(A xB) = fi*(A) • v*(B).
c. Give an example for which (p x v)*(A xB)^ P*(A) * v*(B).
Solution, (a) We have 5 ® £ c .A^ ® Av. Let Ax5 gAm0 Av. Also, let
{An x 5,,} be a sequence of S ® E such that AxK U^li ^ x ^/»- Since (by
Theorem 26.1) p* x v* is a measure on the semiring AM ® Au, it follows from

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS 225
Theorem 13.8 that
OO 00
fji* x v*(A x B) < J^/x* x v*(An x Bn) = ^T/x x v(i4„ x £„).
Consequently, we see that
fi*(A) • v*(B) = /x* x v*(A x £)
oo oo
< infHT/x x v(A„ x £„): (/*„ xjB„)c50S and A x £ c |J An x £„
«=i /i=i
= (/x x u)*(i4 x B).
(b) If 0 < /x*(/\) < oo and 0 < v*(B) < oo, then
(M x v)*(A x 5) = ii*{A) • v*(£)
holds true by virtue of Theorem 26.2. On the other hand, if either fi*(A) = oo or
v*{B) = oo, then—by (a)—the equality holds with both sides equal to oo.
(c) Let X = Y = {0}, S = {0}, and £ = P(y). Also, let ix = 0 on S (the only
choice!) and v = 0 on £. Now, note that /x*(X) • v*(Y) = oo • 0 = 0, while
(M x v)*(X xY) = oo.
Problem 26.2. Let (X, <S, jx) and (Y, E, u) 6e fwo a-finite measure spaces.
Then show that (/x x v)*(A x x9) = /x*(A) • v*(£) holds for each A x B in
AM <S> Ay
Solution. Let {X„} c AM and {yw} c Au satisfy Xn t X, y„ t Y, ti*(X„) <
oo, and v*(Y„) < oo for each n. Using Theorems 15.4 and 26.2, we see that
(AX x v)*(A xfi)= lim U x v)*((A 0 X„) x (B 0 Y„))
= lim L*(M n X„) • v*(B n y„)l
«->ooL J
= n*(A) • i/*(fl)
for each /I x 5 e A^ <8> Av.
Problem 26.3. Let (X, 5, /x) and (Y, T,,v)be two measure spaces. Assume that
A and B are subsets of X and Y, respectively, such that 0 < p>*(A) < oo, and
0 < v*(B) < oo. Then show that Ax B is /x x v-measurable if and only if both A

226
Chapter 4: THE LEBESGUE INTEGRAL
and B are measurable in their corresponding spaces. Is the preceding conclusion
true if either A or B has measure zero?
Solution. If A e AM and B e Av, then by Theorem 26.3, A x B e AMXy. For
the converse, assume that AxB e AMXU. We claim first that (p.xv)*(AxB) < oo.
To see this, pick two sequences [An] C S and {Bm} C £ with A C (J^ An,
5Z~, M*(An) < H*(A) + 1.BC USLi «m, and £~=1 v*(B«) < v*(fi) + 1.
Now, from Ax B £ (J~ , U~=1 A„ x Bm, we see that
OO oo
(M x vT(A xB)< X)l> x v)*Un x fim)
w=l m=l
oooo oo n r °° n
n—\ m=l /j=l m=l
< [ti*{A)+l]>[v*{B) + l] <oo.
Therefore, by Theorem 26.4, (A x B)^ is xi-measurable for v-almost all y. Since
(A x B)y = A holds for all y € £ and v*(5) > 0, it follows that A is
immeasurable. Similarly, B is v-measurable.
Finally, if fJ,*(A) = 0, A ^ 0, and A x £ 6 AMXlM then £ need not be
necessarily v-measurable. An example: Take X = Y = R with /x = v = A. If
£ Q [0i 1] is nonmeasurable, then {0} x E is a jx x v-null set (since {0} x E C
{0} x [0, 1]), and so {0} x E is a /x x v-measurable set.
Problem 26.4. Let (X, <S, /x) arcd (F, E, v) fee ftw a-finite measure spaces, and
let f:X xY -+ Hbe a n x v-measurable function. Show that for fi-almost all x
the function fx is a v -measurable function. Similarly, show that for v-almost all
y the function f} is [i-measurable.
Solution. We can assume /(*, y) > 0 for each (x,y) e X x Y. Since (in
this case) the product measure is a-finite, there exists a sequence {An} of fi x
immeasurable sets with An t X x Y and (/x x v)*(An) < oo for each n.
Now, by Fubini's Theorem, the function (/ A xa„ )x is v-integrable for /x-almost
all x. Since (/ a Xa„)x t /x» it follows that fx is v-measurable for /x-almost all
Problem 26.5. Show that if /(jc, y) = (x2 - y2)/(jc2 + y2)2, with /(0, 0) = 0,
then

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS 227
Solution. Note that
[[I f,,?U<]*y = l [-^ £]„, = -l^y - -|
and
Problem 26.6. Let X = Y = IN, S = Z, =the collection of all subsets of IN,
and [i = v =the counting measure. Give an interpretation ofFubini's Theorem
in this case.
Solution. Let /:NxN —> R be a non-negative fi x u-integrable function.
Then, by Problem 22.3 and Fubini's Theorem, we see that
ffd(n x v)= f [fdvLdv= /"[£] f{n, /n)]dv(/!) = ]T £/(/i, m).
On the other hand, if /: IN x IN —> 1R is a non-negative function such that
CO 00
YlJ2f(n*m>} <o°'
/i=l m=\
then it follows from Tonelli's Theorem that / is \x x v-integrable. Conclusion: A
function f:!NxlN —> R is/xxv-integrableifandonly if J^L{ ]Cm=i l/X^, "01
< oo, and in this case
/OO oo OO CO
Problem 26.7. Establish the following result, known as Cavalieri's Principle.
Let (X, S, fi) and (7, E, v) be two measure spaces, and let E and F be two jtixv-
measurable subsets of X x Y of finite measure. If v*(Ex) = £X*(Fr) holds for
fji-almost all x, then
(/x x v)*(£) = (/x x v)*(F).

228
Chapter 4: THE LEBESGUE INTEGRAL
Solution. By Theorem 26.4, we have
(/x x v)*(E) = f v\Ex)dfi(x) = / v*{Fx)dn(x) = (/x x v)*(F).
Jx Jx
Problem 26.8. For this problem X denotes the Lebesgue measure on R. Let
(X, 5, fi)beaa -finite measure space, and let f:X -» IRbea measurable function
such that f(x) > 0 holds for all x € X. Then show that
a. The set A = {(*, y) e X xlk: 0 < y < f(x)}, called the ordinate set of
f ,is a /x x k-measurable subset of X x R.
b. The set B = {(jc, y) e X x JR: 0 < y < fix)} is a /x x \-measurable
subset ofX x R and (/x x A)*(A) = (/x x *)*(£) /w/rfs.
c. 7Vze gnsp/j o//, z.e., the set G = {(*, /(a*)): a: 6 X),i5fljLtx \-measurable
subset of X x R.
d. /// is fx-integrable, then (/x x X)*(A) = ffdfi holds.
e. /// w fjL-integrable, then (/x x X)*(G) = 0 /20/ds.
Solution. If g: X —> R is an arbitrary positive measurable function, then we
shall write A8 = {(a*, y) e X xR: 0 <y < g(x)\ and
Bg = [(x,y)€XxR:0<y<g(x)}.
Observe that if fn(x) f fix) and hn(x) I f(x) hold for each x e X, then
Bfn t Bf and /4/,n | Af.
Assume first that g is a positive simple function. Let g = X!/'=i fl/Xc, be the
standard representation of g, where a,- > 0 for each 1 < / < n. Then, it is easy
to see that
Ag = (Xx {0}) U (C, x [0, a,]) U (C2 x [0, a2)) U • • • U (C/; x [0, *„]) (*)
and
^ = (Ci x [0, a,]) U (C2 x [0, *2]) U • • • U (C„ x [0, fln]). (**)
By Theorem 26.3, both Af, and BR are /x x A-measurable subsets of X x R.
(a) First assume that / is a bounded measurable function. That is, assume that
there exists some M > 0 such that 0 < f(x) < M holds for all x e X. By
Theorem 17.7 there exists a sequence [\//n} of simple functions with VOiOO t
M — fix) for each x eX. Thus, the sequence [<pn] of simple functions, defined
by (pnix) = M — i//nix), satisfies 4nix) I fix) for each a g X. This implies
A(pn I Af. Since (by the preceding discussion) each A<pn is /x x A-measurable,
we see that in this case Af is likewise a fi x A-measurable set.

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS 229
Now, let / be an arbitrary positive measurable function. For each n, let
fn = / a nl, and note that (by the preceding case) each Ajn is /x x A-measurable.
To infer that Aj is a /x x A-measurable set, observe that A/n f A/ holds,
(b) By Theorem 17.7 there exists a sequence {sn} of simple functions such that
0 < sn(x) t f(x) holds for all x € X. Clearly, B5n | Bf holds. Since each BSn
is /x x A-measurable, it follows that By is likewise /x x A-measurable.
Next, we shall establish the equality (/x x \)*(Af) = (/x x k)*(Bf) by cases.
C4S£ 7. Assume /x*(X) < oo.
Clearly, (/x x A.)*(X x {0}) = /x*(X) • A({0}) = 0. Also, assume that 0 <
f(x) < M < oo holds for all a*. Then, there exist two sequences {(/)„} and [yj/n)
of step functions with 0 < 4>n(x) f /(*) and ij/n(x) I f(x) for all a e X.
Clearly, B<pn t Bf and /\^; I Af. Now, use (•) and (••) in connection with
Theorem 26.3 and the Lebesgue Dominated Convergence Theorem to see that
ffadH = (/x x AT(«J t (M x *)*(*/) = ffd/JL
and
/Vn<*M = (/x x A)*(^„) ; (/x x k)*(Af) = //rfM-
Thus, in this case (/x x X)*(Af) = (/ix X)*(Bf) = // d/x holds.
C4S£2. /Usz/we /x*(X) < oo and that f is a positive pi-measurable function.
For each n let /„ = / a nl. Note that Bfn t #/ and Afn t ^/- By the
preceding case, we have (/x x \)*(Afn) = (/x x A.)*(#yM) for each n. Thus, from
Theorem 15.4, it follows that
(/x x XTiAf) = (|x x X)*(B/) = ^lim //,^/x.
C4S£ J. 77?e general case. Here we shall use the hypothesis that p is a-finite.
Choose a sequence {£„} of measurable subsets of X with En t X and
/x*(£„) < oo for each n. Let g„ = /x£fl» and observe that 5?n t #/ and
i4/fl t ^/- Using the preceding case and Theorem 15.4, we see that
(/x x A.)*(*;)= lim (/x x A)'(£*.) = lim (/x x A)*(A//()=(ax x *)*(*/).

230
Chapter 4: THE LEBESGUE INTEGRAL
Also, it should be noted here that if / is integrable, then
(/x x Xy(Bf) = (/x x XTiAf) = ffdfi.
(c) From the identity G = Aj \ Bj, it follows that the graph G of / is /x x X-
measurable.
(d) The equality follows from the discussion in part (b).
(e) From G = A/\Bf and part (d), we see that
(/x x X)*(G) = (/x x mAf) - Ox x *)*(£/) = 0.
Problem 26.9. Let g: X -+ 1R be a \i-integrable function, and let h:Y -► R &e
# v-integrable function. Define f:XxY-+$lby /(*, y) = g{x)h(y)for each
x and y. Show that f is /x x v-integrable and that
ffd(itxv)=(jgdfi).(jhdvy
Solution. We can assume g > 0 and h > 0. Choose a sequence {0,,} of /x-step
functions and a sequence {^n} of v-step functions with 0„ t g and t/t„ f h-
Then, {0„\M is a sequence of/x x v-step functions such that 4>n^n t £^- The
conclusion now follows from the relation
Problem 26.10. Use Tonelli's Theorem to verify that
I dx = I ( I e~xy sin jc <fo J dy
holds for each 0 < € < r. By letting € —> 0+ an<i r —► oo (and justifying your
steps) give another proof of the formula
f°°SmX , 7T
/ " ^ = 7T-
Jo a* 2

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS 231
Solution. Fix 0 < e < r and consider the function
e~xy if (x, y)e [e,r] x [0, 1]
geAx, y) | e-£y .f (j^ ^ € ^ r] x [u ^
Clearly, the continuous (and hence, measurable) function f(x, y) = e~xy sin.r
satisfies |/(.v, y)| < ge>, (x, y) for all (a, y) e [£, r] x [0, oo). From
r[r&*''""]"^i>'t=r-
— s
— < °°
and Tonelli's Theorem, we see that the function gEJ is Lebesgue integrable over
[£, /*] x [0, oo). So, the function f(x, y) is integrable over [e, r] x [0, oo). Now
Fubini's Theorem guarantees that
/ (/ e~xy sinxdy\dx •=. J / (e~xy sinxdxjdy. (•)
Using the elementary integral
/<
sintdt^-^^-e-0"
and performing the innermost integrations in (*), we get
£**dx = j"[-Z*sg™e-''>\Z]dy
/»oo /»00
__ / ysing-fcosg ^-gy j / vsinr-fcosr f-, y ^
Jo +>" c/0 +>"
and consequently,
/r /»oo /»oo /»oo
^dx = smsJo ^dy+cossj^ gpdy-J^ 2**g$2Le^dy. <**)
We shall compute the limits of the three terms in the right-hand side of (••) as
r -> oo and s -> 0+.
We start by computing lime_*o+ sin e /0°° ^~r dy. To this end, let rj > 0. Since
lim^oo y^r = 0, there exists some yo > 0 such that 0 < j^-j < rj holds for
all y > yo. Now, from lime_>o+ Cosine = 0 and lime_o+ ~^ = 1, we see that

232
Chapter 4: THE LEBESGUE INTEGRAL
there exists some 0 < 8 < 1 such that
0 < e < 8 implies 0 < yo sin £ < r\ and ^~ < 2.
Now, if 0 < e < 8, then (by taking into account that 0 < -—-r < 1 for y > 0)
we infer that
sins
Jo I+>
idy
ryo _ ^ - | poo
sine/ ^dy + sine / ^dy
JO ' ' Jvo
roo
/»oo
JyQ
< yo sin s -f r\ sin £ / e ^ sty
< jo sin e 4- ^ sin e I e~ey dy
Jo
= josine + Ty^i < ^ + 2t? = 3rj.
That is, lime_>o+ sin £ fQ ^pr dy = 0.
For the second limit, note that | -—^ | < j^ holds for each y e [0, oo). Thus,
in view of the Lebesgue integrability of the function h(y) = j^ over [0, oo),
Theorem 24.4 yields
lim cos £ I -T7TT dy = lim [cos s] • lim / -£A dy
For the third limit, note that for each r > 1 and each y > 0, we have
ysinr+cos; -n < 1+y -y < ? -)
and so by the Lebesgue integrability of g(y) = 2e y over [0, oo), it follows from
Theorem 24.4 that
-f
Jo
0dy = 0.

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS 233
Finally, from (*•), we see that
( ™L*dx = lim f ^dx = Yimlsmsf -£^-^1 +
r-*oo
= 0+f + 0=|.
Problem 26.11. 5/iovv r/jaf;/ /(.r, y) = ye~<-]+x')r for each x and y, then
Use the preceding equality to give an alternate proof of the formula
poo
/ e~x2dx =
Jo
<Jtz
T'
Solution. Note that
r»oo_ /»oo
/»oo_ /»oo _ /»oo
and
Since ye-U+fV > 0 holds for all x > 0 and j > 0, Tonelli's Theorem shows
that
(j[".-*r)'.,.
Problem 26.12. Show that
j ( I e xy~ sinxdxj dy = / ( / e xy~ sin* dyj dx

234
Chapter 4: THE LEBESGUE INTEGRAL
holds for all r > 0. By letting r —► oo show that
r°°sin£
2
In a similar manner show that f^^yf dx = \phz/2.
Solution. Since
/oo
M »
it follows from Tonelli's Theorem that e~~xy is integrable over [0, r] x [0, oo).
In view of \e~xy sinx| < e~*r, we see that e~xy sin* is also integrable over
[0, r] x [0, oo), and the stated identity follows from Fubini's Theorem.
Performing the innermost integrations and using the elementary integral
fe'at s'mtdt = -«™h*°" *""',
we get
/»oo /»oo /»r
/ Jl / rsinr+c°'r e~'y2 dv - & I 2^dx M
Jo 1+/ Jo l+yt y 2 Jo * ()
Since |,'y«'IT'| < ^ = /GO holds, and / is Lebesgue integrable
over [0, oo), it follows from Theorem 24.4 that
/»oo /»oo _ _
/»oo /«oo /»oo

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS 235
Now, using the elementary integral
and an easy computation, we see that
/»00 r \~ pOO
/ Jl. — l JtL. -l / Jl. — zdl
Thus, from (•), we see that
f 2Udx= lim [rmdx = 4= -^ =
2 '
Problem 26.13. Using the conclusions of the preceding problem {and an
appropriate change of variable), show that the values of the Fresnel integrals {see
Problem 24.6) are
I sin{x2)dx — I cos{x2)dx =
Jo Jo
V^F
Solution. Using the change of variable x = y/u, we get
f sin(*2) dx = I f ^f rf« and /" cos(jc2) dx = { f ^f </k.
Now, let r -> oo and use the preceding problem.
Problem 26.14. Let X = Y = [0, 1], \x — the Lebesgue measure on [0, 1], and
v = the counting measwe on [0, 1]. Consider the "diagonal" A = {(*, a): x e X]
ofXxY. Then show that
a. A is a fix v-measurable subset ofXxY, and hence, Xa is a non-negative
\i x v-measurable function.
b. Both iterated integrals ffx^dfidv and ff x^dvdfi exist.
c. The function Xa is not fi x v-integrable. Why doesn't this contradict
TonellVs Theorem?
Solution, (a) Consider the two sets
A = {(*, y) 6 X x Y: x > y] and B = {(*, y)eX xY: x < y).

236
Chapter 4: THE LEBESGUE INTEGRAL
FIGURE 4.4.
Note that A = {J[a, b] x [c, d]t where the union extends over all rectangles
[a, b] x [c, d] with rational end points and a > d\ see Figure 4.4.
Clearly, the collection of all such rectangles is countable. Since each rectangle
is fi x v-measurable, it follows that A is \i x v-measurable. Similarly, the set B
is fi x v-measurable. Hence, A = Xxy\/lUB isa/xx v-measurable set.
(b) Note that
jJxAdfidv = jU XA{x,y)dfjL(x)]dv(y)
= / [/ Xbl^U)] <M)0 = / 0 • dv(y) = 0,
and
= J [J X\x) dv(y)] dfx(x) = J 1 • dn(x) = 1.
(c) Fubini's Theorem combined with part (b) shows that xa is not integrable over
X x Y (i.e., (/x x v)*(A) = oo must hold). This does not contradict Tonelli's
Theorem because v is not a or-finite measure.
Problem 26.15. Let f: R —► R be Borel measurable. Then show that the
functions f(x 4- y) and f(x — y) are both X x X-measurable.

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS 237
FIGURE 4.5.
Solution. Since / is the limit of a sequence of step functions, it suffices to
establish the result for characteristic functions of measurable sets of finite measure.
The regularity of the Lebesgue measure allows us to reduce it to the characteristic
functions of open sets of finite measure. Finally, this can be reduced to the case
when / = X(a,b) f°r some finite open interval (<z, b).
The X x A.-measurability of the function g(x, y) = X(a,/>)U + y) follows easily
from the graph shown in Figure 4.5.
The A. x A-measurability of f(x — y) can be proven in a similar manner.

CHAPTER 5
NORMED SPACES AND
Z^-SPACES
27. NORMED SPACES AND BANACH SPACES
Problem 27.1. Let X be a normed space. Then show that X is a Banach space
if and only if its unit sphere [x e X: \\x\\ = 1} is a complete metric space (under
the induced metric d(x, y) = ||.r — y\\).
Solution. Let S = [x e X: \\x\\ = 1}, and note that S is a closed set. Clearly,
if X is a Banach space, then S is a complete metric space.
Conversely, assume that S is complete. Let {xn} be a Cauchy sequence of X.
In view of the inequality
| ll^n II " II JCm II | < II JCn-JT« II,
we see that {||jrn||} is a Cauchy sequence of real numbers. If lim ||jcn|| = 0, then
limxn = 0. So, we can assume that 8 = lim ||jc„|| > 0. In this case, we can also
assume that there exists some M > 0 such that -^-j < M and ||jc„ || < M both
hold for each n. The inequalities
II _*n r^ || _ ll(l|.rmll.vw-||.rJlrm)H
U||.v„|| \\xm II II — l|.t„||||.t„,||
< M2 1 \\xm\\{xn - xm) - (||jf„|| - \\xm\\)xm J
<2M3\\xn-xm\l
show that the sequence {if^ir} is a Cauchy sequence of S. If x is its limit in 5,
then xn = ||.rn|| • j^j —> 8x holds in Xy and so X is a Banach space.
Problem 27.2. Let X be a normed vector space. Fix a e X and a nonzero
scalar a.
239

240
Chapter 5: NORMED SPACES AND LP-SPACES
a. Show that the mappings x i-> a + x and x *-► ax are both homeomor-
phisms.
b. // A and B are wo sets with either A or B open and a and fi are nonzero
scalars, then show that aA + fiB is an open set.
Solution, (a) Observe that \\(a -f x) - (a + y)\\ = ||a — y\\ holds for all a* and
v. This shows that A'^fl + x is, in fact, an isometry.
Also notice that for all a, y € X we have \\ax — ay\\ = |a| • ||a — y\\. This
easily implies that x h» ax is a homeomorphism.
(b) Assume first that B is an open set. Since the mapping x h> a + x is a
homeomorphism, we know that a -f B is an open set for each a e X. This implies
that the set A + B = [Ja(EA(a + B) is an open set for each subset A of X.
Now, assume that B is an open set and that a and fi are nonzero scalars. Since
the mapping x \-+ fix is a homeomorphism, the set fiB is an open set. So, by the
preceding discussion, a A + @B must be an open set.
Problem 27.3. Let Xbea normed vector space, and let B = {a e X: \\x || < 1}
be its open unit ball. Show that ~B = {a e X: \\x\\ < 1}.
Solution. Repeat the solution of Problem 6.2.
Problem 27.4. Let X be a normed space, and let [xn] be a sequence ofX such
that limAn = a holds. If yn = n~](x\ + • • • + xn) for each n, then show that
\imyn = a.
Solution. Let e > 0. Choose some k with \\x„ — a|| < e for all n > k. Fix
some m > k so that ||^[(a'i — a) + • • • -f (xk — x)]\\ < e holds for all n > m.
Thus, if n > m, then
< ||^[(Al -A-)H h(AA -A)]|| + ^[||A^! - X ||H + H^ -A'||]
< e -f e = 2e.
That is, lim y„ = a holds. (See also Problem 4.11.)
Problem 27.5. Assume that two vectors x and y in a normed space satisfy
||a + y\\ = ||a|| -I- \\y\\. Then show that
\\ax + Py\\=*\\x\\+P\\y\\
holds for all scalars a > 0 and fl >"0.
Solution. Assume ||a -f- jy|| = ||a|| + \\y\\ holds for two vectors a and y in a
normed space, and let a > 0 and (5 > 0. Without loss of generality, we can

Section 27: NORMED SPACES AND BANACH SPACES
241
suppose that a > fi > 0. From the triangle inequality, it follows that
\\ax + /iy\\<<*\\x\\+P\\yl
Next, notice that
||e*.r + j8y|| = \a(x + y) + (jB - a)y\\
> \\a(x + y)\\-\\(P-a)y\\
= a\\x + y\\ -(a- fi)\\y\\ = a(\\x\\ + ||y||) - (a - ft\y\
= a||-r||+i8||y||.
Hence, ||orx + £v|| = a||jc|| + £||;y||, as desired.
Problem 27.6. Let X be the vector space of all real-valued functions defined on
[0, 1] having continuous first-order derivatives. Show that \\f\\ = |/(0)| + ||/'||oo
is a norm on X that is equivalent to the norm ||/||oo + ll/'lloo-
Solution. The verification of the norm properties of || • || are straightforward.
From
/W = /(0)+ ff\t)dt,
Jo
we see that |/C*)| < |/(0)| + ll/'lloo holds for each x e [0, 1], and consequently,
ll/lloo< 1/(0)1 +ll/'Hoo holds.
The equivalence of the two norms follows from the inequalities
ll/lloo + ll/'lloo < 1/(0)1 + 2||/'||0O < 2(1/(0)1 + H/'lloo)
= 2II/H < 2(11/1100 + H/'lloo).
Problem 27.7. A series J2T=\X" *n a normed space is said to converge to x if
lim ||„t — ]C/'=i-r/II = 0. As usual, we write x = J2^L{xn. A series X^li*" is
said to be absolutely summable // Y1T=\ II-r« II < oo holds.
Show that a normed space X is a Banach space if and only ifeveiy absolutely
summable series is convergent.
Solution. Let X be a Banach space, and let Y1T=\ x" ^e ^ absolutely summable
series. For each n let sn = ]T]/=i xi- The inequality
n+P n+P
\sn+p-s„l= J J2Xi\ - £ 11^'U
i=n+\ /=«+l
implies that [sn] is a Cauchy sequence, and hence, convergent in X.

242
Chapter 5: NORMED SPACES AND LP-SPACES
For the converse, let [xn} be a Cauchy sequence in a normed space X whose
absolutely summable series are convergent. By passing to a subsequence (if
necessary), we can assume that ||jc„+i — jcJ| < 2~n holds for each n. Put jco = 0, and
note that £~0 ||jc„+i -*„|| < oo. Thus, limn_oo £?~ote+i -*/) = limn->oo*n
exists in X so that X is a Banach space.
Problem 27.8. Show that a closed proper vector subspace of a normed vector
space is nowhere dense.
Solution. Let E be a proper closed subspace of a normed space X. Assume that
E has an interior point a. Then, there exists some r > 0 such that B(a, r) C E.
Now, if y is an arbitrary nonzero element of X, then a + jhu y € Z?(a, r) C £,
and so y = ^^[(a + jjfcji^)"~ a]e &- ^at is, £ = X holds, a contradiction.
Thus, E° = 0.
Problem 27.9. Assume that /: [0, 1] -* R is a continuous function which is
not a polynomial. By Corollary 11.6 we know that there exists a sequence of
polynomials [pn] that converges uniformly to f. Show that the set of natural
numbers
(kN: k = degree of pn for some n]
is countable.
Solution. Let /: [0, 1 ] -> R be a continuous function which is not a polynomial,
and let {/?„} be a sequence of polynomials that converges uniformly to / on [0, 1].
Assume by way of contradiction that the set of natural numbers
K = [k e N: k = degree of pn for some n]
is bounded. This means that there exists some m e IN such that every pn has
degree at-most m. So, if V is the finite dimensional vector subspace generated in
C[0, 1] by the functions {1, jc, x2,..., xm], then [pn] c V holds. Now, a glance
at Theorem 27.7 guarantees that V is a closed subspace of C[0, 1], and thus the
(sup) norm limit / of {/?„} must lie in V. That is, / must be a polynomial of degree
at-most m, contrary to our hypothesis. Hence, K is not bounded, and therefore it
must be a countable set; see Theorem 2.4.
Problem 27.10. This problem describes some important classes of subsets of a
vector space. A nonempty subset A of a vector space X is said to be:
a. symmetric, ifx e A implies — x € A, i.e., if A = —A;

Section 27: NORMED SPACES AND BANACH SPACES
243
b. convex, if xy y e A implies Xx 4- (1 — X)y e Afar all 0 < A. < 1, i.e.,
for every two vectors x, y e A the line segment joining x and y lies in
A; and
c. circled {or balanced) if x e A implies Xx e Afar each |A.| < 1.
Establish the following:
i. A circled set is symmetric.
ii. A convex and symmetric set containing zero is circled.
iii. A nonempty subset A of a vector space is convex if and only ifaA-\-bA —
(a 4- b)A holds for all scalars a > 0 and b > 0.
iv. // A is a convex subset of a normed space, then the closure A and the
interior A0 of A are also convex sets.
Solution, (i) Let A be a circled set. Since | — 1| = 1 < 1, it follows that
—x = (—1)* e A for each x e A. Thus, A is a symmetric set.
(ii) Let A be a convex symmetric set containing zero. Fix x e A and \X\ < 1.
If 0 < X < 1, then Xx = Xx + (1 - A.)0 e A and if -1 < A. < 0, then
Xx = (-X)(-x) 4- (1 + A.)0 e A. So, A is a circled set.
(iii) Let A be a subset of a vector space. Assume first that A is a convex set, and
let a > 0 and b > 0. If x e (a 4- b)A = {{a 4- b)u: u e A), then for some u e A,
we have x = (a + b)u = au -f bu e aA + bA, and so (a 4- b)A C. aA -{- bA
is always true. Now, let x e aA 4- bA. Then, there exist w, v e A such that
x = au + bv. Since A is convex, we have z = -ttU H—fri; € A, and so
x = (tf+&)[^« + ^pH = (0+&)z € (fl+&)A. Therefore, a A+M C (a+Z?)A
is also true, and consequently a A 4- M = (a 4- &)A.
Next, suppose that a A 4- bA = (a 4- fr)A holds true for all a > 0 and b > 0.
Let x, y € A and 0 < X < 1. Letting a = A and Z? = 1 — A., we see that
Xx 4- (1 ~ A)y = ax + by e (a + b)A = A.
This shows that A is a convex set.
(iv) Let A be a convex subset of a normed space. We show first that A is a convex
set. To this end, let jc, y € A and fix 0 < A < 1. Pick two sequences {xn} and [yn]
of A such that xn -> x and yn-* y. Put z„ = Ax„ 4- (1 — X)yn and note that {z„} is
a sequence of A. Since the function f:X —► X, defined by f(u) = Xu + (l— X)u,
is continuous (see Problem 27.2), it follows that z„ -> A..r + (1 — A.)^. This implies
Xx 4- (1 — A.);y G A, so that A is a convex set.
Next, we shall show that A0 is a convex set. Fix 0 < A. < 1. Since A0 is an
open set, it follows (from Problem 27.2) that the set A.A0 4- (1 — A.)A0 is also an
open set, which (since A is convex) is contained in A. Since A0 is the largest open
set contained in A, we infer that A.A0 4- (1 — A)A° C A0. This shows that A0 is
a convex set.

244
Chapter 5: NORMED SPACES AND LP-SPACES
Problem 27.11. This problem describes all norms on a vector space X that are
equivalent to a given norm. So, let (X, || • \\) be a normed vector space. Let
A be a norm bounded convex symmetric subset of X having zero as an interior
point (relative to the topology generated by the norm || • ||). Define the function
pA\X -> IRfry
pA(x) = inf{A. > 0: x e XA}.
Establish the following:
a. The function pA is a well-defined norm on X.
b. The norm pA is equivalent to || • ||, i.e., there exist two constants C > 0 and
K > 0 such that C\\x\\ < pA(x) < K\\x\\ holds for all x e X. _
c. The closed unit ball of pA is the closure of A, i.e., {x € X: pA(x) < 1} = A.
d. Let || | • || | be a norm on X which is equivalent to || • ||, and consider the norm
bounded nonempty symmetric convex set B = {x e X: \\\x\\\ < 1}. Then
zero is an interior point of B and \\\x\\\ = Pb(x) holds for each x e X.
Solution. Assume that A is a norm bounded convex symmetric subset of a
normed space (X, || • ||) such that zero is an interior point of A.
(a) Pick some r > 0 such that £(0, 2r) = [x e X: \\x\\ < 2r] c A. If a* <= X is
a nonzero vector, then t~a* e #(0, 2r) c A, and so x e ^-A. This shows that the
set{X > 0: x e XA] is nonempty and so the formula pA(x) = inf{X > 0: x e XA]
is well defined and satisfies
/M(*)<r||jr|| (*)
for all x e X. Next, we shall show that pA is a norm on X.
Clearly, pA(x) > 0 and pA(0) = 0. Now, if pA(x) = 0, then there exist
a sequence [an] C A and a sequence {Xn} of positive real numbers satisfying
Xn -* 0 and a* = Xnan for each n. Since A is a norm bounded set, it easily follows
that a = lim Xnan = 0. Thus, pA(x) = 0 if and only if a = 0.
Next, we shall show that pA(ax) = \a\pA(x) holds for all a. € IR and all x e X.
Since A is symmetric, we have
[X > 0: Xx e A] = {X > 0: k(-x) e A},
and so for proving pA(otx) = |a|pA(x), we can suppose without loss of generality
that a > 0. Now, note that
pA(ax) = inf{X > 0: ax e XA] = inf{X > 0: a e ^A)
= ainf{£: x e ^A] = a inf{fi > 0: a g fiA]
= Of ^(A).

Section 28: OPERATORS BETWEEN BANACH SPACES 245
Forthe triangle inequality, let x, y e X and fixe > 0. ChooseX > Oand.v g XA
such that X < pA(x) + e. Likewise, pick some /x > 0 such that y G p. A and
\x < PA(y) + €- From Problem 27.10 we know thatx + y G XA + jiA = (A-f fi)A,
and so
Pa(x + y) < X + /x < [pA(*) + <f] + [pa()0 + *] = PaM + P/*0>) + 2€.
Since 6 > 0 is arbitrary, we infer that pA(x 4- y) < PaW + PaOO-
(b) Let x g X and fix some M > 0 such that ||a|| < M holds for all a G A.
Now, if X > 0 satisfies .r g XA, then there exists some y e A such that „r = Xy.
Hence, ||.r|| = A.||y|| < A.M, or X > \\x\\/M. This implies pA(x) > ±\\x\\. Now,
combine this inequality with (*) to establish that pA is a norm equivalent to || • ||.
(c) Assume first that p^OO < 1 and x ^ 0. Then for each n there exist
0 < Xn < 1 + - and an e A such that x = Xnan. By passing to a subsequence, we
can assume Xn -> X. Since x ^ 0 and A is a norm bounded set, it easily follows
that 0 < X < 1. Now, note that the sequence [an] C A satisfies an = -£-x -> ^r,
and so £jc g A. Since /\ is also a convex set (see Problem 27.10), we see that
x = X({x) + (\-X)0eA.
Now, let.v G A. Then, there exists a sequence {.v„} C A such that ||*„—*|| -> 0.
Since || • || is equivalent to Pa, we also have pA(*n — x) -> 0. In particular,
Pa(Xh) -* P/iO0- Now, notice that since xn G A, we have PaOoO < 1 for each n.
This implies p^OO 5 1. Therefore, the closed unit ball of pA is A.
(d) Let || | • || | be a norm on X which is equivalent to || • ||. It is easy to check that
the closed unit ball B of || | • || | is a bounded convex and symmetric set containing
zero as an interior point. We shall show next that |||*lll = PbM holds for each
x eX.
To see this, let x G X be a nonzero vector. Since x/|||jc||| g By we see that
Pa 00/II 1*111 = Pb{x/\\\x\\\) < 1, and sop* 00 < II I* III- On the other hand, there
exist a sequence [Xn] of positive real numbers and a sequence [bn] of B such that
^n -> Pb(*), bn G B and* = Xnbn for each n. Since |||jc||| = A.,,|||fr„||| < Xn, we
easily infer that |||.r||| < p^OO- Hence, pa00 = 111*111 for each* g X.
28. OPERATORS BETWEEN BANACH SPACES
Problem 28.1. Let X and Y be two Banach spaces and let T.X -> Y be a
bounded linear operator. Show that either T is onto or else T(X) is a meager set.
Solution. Assume that T(X) is not a meager set. Then, we have to show that
T(X) = Y holds.
Let V = [x e X: \\x\\ < I}. Since (by assumption) T(X) is not a meager
set, some nT(V) = nT(V) has an interior point. This implies that T(V) has
an interior point. So, there exists some y0 g T(V) and some r > 0 such that

246
Chapter 5: NORMED SPACES AND LP-SPACES
B(yo,2r) C T(V) = -T(V). Note that if y e Y satisfies ||y|| < 2r, then
y-yo = -Cvo -y) e T(Y) and so y = (y - y0) + j0e T(V) + 7(V) c
2T(V). (The last inclusion follows, of course, from the identity V + V = 2V.)
Consequently, we have established that {y e Y: \\y\\ < r] c. T(V). From the
linearity of 7, we infer that
{yeY: \\y\\ < 2~nr] c 2~nTW) = T(2~"V) (**)
holds for each /z.
Next, let y eY be fixed such that \\y\\ < 2~xr = £. From (••), we know
that v € T(2-1Vr). So, for some vector x\ e 2"XV we have \\y-T(x\)\\ <
2~2r. Now, proceed inductively. Assume that xn e2~nV has been
selected such that Iy - J2?=\ T(xi)\\ < 2'"/I"1r. From (••) it follows that y -
E/Li T(Xi) e T(2-n~lV), and so there exists some xR+\ e 2""n_1V such that
I v — X^=/ r(jc/)| < 2~n~2r. Thus, there exists a sequence {jtn} of X such that
||jc„||<2-" and
\,-±T<*>\-[,-T(£*)
l — \ J=l
< 2"n"lr
hold for each az. Now, for each n let sn = jq H h*„ and note that the relation
n+p n+p oo
/=n+l /=n+l i=/i+1
shows that {sn} is a Cauchy sequence of X. Since X is a Banach space, the
sequence {sn} converges; let s = \imsn. Clearly, \\s\\ < X^li Wxn\\ ^ 1 (i-e->
5 € V), and by the continuity and linearity of 7, we see that
T(s) = lim T{sn) = lim Y^7,U/) = j.
That is, y e T(V\ and so [y € 7: ||y|| < 5} £ T(V) c 7(X). Since T(X) is a
vector subspace of 7, the latter inclusion implies that T(X) = Y must hold.
Problem 28.2. Let X be a Banach space, T:X —► X a bounded operator, and
I the identity operator on X. If \\T\\ < I, then show that I —T is invertible.

Section 28: OPERATORS BETWEEN BANACH SPACES
247
Solution. If A, B e L(X, X), then the inequalities
\\ABx\\<\\a\\-\\Bx\\<\\A\\-\\B\\.\\x\\
easily imply that \\AB\\ < \\A\\ - \\B\\. In particular, if a sequence {A,,} of
operators of L(X, X) satisfies lim An = A in L(Xy X) and B e L(X, X), then
the inequality
\\BAn - BA\\ = \\B(An - A)\\ < \\B\\ • \\A„ - A\\
shows that lim BAn = BA. Similarly, \imAnB = AB.
Now, assume T e L{X,X) satisfies ||7|| < 1. In view of the inequality
\\Tn\\ < ||r|r, it follows that
00 oo
£IMI± Elixir = ^n<~-
n=0 /»=0
Thus, YlT=o Tn is an absolutely summable series. Since L(X, X) is a Banach
space, S = YlT=o Tn converges in L(X, X)\ see Problem 27.7. Moreover,
(/ - T)S = lim (/ - T)fV T*) = lim (7 - r'+1) = 7>
i=0
and similarly S(I - T) = /. Therefore, S = (/ - T)"1.
Problem 28.3. On C[0, 1] consider the two norms
ll/lloo =sup{|/U)|: a' € [0, 1]} and ||/||, = / \f{x)\dx.
Jo
Then show that the identity operator /:(C[0, 1], || • ||oo) -> (C[0, 1], || • ||,) is
continuous, onto, but not open. Why doesn't this contradict the Open Mapping
Theorem?
Solution. Clearly, / is onto, and in view of the inequality \\f\\\ < ll/Uoo* w^
see that / is also continuous.
For the rest of the proof, we need to show that (C[0, 1], || • |h) is not a Banach
space. To establish this, consider the sequence {/„} of continuous functions whose
graphs are shown in Figure 5.1.
The inequality \\fn+p — fn\\\ < yt shows that {/„} is a Cauchy sequence for
the norm || • ||i. Assume by way of contradiction that lim \\fn — f\\\ = 0 holds
for some / e C[0, 1].

248
Chapter 5: NORMED SPACES AND LP-SPACES
i
y
i <
\
2 2^2
FIGURE 5.1.
Let a e (0, j). If f(a) ^ 1, then there exist some e > 0 and some 0 < <5 <
min{tf, ^ — #} such that |/(a) — 1| > e holds whenever \x — a\ < 8. Now, note
that 28e < JQ |/„(a) — f(x)\ dx = ||/„ — /Hi for all sufficiently large n, contrary
to lim 11/n - /Hi = 0. Thus, f(a) = 1 holds for all a e (0, i). Similarly,
/(a) = 0 for all a e (|, 1), Now, it is readily seen that / cannot be a continuous
function, contrary to / € C[0, 1]. Thus, {/„} does not converge in C[0, 1] with
respect to the || • || i norm.
Finally, /: (C[0, 1], || • ||oo) —> (C[0, 1], || • \\x) cannot be an open mapping.
Since otherwise, || • ||i and || • Hoc would be equivalent norms, and therefore
(C[0, 1], || • ||i) would be a Banach space.
Problem 28.4. Let X be the vector space of all real-valued functions on [0, 1]
that have continuous derivatives with the sup norm. Also, let Y = C[0, 1] with
the sup norm. Define D:X -+ Y by D(f) = /'.
a. Show that D is an unbounded linear operator.
b. Show that D has a closed graph.
c. Why doesn't the conclusion in (b) contradict the Closed Graph Theorem?
Solution, (a) The standard properties of differentiation guarantee that D is a
linear operator. Now, for each n let fn(x) = xn. Then, /„ € X and ||/„||oo =
sup{|/„(A')|: 0 < x < 1} = 1 for each n. Now, notice that D(fn)(x) = nxn~]
holds for each /?, and from this it follows that
IIOII > ll^(/.)lloc-=sup{/2jcw-1: 0 <X < 1} =/!,
Therefore, ||D|| = oo, and so D is an unbounded operator.

Section 28: OPERATORS BETWEEN BANACH SPACES
249
(b) To see that D has a closed graph, assume /„ -> 0 in X and Dfn — f'n -> g
in Y. That is, {/„} converges uniformly to zero, and {/„'} converges uniformly
to g. We have to show that g = 0.
From f^f'n{t)dt = fn(x) - fn(0) (and Problem 9.16), it follows that
[ g(t)dt = lim C f'n{t)dt = Yim[fn(x)-fn(0)]=0
holds for all x e [0, 1]. Differentiating, we get g{x) — 0 for each x e [0, 1], as
required. (See also Problem 9.29.)
(c) The conclusion in (b) does not contradict the Closed Graph Theorem since
X is not a Banach space. For instance, we know (from Corollary 11.6) that every
function / e C[0, 1] is the uniform limit of a sequence of polynomials. So, if
/ e C[0, 1] is a nondifferentiable continuous function and {/?„} is a sequence of
polynomials that converges uniformly to /, then {/?„} is a Cauchy sequence of X
which cannot converge in X.
Problem 28.5. Consider the mapping T: C [0, 1 ] -> C [0, 1 ] defined by Tf(x) =
x2 f(x) for all f e C[0, 1] and each x e [0, 1].
a. Show that T is a bounded linear operator.
b. ///:C[0, 1] -+ C[0, 1] denotes the identity operator {i.e., 1(f) = f for
each f <=C[0, 1]), then show that ||/+71 = 1 + ||r||.
Solution, (a) From the identities
T(f + g)(.x) = x\f + g)(x) = *2/U) + x2g(x) = (7/ + rs)(.t)
and
7W)(.r) = e*jr/(*) = (aTf){x),
we easily infer that T is a linear operator. For the norm of 7, note that for each
/ eC[0, 1] we have
IIT/Hoo = sup |7Y(a-)| = sup x2|/U)| < sup |/Cr)| = H/Hoo,
\€[0.1] ve[0,l] te[0,l]
and so || r || < 1. On the other hand, for the constant function 1, we have
lir||> 11711100= sup x2 = h
.re[0,l]
Thus, 117*11 = 1, and so T is a bounded operator.

250
Chapter 5: NORMED SPACES AND L^-SPACES
(b) Clearly, || / + T \\ < \\ 11| + || T || = 1 + 1 = 2. Moreover, we have
11/ +71 > ||(/ +r)l||oo = sup (1 -f a-2) = 2,
*€[0.1]
and so ||/ + T\\ = 1 + ||7|| = 2 holds true.
Problem 28.6. Let X be a vector space which is complete in each of the two
norms \\-\\\ and || • l^. If there exists a real number M > 0 such that \\x \\\ < M \\x H2
holds for all x e X, then show that the two norms must be equivalent.
Solution. The identity operator 7:(X, || • H2) —► (X, || • |h) is one-to-one,
continuous, and onto. By the Open Mapping Theorem it is a homeomorphism,
and the conclusion follows.
Problem 28.7. Let X,Y, and Z be three Banach spaces. Assume that T:X -> Y
is a linear operator and S:Y —> Z is a bounded, one-to-one linear operator. Show
that T is a bounded operator if and only if the composite linear operator S o T
(from X into Z) is bounded.
Solution. Assume that S oT is a bounded operator. Let xn —► 0 in X and
T(xn) -+ y in Y. Using that S oT and S are both continuous, we get
S(y) = lim S(T(xn)) = lim S o T(xn) = 0.
Since S is one-to-one, we infer that y = 0, and hence—by the Closed Graph
Theorem—the operator T is continuous.
Problem 28.8. An operator P:V -* V on a vector space is said to be a
projection if P2 = P holds. Also, a closed vector subspace Y of a Banach space is
said to be complemented if there exists another closed subspace Z ofX such that
Y®Z = X.
Show that a closed subspace of a Banach space is complemented if and only if
it is the range of a continuous projection.
Solution. Let Y be a closed subspace of a Banach space X. Assume first that
there exists a continuous projection P: X -> X whose range is Y, i.e., P(X) = Y.
From P2 = />, it follows that Y = {y e X: y = Py).
If /: X ~> X denotes the identity operator, let Z = (/ — P)(X), the range of
the continuous operator I — P. Clearly, Z is a vector subspace of X and in view
of x = Px + (/ - P)(x), we see that r + Z = X. Now, if u e Y n Z, then

Section 29: LINEAR FUNCTIONALS
251
u = z - Pz for some z e Z, and so u = P(w) = P(z - Pz) = P(z) - P2(z) = 0.
This means that Y © Z = X. Finally, to see that Z is also closed, assume that a
sequence [zn] of X satisfies (/ — P)(zn) —> z. Then, the continuity of P implies
0 = P(I - P)(zn) -» Pz, and so Pz = 0. Hence, z = (/ - P)(z) e Z, proving
that Z is also closed. Thus, Y is a complemented closed subspace.
For the converse, assume that Z is a closed subspace such that Y © Z = X. So,
for each x e X there exist y £ Y and zeZ (both uniquely determined) such
that x = y + z. Define an operator P: X -> X via the formula P(jc) = y, where y
satisfies x — y e Z. We claim that P is a continuous projection whose range is Y.
Notice first that P is a linear operator. Also, P2(x) = P(y) = y = P(x) holds
for each x e X, so that P is a projection. Clearly, the range of P is Y. To finish
the proof, we must show that P is also continuous. For this, it suffices to show (in
view of the Closed Graph Theorem) that P has a closed graph.
To this end, assume that a sequence^} of X satisfies xn -> x and P(xn) -> y
in X. For each n let xn = yn -f z„, where y„ g 7 and zn e Z. Clearly, yn = P(a-„)
for each n. Since K is a closed subspace, it follows that y e Y. Now, from
zn = xn — yn -> x — y and the closedness of Z, we infer that z = x — y e Z.
Thus, A' = y + z, and so y = P(x). This shows that P has a closed graph, and we
are done.
29. LINEAR FUNCTIONALS
Problem 29.1. Let f:X->]Rbea linear functional defined on a vector space
X. The kernel of f is the vector subspace
Ker/ = /-1({0}) = {.rGX: /(*) = 0}.
IfX is a normed space and /: X —► R is nonzero linear functional, establish the
following:
a. / is continuous if and only if its kernel is a closed subspace ofX.
b. / is discontinuous if and only if its kernel is dense in X.
Solution, (a) Clearly, if / is continuous, then its kernel /_1({0}) is a closed
set. For the converse, assume that / ^ 0 and that f~l({0}) is a closed set. Pick
some e e X with f(e) = 1.
Suppose by way of contradiction that ||/|| = oo. Then, there exists a sequence
[xn] of X with ||a-„|| = 1 and |/Cx„)| > n for each n. Note that the sequence
{y„}, defined by yn = ^ -—^, satisfies yn e /_I({0}) foreach n and yn —> e.
Since the set /_1({0}) is closed, it follows that e e f~l({0}\ and so f(e) = 0,
which is a contradiction. Thus, / is a continuous linear functional.

252
Chapter 5: NORMED SPACES AND LP-SPACES
(b) If Ker / is dense in X, then Ker / is not closed and hence, by part (a), / is not
continuous. For the converse, assume that / is a discontinuous linear functional,
i-e*> 11/11 = °°. This implies (as in the previous part) that there exists a sequence
{xn} of X satisfying ||a:,,|| = 1 and |/(a'„)| > n for each n. Now, if x e X and
yn = x — TfH*/!, then {yn} is a sequence in Ker/ and satisfies yn -> x. This
shows that ker / is dense in X.
Problem 29.2. Show that a linear functional f on a normed space X is
discontinuous if and only if for each a e X and each r > 0, we have
/(£(*,/•)) = {/(*): ||fl-x||<r}=R.
Solution. Let / be a linear functional on a normed space X and let B =
£(0, 1) = [x e X: \\x\\ < 1}. Assume that / is discontinuous. Fix a e X and
r > 0. From the relation B(a, r) = a -f /\#(0, 1) = a •+- rB and the linearity of
/, it follows that f(B(a, r)) = R holds if and only if f(B) = SR.
We claim first that f{B) is unbounded from above in R. To see this, assume
by way of contradiction that there exists some M > 0 such that f(x) < M holds
for each x e B. Note that if x e X satisfies ||a|| < 1, then ±^x e B, and so
from
±l/(A-) = /(±^)<f,
we see that \f(x)\ < M holds for all x e X with ||a|| < 1. That is, ||/|| =
sup{|/(A')|: ||a"|| < 1} < M < oo, and so / is a continuous linear functional, a
contradiction. Thus, f(B) is unbounded from above in R. Now, let a > 0 bean
arbitrary positive real number. By the above, there exists some x e B satisfying
/(a*) > a. Now, note that the element y = -—ta* € B satisfies f(y) = or (and,
of course, —yeB satisfies /(—y) = —a). Consequently, f(B) = R.
For the converse, assume that f(B(a,r)) = R holds for each a e X and each
r > 0. In particular, from
oo = sup{|/(A)|: IIa-H < ^} < sup{|/(A)|: ||a|| < l} = ||/||,
we see that ||/|| = oo. Thus, / is unbounded and so (by Theorem 28.6) / is a
discontinuous linear functional.
Problem 29.3. Let /, /i, /2,..., /„ be linear functionals defined on a com-
mon vector space X. Show that there exist constants X\,..., Xn satisfying f =
£?=i hfi (i.e., f lies in the linear span of f\,..., fn) if and only if HJLi Ker /• c
Ker/.

Section 29: LINEAR FUNCTIONALS
253
Solution. If / = £/'=i hfi holds, then clearly fl/Li Ker// £ Ker/- For the
converse, assume (~)"=l Ker/,- C Ker/. Let
V = {ye R": ].veX such that y ^ (/,(*), /2(*),..., /„(*))}.
It is easy to verify that V is a vector subspace of IR". Now, define the linear
functional g: V -r R via the formula
*(/i (-*),/>(*) /«W) = /W.
Notice that g is well defined. To see this, assume
(/i(-v), Hx\ .... f„(x)) = (/,(y), /2(y) /„()>)).
Then, /,(x — y) = 0 for each /, and so x — y e H/Li Ker /< • From our hypothesis,
it follows that a* — ye Ker /, which means that f(x) — f(y). Now, it is a routine
matter to verify that g is linear.
Denote by g again a linear extension of g to all of IR". This implies that
there exist scalars \\,..., A.,, such that g(z\,..., z„) = YH=\ ^<z' holds for all
(zj,..., z„) e R". In particular, we have
n
ft*) = g{f\W, /2(-V), ■ . - , fnW) = ^ ^W
/=1
for all x G X, as desired.
Problem 29.4. Prove the converse of Theorem 28.7. That is, show that ifX and
Y are (nontrivial) normed spaces and L(X,Y) is a Banach space, then Y is a
Banach space.
Solution. Let {yn} be a Cauchy sequence of Y. Pick some / in X* with
/ 7^ 0, and then consider the sequence of operators {Tn} of L(X, Y) defined by
Tn(x) = f(x)yn. The inequality
||T„Cv) - rw(.v)|| = \\f(x)(yn - ym)\\ < \\f\\ • \\yn - ym\\ • ||jc||,
shows that ||7„ — Tm\\ < \\f\\ • \\yn — ym\\, and so {Tn} is a Cauchy sequence of
L(X, Y). By the completeness of L(X, Y), there exists some T e L(X, Y) with
lim Tn = T. Now, pick some e e X with /(e) = 1, and note that
lim Tn(e) = lim yn = T(e).
n-+oo n-+oo

254
Chapter 5: NORMED SPACES AND LP-SPACES
Problem 29.5. The Banach space B(JN) is denoted by l^. That is, l^ is the
Banach space consisting of all bounded sequences with the sup norm. Consider
the collections of vectors
Co = {x = U'i, a2, A'3,...) e l^ xn -* 0}, and
c = [x = (x\, *2, A3,...) € ^oo: lim a,, emto in JR}.
Show that cq and c are both closed vector subspaces of l^.
Solution. It should be obvious that c$ and c are vector subspaces of l^ (and
that Co is a vector subspace of c). What needs verification is their closedness. To
see that cq is closed, assume that a sequence [xn] of Co, where xn = (a", aJ ,...)»
satisfies ||jc" — A'||oo -* 0. If a* = (a'i, a*2. • • •)» we must show that lim a,, = 0.
To this end, let 6 > 0. Fix some k such that ||aw — x\\ < 6 for all n > k\ clearly
\x" —A/| < 6 also holds for all n > k and all /. Since lim/^oo xf = 0, there exists
some m > k such that |a* | < 6 holds for all / > m. Now, notice that if / > m,
then
\Xi\< \Xi-X$\ + \x$\ < 6+6 =26.
This shows that lim xn = 0, as desired.
Next, we shall establish that c is closed. For simplicity, for a sequence x =
(aj, A2l ...)6cwe shall write a^ = limA,,. Now, assume that a sequence {a"}
in c satisfies xn -> a = (aj , a'2, ...) € ^oo. We must show that lim xn exists in JR.
Start by fixing some 6 > 0. Then, there exists some /: such that
\\xn - A'Hoo < * holds for all n > k. (•)
This implies \\xn - jcm||oo < 26 for all n, m > k, and so \x? - xf\ < 26 for all
n,m > k and each i. Consequently, |a£, — a™ | < 26 for all n, w > /:. This shows
that {a^q} is a Cauchy sequence of real numbers. Let Aqq = lim a^ and note that
l*£> - Aool < 2€ holds for all n > k.
We claim that a„ -> a^. To see this, let again 6 > 0 and choose £ so that (•)
is true. Next, fix some r > k such that |a* — a£J < 6 holds for all a > r. Now,
note that if n > r, then
l*/i ~ A'ool < |An ~ A*| + \xkn - A^| + |a£, - A^ < 6 + 6 +- 26 = 46.
This shows that a„ -> Aoo,andsoA-e c. Therefore, c is a closed subspace of t^.
Problem 29.6. Let c denote the vector subspace of £oo consisting of all
convergent sequences (see Problem 29.5). Define the limit functional L:c -> R

Section 29: LINEAR FUNCTIONALS
255
by
L(x) = L(x\,x2l...) = lim xnt
n-+oo
and p\ In -+ IRby p{x) = p{x\,x2,...) = lim sup xn.
a. Show that L is a continuous linear functional, where c is assumed equipped
with the sup norm.
b. Show that p is sublinear and that L{x) = p(x) holds for each x e c.
c. By the Hahn-Banach Theorem 29.2 there exists a linear extension of L to
all of t^ {which we shall denote by L again) satisfying L(x) < p{x)for
all x e too- Establish the following properties of the extension L:
i. For each x € loo, we have
lim inf *„ < L(x) < lim sup xn.
ii. L is a positive linear functional, i.e., x > 0 implies L{x) > 0.
iii. L is a continuous linear functional {and in fact \\L\\ = 1).
Solution, (a) Clearly, L is a linear functional. Moreover, if x = (*i, x2, • • •) £ c,
then|xn| < ||jc ||oo = supm |.rm| for each/2, and so |L(jc)| =lim|^| < H^Hoq. This
shows that L is a continuous linear functional. (Since L{ 1, 1, 1,...) = 1, it is easy
to see that ||L|| = 1.)
(b) The sublinearity of p follows immediately from Problem 4.7. The equality
p{x) = L{x) = lim*„ for each x e c should be also obvious.
(c) We shall establish the stated properties.
(i) If x € loo, then notice that
—L{x) = L{—x) < limsup(—xn) = — liminf*,,,
and so lim inf xn < L{x) < lim supxn holds true.
(ii) If x = {x\, X2,...) > 0 (i.e., if xn > 0 for each n), then it follows from
Problem 4.8 and the preceding conclusion that L{x) > lim inf xn > 0. That is, L
is a positive linear functional.
(iii) If H^Hoo < 1 (i.e., if \xn\ < 1 for each «), then it follows from part (i) and
Problem 4.8 that
-1 < lim inf *,, < L{x) < lim sup xn < 1,
and so \L{x)\ < 1. This implies ||L|| = sup{|LU)|: ||jc||oo < 1) < 1. Since
L(l,l, 1,...)= 1, we easily infer that || L || = 1.

256
Chapter 5: NORMED SPACES AND LP-SPACES
Problem 29.7. Generalize Problem 29.6 as follows. Show that there exists a
linear functional Cim: £oq —> JR. (calle d a Banach-Mazur limit) with the following
properties.
a. Cim is a positive linear functional of norm one.
b. For each x = (x\, a'2,...) € £«>, we have
hmmf < £im(x) < limsup .
«-+«> n w-oo n
In particular, Cim is an extension of the limit functional L.
c. For each x = (x\, a*2, ...) € l^, we have
Cim(x\, a*2, A'3,...) = Cim(x2, A3, A4,...).
Solution. For each a = (x\, A2,...) 6 £00, let
At \ ( A'l+*2 A'i +A2-f-"+An \
be the sequence of averages of a. If we define p: £& -► R via the formula
p(x) = lim sup A(x) = hm sup ,
n—nx) n
then a glance at Problem 4.7 guarantees that p is a sublinear functional. Moreover,
it is easy to see that if a g c, then
L(a) = lim a„ = p(x).
n-*oo
Now, by the Hahn-Banach Theorem 29.2, L has an extension Cim: £oo -» R
satisfying Cim(x) < p(x) for each x e t^. Properties (a) and (b) can be established
exactly as in the solution of Problem 29.6.
To verify (c), let a = (aj , A2,...) € £00 and put y = (A2, A3,...). Then, an easy
computation shows that
/ Aj — A3 A] — A4 A] — A„+i \
AOc-y)-^-*,—;-,^-,...,——,...).
Since a = (aj , A2,...) is a bounded sequence, the latter implies
pU — y) = Hm sup — = 0.

Section 29: LINEAR FUNCTIONALS
257
Hence, Cim(x — y) < p(x — y) = 0. Similarly, L(y — x) < 0, and so Cim(x) —
Cim(y) = Cim(x - y) = 0. Thus, Cim(x) = Cim{y), as desired.
Problem 29.8. Let X be a normed vector space. Show that if X* is separable
(in the sense that it contains a countable dense subset), then X is also separable.
Solution. Let {/, fa,...} be a countable dense subset of X*. For each n
choose some xn e X with \\xn\\ = 1 and \fn(xn)\ > ^||/„||, and let Y be the
closed subspace generated by {-t|, *2,...}. We claim that Y = X.
To see this, assume by way of contradiction that Y ^ X. Fix some a £ Y with
\\a\\ = 1. By Theorem 29.5, there exists some / € X* with f(y) = 0 for all
y e Y and f(a) ^ 0. Given e > 0 choose some n with \\f — fn\\ < £, and
note that
\Ma)\ < \\fn\\ < 2\fn(xn)\ = 2|(/„ - /)(*„)! < 2||/„ - f\\ < 2e.
Thus, |/(a)| < |/(a) - fn(a)\ + \f„(a)\ < 3e holds for all e > 0, and so f(a) =
0, a contradiction. Therefore, Y = X holds.
Now, note that the collection of all finite linear combinations of the countable
set {a*i , a'2, ...} with rational coefficients is a countable dense subset of X.
Problem 29.9. Show that a Banach space X is reflexive if and only if X * is
reflexive.
Solution. Assume that X* is reflexive. If X ^ X**, then by Theorem 29.5
there exists some nonzero F e X*** with F(x) = 0 for each x e X. Since X*
is reflexive, there exists a nonzero x* e X* so that F(f) = f(x*) holds for all
/ e X**. In particular,
x*(x) = x(x*) = F(x) = 0
holds for all x e X, and so x* = 0, a contradiction. Therefore, X must be a
reflexive Banach space.
Problem 29.10. This problem describes the adjoint of a bounded operator. If
T:X —>• Y is a bounded operator between two normed spaces, then the adjoint
ofT is the operator T*: Y* -> X* defined by (T*f)(x) = f(Tx) for all f eY*
and all x e X. (Writing h(x) = (a\ h), the definition of the adjoint operator is
written in "duality" notation as
(Tx,f) = (x,T*f)

258
Chapter 5: NORMED SPACES AND Lp-SPACES
for all f eY* and all x eX.)
a. Show that T*:Y* —► X* is a well-defined bounded linear operator whose
norm coincides with that ofT, i.e., \\T*\\ = ||jT||.
b. Fix some g e X* and some u eY and define S:X —► Y by S(x) = g(x)u.
Show that S is a bounded linear operator satisfying \\S\\ = ||g|| • \\u\\. (Any
such operator S is called a rank-one operator.)
c. Describe the adjoint of the operator S defined in part (b).
d. Let A = [ajj] beanm x n matrix with real entries. As usual, we consider
the adjoint A* as a (bounded) linear operator from R" to Rm. Describe
A*.
Solution. As usual, we shall denote from simplicity T(x) by Tx.
(a) Fix / e Y*. Then, for a, y e X and a, fi e R, we have
(T*f)(ax + fiy) = f(T(ax + fly)) = f(aTx + 0Ty)
= af(Tx) + fifffy) = a(T*f){x) + P(T* f)(y\
so that T*f is a linear functional on X. To see that T* f is also continuous, notice
that
|(rV)(*)| = |/(rjc)| < ii/ii • nr(A)|| < ||/|| • urn • i|a||
holds for all x e X. This shows that T*f is a bounded (and hence, continuous)
linear functional and that || T*f \\ < \\ T\\ • ||/|| holds true for each f eY*.
The last inequality also shows that T*: Y* ~> X* is a bounded operator and
that ||r*|| < ||y ||. For the reverse inequality, let a e X satisfy ||a|| < 1. By
Theorem 29.4 there exists some h e Y* satisfying \\h\\ = 1 and h(Tx) = \\Tx\\.
So,
lir*n > iir/iii > \\t*kx)\\ = \\h(Tx)\\ = faii,
for each a e X with ||a|| < 1. This implies || T \\ =sup{||rA||: ||a|| < 1} < ||r*||.
Hence, 1171 = 1171.
(b) It is a routine matter to verify that S is linear. From
HS(A)|| = \\gWu\\ = \g(x)\.\\u\\
< 11*11 •IWI-II«II = (II«II-II«II)IIj:|I.
we see that S is a bounded operator and that ||5|| < ||g|| • \\u\\. Now, if a e X

Section 30: BANACH LATTICES
259
satisfies ||„r|| < 1, then we have
\\S\\>\\S(x)\\ = \\g(x)u\\>\g(x)\-\\u\U
andso ||5|| > sup{\g(x)\ • \\u\\: x e X and ||jr|| < 1} = \\g\\ • ||u||. The preceding
show that ||5|| = ||£||.||u||.
(c) Note that for each / eY* and each xeXwe have
(S*f)(x) = f(Sx) = f(g(x)u) = f(u)g(x) = [f(u)g]{x).
So, S*f = f(u)g holds for all / e Y*.
(d) Let A = [fl/y] be an m x « real matrix. Note that the norm dual of Rn is
again ]R/\ where every y = (y\ y„) G R" defines a linear functional on R"
via the formula
n
y(x) =(x,y)=x-y = Y^Wi-
This easily implies that the adjoint A* of A is an n x m matrix B = [bij] that
satisfies the duality identity (Ax, y) = (.r, A*y), or Ax • y = „t • £y. That is, the
elements of B satisfy the equation
m n n m
Y H w^t = Yl Jl bJiyjXi
7=1 i = l i = l y = l
for all x e R" and all y e Rm. This easily implies bij = ay/ for all / and j.
Therefore, A* is the transpose of A, i.e. A* = A*.
30. BANACH LATTICES
Problem 30.1. Let Xbea vector lattice, and let f: X+ -> [0, oo) be an additive
function (that is, f(x -f y) = f(x) -f /(y) holds for all x, y e X+). Then show
that there exists a unique linear functional g on X such that g(x) = f(x) holds
forallxeX+.
Solution. Note first that if x > y > 0 holds, then
m = f{y + (* - j)) = /OO + /(* - y) > /OO-
Also, the arguments of the proof of Lemma 18.7 show that f(rx) = rf(x) holds
for all x e X+ and all rational numbers r > 0.

260
Chapter 5: NORMED SPACES AND LP-SPACES
Now, let a > 0 and x > 0. Pick two sequences {/*„} and [tn] of rational
numbers with 0 < rn f a and /„ | a. Then, the inequality r„x < ax < tnx
implies
>'nfW = fO'nX) < f(ax) < f(tnx) = /„/(*),
from which it follows that a/(a) — f(ax) holds.
Now, define g: X —> IR by
g(x) = /(*+) - /(A*").
Note that if x = y — z holds with y, z € X+, then the relation a+ + z = y + a~,
coupled with the additivity of / on X+,showsthat /(a+)+/(z) = /(y)+/(A~).
That is,
Six) = /(*+) - f(x') = f(y) - f(z).
In particular, for a, y e X we have
g(A- + y) = g(x+ + y+ - (A~ + y")) = /(*+ + y+) - /(a" + y")
= /U+) + /(y+)-/(A-)-/(y-)
= [/U+) - /(A")] + [f(y+) - /(y-)]
= gW + giy).
Moreover, for a. > 0 we have
*(«*) = f{ax+) - /(«r) = c*[/(a+) - f{x-)} = <**(*),
and if a < 0, then
g(aA) = -Q!g(-A) = ~a[g(x- - A+)] = -Ci[f(x~) - f(x+)] = Ofg(A).
Thus, g is a linear functional on Xt which is clearly a unique extension of /.
Problem 30.2. A vector lattice is called order complete ifeveiy nonempty subset
that is bounded from above has a least upper bound (also called the supremum of
the set).
Show that if X is a vector lattice, then its order dual X~~ is an order complete
vector lattice.

Section 30: BANACH LATTICES
261
Solution. Let A be a nonempty subset of X" that is bounded from above by
some g e X~~. By replacing A with the set [g — f: f e A], we can assume that
A c X+. Let B denote the collection of all finite suprema of A, i.e., / e B if
and only if there exist f\,..., /„ g A with ./ = V/'=i //• Clearly, f < g also
holds for all f e B. Next, define h:X+ —> R+ by
/*(*) = sup{/(x): /65)
for each a g X+. Clearly, 0 < /z(a) < #(a) holds.
Let a, y G X". Since /(a + y) = /(a) + /(y) < M*) + Ky) holds for all
/ G 5, we see that
A(* + y) < A(-r) -f h(y).
On the other hand, given s > 0 choose /i, fceB such that /z(a) — e < /j(a)
and /?(y) — e < fiiy). Taking into account that f\ v f2 G £, we see that
/Ka) + /Ky) - 2s < /,(a) + My) </|V /2(a) + /, v /2(y)
= /i v/2(A + y)</z(A + y)
holds, for all e > 0. Thus,
/i(a) + /z(y) < h(x + y)
also holds, and so h{x + y) = /z(a) 4- /z(y).
By the preceding problem, h extends uniquely to a positive linear functional.
Clearly, f < h holds for all / G A. On the other hand, if / < 0 holds for all
f £ A, then / < 0 also holds for all f € B. This easily implies h < (j>. That
is, /i = sup/l holds in X~.
Problem 30.3. Show that the collection of all bounded functions on [0, 1] is an
ideal oflR[0A]. Also, show that C[0, 1] is a vector sublattice o/lR[(U1 but not an
ideal.
Solution. Let /: [0, 1] —> 1R be a bounded function. If |/(a)| < M holds for
all a g [0, 1] and g e R[0A] satisfies |#| < |/|,then \g(x)\ < M also holds for
all a G [0, 1]. This implies that the space of all bounded functions is an ideal of
R[0'lj.
The function x^) ls not a continuous function and satisfies 0 < X(o,^) - 1»
where 1 denotes the constant function one on [0, 1]. Hence, C[0, 1] is not an
ideal of Rl0J].

262
Chapter 5: NORMED SPACES AND LP-SPACES
Problem 30.4. Let Xbea vector lattice. Show that a norm || • || on X is a lattice
norm if and only if it satisfies the following two properties:
a. // 0 < x < y, then \\x\\ < \\y\\f and
b. ||a|| = || |*| || holds for all x e X.
Solution. Assume that || • || is a lattice norm. Clearly, 0 < x < y implies
||j?|| < |M|. Also, \x\ = ||jc|| holds, and so \\x\\ < \\\x\\\ < \\x\\.
Conversely, assume (a) and (b) to be true. If |a| < \y\, then
ll^ll = |||jc|||<||lyl|| = lly|l
so that || -|| is a lattice norm.
Problem 30.5. Show that in a normed vector lattice X its positive cone X+ is a
closed set.
Solution. From Theorem 30.1(3) we see that
\x--y-\ = \(-x)+-(-y)+\<\x-y\.
This implies that the function x i—> x~ from X into X is (uniformly)
continuous. Thus, X+ = {x e X: x~ = 0} is a closed set.
Problem 30.6. Let Xbea normed vector lattice. Assume that [xn} is a sequence
ofX such that xn < a„+i holds for all n. Show that if limA*n = x holds in X,
then the vector x is the least upper bound of the sequence [xn] in X. In symbols,
xn f a* holds.
Solution. Assume that {*„} satisfies xn < xn+\ for each n and limjc„ = x.
Then, xn+p — xn > 0 holds for all n and all p and lim/,_>00(A„+;? — xn) = x — xn.
Since (by Problem 30.5) the positive cone X+ is closed, we see that x — xn > 0,
or x > xn for each n. This shows that x is an upper bound for the sequence [xn}.
To see that x is the least upper bound for the sequence [xn}, assume that y > xn
holds for each n. So, y — xn > 0 holds for all n and lim(y — xn) = y — x. Using
once more that X+ is closed, we get y — x e X+. That is, y — x > 0, or y > x.
Therefore, x = sup{An}, or xn t a holds true, as desired.
Problem 30.7. Assume that xn -+ x holds in a Banach lattice and let {€„} be a
sequence of strictly positive real numbers. Show that there exists a subsequence
{xicj of{xn] and some positive vector u such that |^n —x\ < €nu holds for each n.
Solution. An easy inductive argument guarantees the existence of a subsequence
[x^} of {*„} satisfying H*^ — *|| < €n2~n foreachn. Now, notice that the series of

Section 30: BANACH LATTICES
263
positive vectors X^i^")-11*** — jc | is absolutely summable. Since X is a Banach
space, u = X^lita*)1 \xk„ —x\ exists in X. Now, a glance at Problem 30.6 shows
that CO"1!**,, — x\ < u for each n. Thus, \xkn — x\ < €nu holds for each n, as
desired.
Problem 30.8. Let T:X -> Y be a positive operator between two normed vector
lattices, i.e, x > 0 in X implies Tx>0inY.IfX is a Banach lattice, then show
that T is continuous.
Solution. Let T: X —► Y be a positive operator, where X is a Banach lattice
and Y is a normed vector lattice. Assume by way of contradiction that T is not
continuous. Then, there exist a sequence {xn} of X and some € > 0 satisfying
xn -+ 0 and \\Txn\\ > € for each n. By Problem 30.7 there exists a subsequence
{yn} of {*„} and some « e X+ satisfying \yn\ < ^u for each n. Now, notice that the
positivity of T implies \Tyn\ < T\yn\ < jTu for each «, and so ||ry„|| < ^||rw||
for each n. Since ~||7w|| -> 0, it follows that ||ry„|| -> 0 contrary to \\Tyn\\ > e
for each n. Thus, T is a continuous operator.
Problem 30.9. Show that any two complete lattice norms on a vector lattice
must be equivalent.
Solution. If || • ||i and || • ||2 are two complete lattice norms on a vector lattice
X, then, by Problem 30.8, the identity operator /: (X, || • |h) —► (X, || • ||2) is a
homeomorphism. That is, || • |h and || • H2 are two equivalent norms.
Problem 30.10. The averaging operator A: l^ -> l^ is defined by
a, x / *l+*2 *l+*2+.*3 JC1+X2+ •••+*„ \
for each x = (x\, xo,...) 6 ioo- Establish the following:
a. A is a positive operator.
b. A is a continuous operator.
c. The vector space
V = {.v = (jc,, jt2,...) € £00: {*'+*+•"+*"} converges in R}
is a closed subspace oftoo. IsV = ^oo?
Solution, (a) If* = (*i, jc2,...) > 0, then */ > 0 for each / and so *i+*-+-+*" >
0 for each n. This implies A(x) > 0, and so A is a positive operator.

264
Chapter 5: NORMED SPACES AND LP-SPACES
(b) By Problem 30.8 every positive operator on a Banach lattice is continuous.
Therefore, A (as a positive operator) is continuous.
(c) We know from Problem 29.5 that the vector space of all convergent sequences
c = {a* = (a'i , Jt'2,...) € loo', lim xn exists in R}
n—>oo
is a closed subspace of ^oo. Clearly, V = A~](c). Since A is continuous, the latter
guarantees that V is a closed subspace of l^.
There are bounded sequences having divergent sequences of averages. Here is
an example:
(1,-1,1,1,-1,-1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,...).
Hence, V is a proper closed subspace of £qo-
Problem 30.11. This problem shows that for a normed vector lattice X its norm
dual X* may be a proper ideal of its order dual X^. Let X be the collection of all
sequences [xn} such that xn = 0 for all but a finite number of terms (depending
on the sequence). Show that:
a. X is a function space.
b. X equipped with the sup norm is a normed vector lattice, but not a Banach
lattice.
c. Iff: X ->R« defined by /(a) = J^L\nxn for each x = [xn] e X, then
f is a positive linear functional on X that is not continuous.
Solution, (a) Routine.
(b) If x„ = (1, |, j,..., £,...), then {x„} is Cauchy sequence of X that does
not converge in X.
(c) Clearly, / is a positive linear functional. If e„ denotes the sequence whose nlh
component is one and every other zero, then HeJIoo = 1 and n = /(e„) < ||/||.
That is, H/ll =oo.
Problem 30.12. Determine the norm completion of the normed vector lattice of
the preceding problem.
Solution. Let
co = {x = (ai,a2, .-.) 6£oo: lim xn =0}.
Clearly, cq is a vector sublattice of l^. Also, it is not difficult to see that Co is
a closed subspace, and so cq is a Banach lattice (with the sup norm). We claim
that Co is the norm completion of the normed vector lattice X of the preceding
problem.

Section 30: BANACH LATTICES
265
To see this, note first that X is a vector sublattice of c0. Now, let x =
(a'i, a'2, ...) e Co and let s > 0. Choose some n with |a*| < e for all k > n,
and note that the element y = (ai, ..., a„, 0, 0,...) e X satisfies ||a —y||oc < £.
Thus, X is dense in Co, and our claim follows.
Problem 30.13. Let CC(X) be the normed vector lattice—with the sup norm—of
all continuous real-valued functions on a Hausdoiff locally compact topological
space X. Determine the norm completion of CC(X).
Solution. Consider the vector space of functions
c0(X) = {/ €C(X): Ve> 0 3 K compact with |/(a)| <eforA i K}.
Clearly, co(X) is a vector sublattice of B(X). We claim that Cq(X) is a closed
subspace. To see this, let {/„} c co(X) satisfy fn —> f in B(X), and let e > 0.
By Theorem 9.2, / e C(X). Pick some n with ||/ — /„||oo < ^» and then select
a compact set K with |/„(-v)| < e for a* $ K. Thus,
\fM\<\f(x)-ftl(x)\ + \fn(x)\ <2e
holds for each x £ Ky and so / € c0(X). Therefore, co(X) (with the sup norm)
is a Banach lattice.
Clearly, CC(X) is a vector sublattice of co(X), and we claim that CC(X) is
dense in Co(X). To see this, let / e cq(X) and let e > 0. Choose some compact
set K with |/(a)| < £ for all x £ K, and then use Theorem 10.8 to pick some
g e CC(X) with g(x) = 1 for all x e K and 0 < g(.t) < I for x i K. Then,
/^ e CC(X) and ||/g - /!loo <e holds, proving that CC(X) = c0(X). Thus,
co(X) is the norm completion of CC(X).
Problem 30.14. Let X and Y be two vector lattices, and let T:X -> Y be a
linear operator. Show that the following statements are equivalent:
a. T(x vy) = 7(a) v T(y) holds for all a, y eX.
b. T(x A v) = 7(a) a T(y) holds for allx,y eX.
c. 7(a) a T(y) = 0 holds in Y whenever x A y = 0 /20/dy 7/1 X.
d. |7(a)| = 7(|a|) holds for all x e X.
(A linear operator T that satisfies the preceding equivalent statements is referred
to as a lattice homomorphism.)
Solution. (1) =» (2) From the identity (a) of Problem 9.1, we get
7(a a v) = 7(a + y-xvy) = 7(a) + T(y) - 7(a v y)
= 7(a) + 7()0 - 7(a) v T(y) = 7(a) a T(y).

266
Chapter 5: NORMED SPACES AND LrSPACES
(2) => (3) If x A y = 0, then
T(x) a r(y) = T(x A y) = 7(0) = 0.
(3) =» (4) Using the identity (e) of Problem 9.1, we see that
|7(a-)| = \T(x+) - T(x-)\ = T(x+) v T(x~) - T(x+) a T(x~)
= r(A-+) v t(x") = ra-+) + r(jc~) - r(A-+) a t(a-)
= T(x+) + no = t(a+ + o = r(|A|).
(4) ==> (1) From the identity (f) of Problem 9.1, we get
T(x v y) = r(i[jr + y + |a- - y|]) = i[r(x) + T(y) + T(\x - y|)]
= I[r(A-) + T(y) + \T(x) - T(y)\] = T(x) v T(y).
Problem 30.15. Lef ■£«, be f/ie Banach lattice of all bounded real sequences;
that is, too = B(JN), and let (n, /'2,...} be an enumeration of the rational
numbers of [0, 1]. Show that the mapping T:C[0, 1] -> ix defined by T(f) =
(/0*i)i fQ'i), • • •) w a lattice isomewy that is not onto.
Solution. Clearly, T is a linear operator. Let / e C[0, 1]. Since / is a
continuous function and the set of all rational numbers of [0, 1 ] is a dense set, it
easily follows that
lr(/)|eo = sup{|/(rfl)|:n = l,2,...}
= sup{|/(jr)|:jre[0,l]} = ||/||ac.
In addition, note that
|7"(/)| = (|/(r,)|, 1/0-2)1,...) = (|/|(r,), |/|(r2)....) = T(\f\),
which shows that T is a lattice isometry.
To see that T is not onto, note that
nf)± (0,1,0,1,...)
holds for each / € C[0, 1].

Section 30: BANACH LATTICES
267
Problem 30.16. Let X be a normed vector lattice. Then show that an element
x G X satisfies x > 0 if and only if f(x) > 0 holds for each continuous positive
linear functional f on X.
Solution. If x > 0 holds, then clearly f(x) > 0 also holds for each 0 < / g
X".
For the converse, assume that x is fixed and satisfies f(x) > 0 for each
/ e X+. Let 0 < / G X" be fixed. Since -g(.t) < 0 holds for all 0 < g < /,
it follows from Theorem 30.3 that
0 < f(x~) = sup{-g(jt): g g X~and 0 < g < /} < 0.
That is, /CO = 0 holds for all 0</gX" and consequently /(*-) = 0 for all
/ G X*. From Theorem 29.4, we see that jc" = 0. Thus,* = x+-x~ = x+ > 0,
as required.
Problem 30.17. Let X be a Banach lattice. 7/0 < x G X, then show that
||*|| = sup{/U): 0<feX* and \\f\\ = 1}.
Solution. Let x > 0. In view of the inequality |/(jc)| < \f\(x) < \\f\\ • ||*||,
we have
||*|| = sup{|/(*)|: /gX* and ||/|| = 1}
< sup{|/|(.r): feX* and ||/|| = 1}
= supj/(.r): 0 < / G X* and ||/|| = 1} < ||*||,
and the conclusion follows.
Problem 30.18. Assume that <p: [0, 1] -> 1R /£ a strictly monotone
continuous function and that 7":C[0, 1] -> C[0, 1] zs a continuous linear operator. If
T(cpf) = cpT(f) holds for each f e C[0, 1] (where (pf denotes the pointwise
product of(p and /). S/jovv z7zaf //zere exzsta a unique function h G C [0, 1 ] satisfying
T(f) = hf for all feC[0, 11
Solution. Taking / = 1, the constant one function, and letting h = 71, we
obtain T(cp) = h(p, and by induction T(cpn) = /z<pn for each « > 0. Hence, by the
linearity of 7, we see that
T{P(<p))=hP(cp)
(*)

268
Chapter 5: NORMED SPACES AND LP-SPACES
for each polynomial P of one variable. Since the function (p is strictly increasing,
the algebra A = [P(cp): P polynomial} separates the points and contains the
constant function 1. Consequently, by the Stone-Weierstrass Theorem 11.5, A is
dense in C[0, 1]. From (•), it easily follows that T(f) = hf for each / e C[0, 1].
Problem 30.19. /// € C[0, 1], then the polynomials
where (") is the binomial coefficient defined by (") = kl,jjLky, are known as the
Bernstein polynomials off.
Show that if f e C[0, 1], then the sequence [Bn] of Bernstein polynomials of
f converges uniformly to f.
Solution. Let [Tfl] be the sequence of positive operators from C[0, 1] into
C[0, 1] defined by
for all / g C[0, 1] and each / e [0, 1]. We must show that
lim||rn/-/||oo=0
holds for each / e C[0, 1]. By Korovkin's Theorem 30.13, it suffices to establish
that lim \\T„f - /||oo = 0 holds for / = 1, x, and x2.
To do this, we need some elementary identities. First note that by the binomial
theorem
^C)r*(l-0"-l = [r + (l-0]" = l (*)
k=0
holds for all /. Differentiating (•), we get
£ {nk)[ktk-\l - t)n~k - (/i - k)tk(\ - t)n-k~l]
k=Q
n
= Z) CD'*"'*1 - 0""*"'(* - nt) = 0.
k=0

Section 30: BANACH LATTICES
269
Multiplication by t(\ — t) yields
n
J^CK(i-o#,"A(*-«o = of
Jfc=0
and by using (•), we see that
EC)**!-')""* = '• (*+)
Differentiating (••) yields
EC)^""ld-0,,-4-1(ik-/i/)=l,
and multiplying by t(\ — t), we get
J2{l)ttk(\-t)k-"(k-nt) = t(\-t).
That is,
E C)(*)2'4o - »k- -1E (D^o - o»- = ^.
and by taking into account (••), we see that
EO^d-o"^-.-^. (***)
The identities (•), (••), and (• * •) can be rewritten as follows:
7„1 = 1, Tnx=x, and [Tnx2 - x2](t) = ^.
Now, note that these identities readily imply that lim 117*,,/ — /||oo =0 holds for
/ = 1, -v, and ,v2.
Problem 30.20. Let T: C[0, 1] -> C[0, 1] 6e a positive operator Show that if
Tf — f holds true when f equals 1, x, and x2, then T is the identity operator
(that is, Tf = / holds for each f e C[0, 1]).

270
Chapter 5: NORMED SPACES AND L^SPACES
Solution. For each n, let Tn = T. Clearly, \imTnf = / holds in C[0, 1] when
/ = 1, A', and a*2. By Korovkin's Theorem 30.13, we have Tf = lim T„f = /
for each / e C[0, 1].
Problem 30.21 (Korovkin). Let {Tn} be a sequence of positive operators from
C[0, 1] into C[0, 1] satisfying Tnl = 1. If there exists some c € [0, 1] such that
\in\Tng = 0 holds for the function g(t) = (f — c)2, then show that \imTnf =
f{c). 1 holds for all f e C[0, 1].
Solution. Let / 6 C[0, 1] and let e > 0. It suffices to show that there exist
constants C\ and C2 such that
|rB/-/(c).i|00<c + cI||rl.i-i||00 + c2flr1,«800
holds for all n.
Set M = ||/1| oq. By the continuity of / at the point c there exists some 8 > 0
such that — e < /(/) — f(c) < £ holds whenever f e [0,1] satisfies \t — c\ < 8.
Next, observe that
-e - ^(r - c)2 < /(;) - /(c) < e + 2j£(r - c)2 (a)
holds for all t e [0,1]. (To see this, repeat the arguments in the proof of
Theorem 30.13.) Since each T„ is positive and linear, it follows from (a) that
-eT„l - yfTng < T„f - /(c) • r„l < eTn\ + ^-Tng.
Put C = ^r, and note that
|7"n/ - /(c) • r„l| < er„l + Cr„g = el + e\T„l - l| + CT„g.
Consequently,
\T„f - /(c) • i| < \T„f - /(c) • r„i| + |/(c)| • |r„i -1|
<el + (e + |/(c)|)|r„l-l|+Cr„g,
and so
||r„/ - /(c) • lj <e + (e + i/(c)i)||rni - nioc + cllr^iioo.

Section 31: L^-SPACES
271
31. Lp.SPACES
Problem 31.1. Let f e Lp(n), and let € > 0. Show that
UL'ilx e X: \f(x)\ > €)) < e~p J \f\p dp.
Solution. Consider the measurable set E = {x e X: \f(x)\ > s), and note that
E = {x eX: \f(x)\p > ep}. Thus,
f\f\pdfi>jxE\f\pdfi>J
\f\pdii> XE\f\pdn> epXEdvi = Epii*{E).
Problem 31.2. Let {/„} be a sequence of some Lp(fx)-space with 1 < p < oo.
Show that if lim \\fn - f\\p = 0 holds in Lp(fi), then {/„} converges in measure
tof.
Solution. From the preceding problem, we see that
/z*({a- € X: \fn(x) - f(x)\ > e)) < e'pf\fn - f\P dn
holds. Clearly, this inequality shows that /„ —> f holds whenever lim ||/„ —
f\\P = o.
Problem 31.3. Let (X, <S, /x) be a measure space and consider the set
E = [xa- A € AM with p,*(A) < oo}.
Show that E is a closed subset of L \ (/x) (and hence, a complete metric space in
its own right). Use this conclusion and the identity
H(AAB) = J\Xa ~ XB\dfM = ||xA - Xfllli
to provide an alternate solution to Problem 14.12(c).
Solution. Assume that {xa„ } is a sequence of E such that J\xa„ — f\dfj. -+ 0
holds for some / e L\(ii). By Lemma 31.6 there exists a subsequence {XALn} of
{XAn} sucn tnat Xal„ -> f a.e. This implies (how?) that / = xa a.e. for some
A € A,* with n*(A) < oo. Thus, / e E and so E is a closed subset of L\((i).

272
Chapter 5: NORMED SPACES AND Lp-SPACES
Problem 31.4. Show that equality holds in the inequality
a'b1'' < f<z + (1 - t)b, 0 < f < 1; a > 0; b>0
if and only if a — b. Use this to show that if f e Lp(n) and g e Lq(fi), where
1 < p, q < oo and 1 + 1 = 1, then f \fg\dfji = ||/||p • \\g\\q holds if and only
if there exist two constants C\ and Ci (not both zero) such that C\\f\p = C2\g\q
holds.
Solution. Clearly, if a = b > 0, then a'b1'1 = ta + (1 - t)b = a holds.
For the converse, let a'b]"~' = ta + (1 — t)b hold for some a, b > 0. Put
y = |, and rewrite the given equality as 1 — f + ty — yr = 0. Since the function
/(a) = l—t + tx—x* for x > 0 (and some fixed 0 < f < 1) attains its minimum
when jc = 1 (see the proof of Lemma 31.2), it follows that y = | = 1, and so
a = Z?. Thus, a'b1-' = /a + (1 - t)b holds if and only if a = b.
For the second part, assume first that there exist two constants C\ and Ci
(which are not both zero) such that Cj|/|p = Ci\g\q. We can assume C\ > 0
and Ci > 0. Then, we have
J\f8\d» = f(%y\g\>\8\^ = (cc;)> j\8\<d»
= [/(§t)MH'-[/lslH'
= (J\f\'dtf .(fwdrf = [f\p.\g\r
For the converse, assume f\fg\dfi = \\f\\p • llgll^. If either / or g is zero,
then the conclusion is trivial. (If / = 0, then put C\ = 1 and Ci = 0.) So, we
can assume / # 0 and g ^ 0. Taking f = i, a = {jj^Y* and Z> = (1gi)<7,
the inequality a'/?1"' < ftf + (1 — f)b gives
Integrating (and using our hypothesis), we get
0< f\l(UM\p + i/risS£)lV -. M£MLlLl ju(A»)
- I 4. I _ /l/(*)gU)|</M(x) _ i i _n
"" /> "*" 9 ll/lfp ll*U, "*

Section 31: LP-SPACES
273
Consequently,
l/(Q«?(.r)l _ 1 (\f(*)\\P , 1 (\tj(x)\\q
WfWp hi, ~~ p \ ll/llp ) ^ q\ lU'll./ /
holds for almost all a, and by the first part of "the problem, we see that (^4ff-)P =
0-f&)q holds, so that
v llftllv '
(H*B,)'|/C*)|P = (ll/llP)PkW|"
holds for almost all a\ as required.
Problem 31.5. Assume that n*(X) = 1 and 0 < p < q < oo. /// is in Lq{p)f
then show that \\f\\p < \\f\\q holds.
Solution. Assume /x*(X) = 1 and 0 < p < q < oo. Let / g Lq(fi). From
Theorem 31.14, we know that Lq{p.) C Lp(/i), and so / € Lp{n).
Put r = - > 1, and then choose 5 > 1 so that £ -f- -j- = 1 holds. Since
\f\p e Lr(p,) and 1 e L5(/x), it follows from Holder's inequality that
{mP)' = f\f\'d» = f\f\p-ldv<(J\f\p'dli)i.(jl'dv)i
. =^\f^dlx)' = [j\f\u^ = {\\f\\qy.
Consequently, \\f\\p<\\f\\q holds.
If q = oo, then
"/Up=(f\f\"d^y < (f {uuooY duf=ii/Hoo=H/n,
also holds in this case.
Problem 31.6. Let f e L\(/jl) n LooO-0- 77ze/2 s/zovv r/zar
a. / e Lp{ii) for each 1 < p < oo.
b. 7/V(X) < oo, //ie/2 lim^oo H/llp = H/Hoo Ao/&.
Solution, (a) If M = ||/||oo, then the inequality
i/i'=i/r!-i/i <*'-'• i/i
shows that / 6 L^m) for each 1 < p < oo.

274
Chapter 5: NORMED SPACES AND LrSPACES
(b) Let {pn} be a sequence of positive real numbers satisfying pn > 1 for each
n and lim pn = oo. From the inequality
it follows that
limsup||/||pn < ||/||to.
Let 0 < e < M. Then, the measurable set
E = {xeX: \f(x)\ > H/IU - e\
satisfies p*{E) > 0. From (||/||oo - e)Pnxe < \f\p\ we see that (ll/IU - e)
[li\E)]» < ||/||^,andso \\fU -* < Hminf \\f\\Pn holds for all 0 < e < M.
That is,
||/||oo< liminf 11/11,..
Thus, lim sup ||/||p„ < ||/||oo < liminf ||/||P/( holds. This shows that lim \\f\\Pn =
H/lloo, and from this it follows that lim^oo \\f\\p = H/Hoo-
Problem 31.7. Let f e L2[0, 1] satisfy ||/||2 = 1 and f* f(x)dk(x) > a > 0.
Also, for each 0 e R let Efi = {a- g [0, 1]: /(a) > j8}. If0<fi< a, show that
UEfi)>{fi-ct)2.
(This inequality is known in the literature as the Paley-Zygmund Lemma.)
Solution. Assume / € L2[0, 1] satisfies the stated properties and let 0 < ft < a.
Then, note that
/-jB<(/-j8)x£„ </xe,.
and so, from Holder's inequality, it follows that
0<a-p< f f{x)dkW-p = / [f{x)-p]dk{x)< f fWxE,Wdk(x)
Jo Jo Jo
< \\fh-[HEp)]'=[X(Ep)]K
This implies \(Ep) > (or - /S)2.

Section 31: Lp-SPACES
275
Problem 31.8. Show that for 1 < p < oo each tp is a separable Banach lattice.
Solution. Let en denote the sequence whose nth component is one and every
other is zero. Also, denote by E the set of all finite linear combinations of
(ei, £2, • • •} with rational coefficients. Clearly, E is a countable set, which we
claim is also dense in lp whenever 1 < p < oo.
To see this, let x = Ui,*2,...) e lp (1 < p < oo), and let e > 0. Fix
some natural number n with J2h=n+\ \x'i\p < T* Then» Vlc^ rational
numbers n, r2,..., rn with Xl?=i l-r< ~" ri\p < T* ^ note *at tne e^ement a =
(/"i1r2,...,r/,,010,...)='*i^i +^"2^2H \-rnen e E satisfies
Ik -*lip = (I> -*l' + E W)* < (* + *)* = s.
That is, the countable set E is dense in lp, and so each ip (1 < p < oo) is a
separable Banach lattice.
Problem 31.9. Show that £«, is not separable.
Solution. Let E = [x\1X2,...} be a countable subset of too- Write xn —
Up jtj, • • •) f°r eacn ^ Now, define
jo if K\>\
y" " }2 if |.r^| < 1 .
Clearly, y = (y\, y2» • • •) € £oo and
lb--tn||0O>bn-<|>|3'n-Kni|>i
holds for each w. Thus, B(y, 1) fl £ =0, and this shows that no countable subset
of too can be dense.
An alternate way of proving that ^ is not separable is as follows: Consider
the set F of all sequences whose coordinates are zero or one. By Problem 2.8,
the set F is uncountable, and it is not difficult to see that ||jc — y||oo = 1 holds
for each pair xy y e F with x ^ y. It follows that {#U, 1): x e F] is an
uncountable collection of pairwise disjoint open balls. This easily implies that
every dense subset of £<» must be uncountable.
Problem 31.10. Show that Loo([0> 1]) (with the Lebesgue measure) is not
separable.

276
Chapter 5: NORMED SPACES AND LP-SPACES
Solution. Write fx = X[o,aJ» 0 < a* < 1. Since ||/A — /Jloc = 1 holds
whenever x ?£ y, it follows that {B(fx, 1): x e (0, 1)} is an uncountable collection of
pairwise disjoint open balls of Loo([0, 1]). This easily implies that every dense
subset of Loo([0, 1]) must be uncountable, and so Loo([0, 1]) is not a separable
Banach lattice.
Problem 31.11. Let Xbea Hausdorff locally compact topological space, and fix
a point a e X. Let p be the measure on X defined on all subsets ofX by p(A) = 1
if a e A and p(A) = 0ifa £ A. In other words, p. is the Dirac measure (see
Example 13.4). Show that p is a regular Borel measure and that Supp p = [a].
Solution. The regularity of p. will be established first.
1) Clearly, p(A) < 1 holds for each KX.
2) Let B CX. If a e B, then
1 = M(£) < inf{/x(0): O open and B c O) < p(X) = 1.
On the other hand, if a £ B, then use the open set X \ {a} to see that
0 = p(B) < inf{/z(e>): O open and B CO] <p(X\ {a}) = 0.
3) Let B C X. If a g B, then each subset C of B satisfies p(C) = 0, and so
0 = sup{/x(AT): K compact and K c B) < p(B) = 0.
Now, if a e B, then using that [a] is a compact subset of B we see that
1 = p,({a}) < sup{/x(AT): K compact and K C B\ < p(B) = 1.
Thus, /x is a regular Borel measure. Since p(X \{a}) = 0, it is easy to see that
Supp/x = {a} holds.
Problem 31.12. Ifg eC][a, b] and f eLx[a, b], then
a. show that the function F: [a, b] ~-> R defined by F(x) = J* fit) dX(t) is
uniformly continuous, and
b. establish the following "Integration by Parts" formula:
f g(x)f(x)d\(x) = g(x)F(x)\h- f g\x)F{x)dx.
Ja lfl Ja

Section 31: LP-SPACES
277
Solution, (a) The uniform continuity of F follows immediately from
Problem 22.6.
(b) Start by choosing some constant C > 0 such that \g{x)\ < Cand|g'U)| < C
hold for each .r e [a,b]. Now, by Theorem 25.3 there exists a sequence of
continuous functions {/„} satisfying lim/a \f — fn\dX = 0. From Lemma 31.6,
we can suppose (by passing to a subsequence if necessary) that there exists some
function 0 < h e L\ [a, b] satisfying \fn\ < h a.e. for each n and /„ —> f a.e. Let
Fn(x) = f*fn(t) dt, and note that by the "standard" Integration by Parts Formula
we have
[ g(x)fn(x)dX(x) = f g(x)fn(x)dx = g(x)Fn(x)\h- f g,(x)Fn(x)dx.
J a J a ,a J a
(*)
From |g/„| < Ch e L\[ayb]y gf„ —>» gf a.e., and the Lebesgue Dominated
Convergence Theorem, it follows that
lim ['g{x)f„{x)dx= [ g(x)f(x)dk(x).
Likewise, the Lebesgue Dominated Convergence Theorem implies
FnW= ffn(0dt—> fXf(t)dX(t) = F(x).
J a J a
for each x e [a, b]. Observing that \g'F„\ < C fahdX and g'Fn -> g'F, the
Lebesgue Dominated Convergence Theorem once more yields
lim f g'(x)Fn(x)dx= [ g\x)F(x)dx.
"-co J a J a
Finally, letting n -» oo in (•), we obtain
/ g(x)f(x)dk(x) = S(x)F(x)\h- f g\x)F(x)dx1
Ju ,a Ja
as desired.
Problem 31.13. Let \x be a regular Borel measure on R". Then show that
the collection of all real-valued functions on R" that are infinitely many times
differentiable is norm dense in Lp{^x)for each 1 < p < oo.

278
Chapter 5: NORMED SPACES AND LP-SPACES
Solution. Let <S be the semiring consisting of the sets of the form n?=i[fl' ♦ ^")-
By Theorem 15.10, the outer measure generated by (R",S,/i) agrees with /x on
the cr-algebra B of all Borel sets of JRn. Thus, what needs to be shown is that given
/ = YYi=\[ai* °i) ^d £ > 0» there exists some C°°-function / with compact
support such that \\xi — f\\p < e-
To this end, let / = fj"=1 [aiy bj) and let s > 0. The arguments of the first part
of the solution of Problem 25.6 show that there is a C°° -function /:R" —> [0, 1]
satisfying / \xi - f\d\x < 2~pep. Since |x/ — f\<2 holds, it follows that
\x,-f\p = {J\x,-f\'dli)l' = (J\xl-fr}-\x,-f\dv)1'
<2(f\xi-f\dn)L' <2-2~1e = £,
Problem 31.14. Let (X, 5, j±) be a measure space with ijl*(X) = 1. Assume that
a function f e Li(/x) satisfies f(x) > M > Ofor almost all a\ Then show that
ln(/) € L,00 and that fln(f)dti < Hffdfi) holds.
Solution. The function g(t) = / — 1 — lnr, t > 0, attains its minimum value
at / = 1. Thus, 0 = g(l) < g(t) = t - 1 - In/ holds for all t > 0, and so
In/ < / — 1. Replacing / by j, the last inequality yields 1 — j < In/. Therefore,
1-7 <lnr </~l (*)
holds for each / > 0.
Since the function In a* is continuous on (0, oo) and / is a measurable function,
it follows that ln(/) is a measurable function. (See the solution of Problem 16.8.)
Replacing / by jjj- in (•), we see that
i-W<K/(*))-mi/ii.)<#ft-i (**>
holds for almost all x. From our assumptions, it is easy to see that both functions
1 - JL0J- and jj£ are integrable. Thus, from (*•) and Theorem 22.6, it follows
that ln(/)€Li(/x).
Finally, integrating the right inequality of (••) (and taking into account that
/z*(X) = 1), we see that
f ln(f)dfj. - ln(||/||,) < Jjfa dn-l=Q.

Section 31: L„-SPACES
279
That is,
y ln(/)rfM < ln(||/||,) = ln(//rfAt)
holds, as required.
Problem 31.15. Theorem 31.7 states that: If \ < p < oo, f in Lp(fi), {/„} c
Lp(n), /„ —> / a.e., and lim ||/fl||p = \\f\\p, then lim ||/„ - f\\p = 0.
Show with an example that this theorem is false when p = oo.
Solution. Consider the sequence {fn} of Loo([0, 1]) defined by /„ = X(i,n-
Then fn —> 1 a.e., and H/J*, = 1 -> 1 = HlHoo. However, ||/„ - l||oo = 1
holds for each n.
Problem 31.16. This exercise presents a necessary and sufficient condition for
the mapping g i—► Fg from Lqo(m) into L*(/x) {defined by Fg(f) = / fgdfi)
to be an isometry.
a. Show that for each g e LqoCaO the linear functional Fg(f) — f fgdfi,
for f e L\{fi), is a bounded linear functional on L\(fi) such that \\Fg\\ <
llglloo holds.
b. Consider a nonempty set X and fi the measure defined on eveiy subset of
X by fi{0) = 0 and fi{A) = oo if A ^ 0. Then show that L\{fi) = {0}
and Loo{fi) = B(X) [the bounded functions on X] and conclude from this
that g e Loo(fJL) satisfies \\Fg\\ = HgH*, if and only if g = 0.
c. Let us say that a measure space (X, <S, fi) has the finite subset property
whenever every measurable set of infinite measure has a measurable subset
of finite positive measure.
Show that the linear mapping g i—> Fg from Lqo(m) into L\{fi) is a
lattice isometry if and only if (X, S, fi) has the finite subset property.
Solution, (a) Let g e LooCaO- Then, for each / e L\(fi), we have |/g| <
IUIIoo-1/l.andso
That is, Fg is a bounded linear functional on L\(fi), and ||F5|| < ||g||oo holds,
(b) Since every nonempty set has infinite measure, it is easy to see that there is
only one step function. Namely, the constant function zero. That is, L\{fi) = {0}
holds. On the other hand, since every one-point set has infinite measure, each
equivalence class of L^fi) consists precisely of one function. This implies that
LooOO = B(X).

280
Chapter 5: NORMED SPACES AND LP-SPACES
Finally, note that in view of L*(/x) = {0}, we must have Fg = 0 for each
g e Looifi). Thus, \\F8\\ = HgHoo holds if and only if WgWn = 0 (i.e., if and only
if * = 0).
(c) Assume that a measure space (X, <S, p) has the finite subset property. Let
0 < g e Loo(fi) and let 0 < e < ||g||oo. The set
E = {xeX: \g(x)\ > \\gU - e)
is measurable and /x*(£) > 0 holds. By the finite subset property, there exists a
measurable set F with F C E and 0 < /z*(£) < oo. Put / = s*n.g*F e Lx(p),
and note that ||/||i = 1. Therefore,
tot > |w| = \ffgd»\ = fF[^]dd > !*!„-*.
Since 0 < e < ||g||oo is arbitrary, \\Fg\\ > \\g\\oo holds. Now, using part (a), we
see that ||£g|| = ||g||oo holds for all g e Loo(aO- Therefore, g i—> Fg is a lattice
isometry.
For the converse, assume that g i—> Fs is a lattice isometry, and let E be
a measurable set with /x*(£) = oo. Then g = xe £ £oo(m)> and so ||£j,|| =
||«||oo = l. Pick some 0 < / € L,(/x) with Fg(f) = f fgdfi = fE f dfi > {.
It is easy to see that there exists a step function 0 < <p < /xe with J <t>dp > ^.
From this, it easily follows that there exists a measurable set F c E with 0 <
fi*(F) < oo.
Problem 31.17. Let (X, 5, /x) 6e a measure space. Assume that there exist
measurable sets E\%..., En such that 0 < /z(£,) < oo for 1 < / < n, X =
U/Li^m and each Ej does not contain any proper nonempty measurable set.
Then show that L^ifx) = L\(fi); that is, show that g \-± Fg from L](/x) to
Llchi) is onto.
Solution. From our assumptions, we see that £/ n Ej = 0 holds whenever
/ ^ j. For each 1 < / < n fix some jc,- 6 £/ and note that /x*({*/}) > 0. If / is
a measurable function and or,- = /(*/), then the set f~] ({a/}) f) £; is nonempty
and measurable. Thus, by our hypothesis, /^({a,}) n £/ = £, holds, and
therefore, / must be constant on each £,-. In other words, / = X)?=i /(A'/)X£,
holds for each measurable function /.
To see that g i—> FR from Lj(/x) to L^(/x) is onto, let £ be an
arbitrary functional in £^(/x). Put cx = £(x£.) for 1 < / < w, and then let g =

Section 3i: L^-SPACES
281
^=i[^)]xE,eL1(/x). Note that
FsiXE,) = / XEigdfi = /[jp£L.]X£| d\i = c/ = F(xe,.).
Consequently,
w> = ^(E/(-v,)x£,) = itnxdFM
1=1 /=i
= £/(.v,)F(X£,) = f(£/(jc,-)Xe,) = f(/)
1 = 1 1 = 1
holds for all / e Li(/x), and so F = F?. That is, g i—> FR is onto.
Problem 31.18. Let (X, <S, /x) be a measure space, and let 0 < p < 1.
a. S/zcw iby a counterexample that \\ -\\p is no longer a norm on Lp{p).
b. /weac/7 /, £ g Lp(/x) ter </(/, *) = f\f-g\p dfi = (||/ - g\\p)p. Show
that d is a metric on Lp(fi) and that Lp{p,) equipped with d is a complete
metric space.
Solution, (a) Let 0 < p < 1 and consider the space L/7([0, 1]). Take / =
X(o^) and g = X(i,i)» and note that
\\f + g\\p = l>2^ = (l)l> +(tf = \\f\\p + \\8\\p.
That is, || • ||p does not satisfy the triangle inequality,
(b) If a > 0 and b > 0 (and 0 < p < 1), then
(a + b)p = (a + b){a+b)p-l=a{a + b)p-{+b{a+b)p-]
< a-aP~] + b-bp~l = ap+bp.
Thus, (a 4- b)p <ap+bp holds for each a > 0 and each b > 0. This inequality
easily implies that d(/, g) = f \f — g\P dp, is a metric on Lp(jjl).
For the completeness, let {/„} be a Cauchy sequence in the metric space
(Lp(p),d), where 0 < p < 1. By passing to a subsequence, we can assume
that / \fn+\ — fn\p dp, < 2~n holds for each n. We shall establish the existence
of some / G Lp(fi) such that lim ||/„ — f\\p = 0.

282
Chapter 5: NORMED SPACES AND LP-SPACES
Set si = 0 and sn = |/,| + |/2 - /i| + ■ • • + |/„ - /„-il for n > 2. Clearly,
0 < gn t and
holds for each n. By Levi's Theorem 22.8, there exists some g e Lp(ii) such
that 0 < gn t g a.e. From
n+k n+L
\fn+k - /n| = £ (// - fi-l) I < 2] U' "" ft'* I = *»+* " *"•
/=/i+l i=n+l
it follows that {/„} converges pointwise (a.e.) to some function /. Since |/n| =
|/i + Y!i=i(fi - fi-\)\ < gn < g hold a.e., we see that \f\<g a.e. also holds.
Therefore, / e Lp(/x). Now, note that \f„ -f\<2g and |/n - f\? —► 0 hold,
and so by the Lebesgue Dominated Convergence Theorem, we see that
d(fn,f)=yv„-/ip<//x-+o.
Therefore, (Lp(/x), d) is a complete metric space.
Problem 31.19. // (X, 5, /x) zs a finite measure space, then show that the vector
space of all step functions is norm dense in Loo(/x).
Solution. Let / € Loo(aO and let e > 0. Choose some C > 0 such that
|/(jc)| < C holds for almost all *, and then pick a partition —C = a0 < ^ 1 <
... < an = C of [—C,C] with o, — fl/_i < £ for each 1 < / < n. Let
£, = /^([ffi-i.fl/)). and note that (since /z*(X) < 00) the simple function
0 = IX=i #/X£, is a step function satisfying ||/ — 0||oo < £.
Problem 31.20. If K is a compact subset of a metric space X, then show that
there exists a regular Borel measure fi on X such that Supp \x = K,
Solution. Let K be a compact subset of a metric space X. Pick a countable
dense subset {jcj, a*2, ...} of AT (see Problem 7.2) and then for each /z consider the
Dirac measure 8Xn supported at the point xn (see Example 13.4). Now, consider
the measure fi:V(X) —> [0, 1] defined by
00
/x(^) = ^2-^AnM) = ^2-w,
1=1 neA
where A = {/2 e IN: xn £ A}.

Section 31: L^-SPACES
283
Clearly, n{X \ K) = 0. On the other hand, if O is an open subset of X
satisfying O C\ K ^ 0, then for some n we have x„ e O, and so ji{0 C\ K) >
It remains to be shown that p. is a regular Borel measure. To this end, let
i4CXbe fixed. Note first that if Cn = (x\,..., xtl] D A C A, then C„ is a finite
set (and hence, a compact set) and, moreover, p{Cn) t P-{A) holds. Therefore,
fi{A) — sup{/x(C): C compact and C c A}.
In the other direction, note that if for each n we consider the open set
On=X\ [xr. 1 < i < n and x( £ A],
then A c On and ^(Ow) | p(A) (why?). Therefore,
/x(i4) = inf{/x(0): O open and A c (9}
also holds, proving that fi is a regular Borel measure.
Problem 31.21. // {/„} /s cr worm bounded sequence of L2{p), f/ze/? s/jow r/?ar
/„//i -> Oa.e.
Solution. Assume that a sequence {/„} C L2(p) satisfies J{fn)2dpL < C for all
«, where C > 0 is a constant. Then,
CO » 00
L/(£)2^<cX><oo
holds. By the series version of Levi's Theorem 22.9, we know that the series
YlfnL) in) defines an integrable function. Therefore, ^ —> 0 a.e. must hold.
Problem 31.22. Let (X, <S, p) be a measure space such that p*(X) = I. If
f,g€ L\(p) are two positive functions satisfying f(x)g(x) > 1 for almost all x.
then show that
(ffdp)-(Jgdp)>l.
Solution. Note that the functions VJ an<3 •%/# both belong to L2(p) and
satisfy «Jf{x)*Jg{x) > 1 for almost all x. Applying Holder's inequality, we

284
Chapter 5: NORMED SPACES AND LP-SPACES
see that
i=jidn < f/fjgdv < (f{/f)2d!J.y. (j^fditf.
Squaring, we get (ffdfji) • (fgdfji) > 1.
Problem 31.23. Consider a measure space (X, <S, /x) with /i*(X) = 1, and let
f,g€ L2(m). // // dp, = 0, then show that
(ffgdpf <[jg2dn-(fgdfi)2]ff2dn.
Solution. Put a = fg dp. Then, using Holder's inequality', we get
\ffgdfi\ = \j{fg-oif)d^\ = \j f{g-a)dn\
< f \f\\g ~ot\dn< (j fdti)1 ■ (J{g - af)-
= (jf2 dfi)" (fg2 dn - 2a jg dfi + a2) *
= (Jf2dix)-[jg2dv-2(jgdv)(Jgdii) + (jgdij)2]-
= (jf2dfJ.f[fg2d(.-(fgdf,)2]\
and our inequality follows.
Problem 31.24. If two functions f,g€ L3(/x) satisfy
\\fh = \\gh = ff2gdfi=\,
then show that g = | /1 a.e.
Solution. Let p — | and (7 = 3, and note that - 4- - = 1. Clearly, f2 G
^/?(m) = £j(aO. and since g € L3(/z), we see that f2g e L](fi).

Section 31: Lp-SPACES
285
Now, using Holder's inequality, we obtain
1 =
jfhd»\<jfM*^\\f2\\P-\\s\\q
= [/(/2)* ^]§ ■ i*e3 = (i/i3)2 ■ i*i3 = i-
and so ff2\g\dfi = ||/2|| • \\g\\q = 1. By Problem 31.4, there exists a constant
C > 0 such that C\f2\P = |g|«, or C|/|3 = \g\\ From ||/||3 = \\gh = 1, we
infer that C — 1, and so |/|3 = |g|3 holds. Therefore,
bigl\f\ = \g\ a.e. (*)
From the relation
Jf2(\g\~g)dfi = jf2\g\dn - jf2gdn = j
\fVdii - 1 = 1 - 1 = 0
and /2(|g| — g) > 0 a.e., we conclude that /2(|g| — g) = 0 a.e. Taking into
account (•), the latter easily implies that g = \g\ = |/| a.e. holds.
Problem 31.25. For a function f € L \ (/x) n Ln{jJi) establish the following
properties:
a. f e Lp{ii) for each 1 < p < 2, and
b. lim^I+||/||p = ||/||,.
Solution. Let / e Lj(/x) fi L2(/x); we can assume that f(x) e IR for each
x € X. Consider the measurable set A = {* € X: |/(jc)| > 1} and then define
the function g: X —► R by
„(v)=(l/WI2 ifxeA
SK' \ \f(x)\ ifx£A,
i.e., g = /2X/i + /Xac- From our hypothesis, we see that g € L\bx).
(a) Let 1 < p < 2. Then, the inequality
l/tol" < I I/(X)|2 if'rGi4 =S(r), (*)
implies / G Lp(ix) for each 1 < p < 2.

286
Chapter 5: NORMED SPACES AND Lp-SPACES
(b) Let a sequence [pn] of the interval [1,2] satisfy pn —> 1. From (•), we
see that \f\Pn < g holds for each n. Now, from \f\Pn —> \f\ a.e. and the
Lebesgue Dominated Convergence Theorem, we infer that
The preceding easily implies that lim ||/||p = ||/||i holds.
Problem 31.26. Assume that the positive real numbers a\,...,an satisfy 0 <
of/ < lfor each i and ]£"=ia/ = 1. If f\,..., fn are positive integrable functions
on some measure space, then show that
a- f?,f?---f?eLiM,and
b- //r/2°2 • • • /„"" dfx < (||/, \Ur(\\f2hr- ■ ■ • (H/J1)0".
Solution. We shall establish the result by using induction on n. For n = 1
the result is trivial. For n = 2, note that /."' € Lx(m) and /,"2 € L±(fj.).
Since (^-) + (^-) = a\ + a-i = 1, it follows from Holder's inequality that
tf'f? s't,(/i) and that
=(ii/.ii,r(»^».r-
For the inductive argument, assume that the result is true for some n. Let
/i. • • • i /«. fn+\ be n 4- 1 integrable positive functions and let ori,..., a„, an+\
be positive constants such that Y^!l=\ <*/ = 1. Put a — XX i ai > 0, and note that
XX i 77 = *• Now, by our induction hypothesis, we have fxa • • • f„a € Li(/x).
Also, applying the case /2 = 2 for a and 1 — a = a„+i, we see that
//r • • • ccr ^ = /(/,- • • • /»-)>«' ^
a a £fl. \a / f \an+\
/,« ■■/„" dll) (] fn+idll)
<[(»/.!.)•-(I/.I.)-]B(|/I+.|1)
-(i/.i.r-(i/.i,ro/-+.i.r.
and the induction is complete.
Qn+1

Section 31: Lp-SPACES
287
Problem 31.27. Let (X, <S, p) be a measure space and let [An] be a sequence
of measurable sets satisfying 0 < p*(An) < oo for each n and \\mp*{An) = 0.
Fix 1 < p < oo and let gn = [p*(An)]~^XAn (n = 1,2,...), where 1 + 1 = 1.
Prove that lim ffgn dp, = 0 for each f e 'Lp{p).
Solution. Pick 1 < q < oo such that 1 + 1 = 1 and let / e Lp(p). Then, by
Holder's inequality, we have
J I fgn dp J = J / (/X/\„)^^|
< (f\fx^pdp)ip(f\gn\«dp)< = (/J/I^m)".
From Problem 22.6, we know that lim/A \f\p dp. = 0, and therefore lim
//#„ dp = 0 likewise holds.
Problem 31.28. Lef (X, <S, /x) be a measure space such that p*(X) = 1. For
each I < p < oo define the set
Sp = [feLl(p):f\f\dp=l and J l/I'd/z = 2 |.
Show that for each 0 < € < 1 there exists some Sp > 0 shc/i r/?ar
H*({x e X: |/U)| > 6}) > <5p
for each f € £p.
Solution. Fix 0 < s < 1. For each f e £p put
£/ = {* e X: |/(x)| > e} and F, = X\Ef = {* 6 X: |/(*)| < e}.
From \f\xFf < £X/7» it follows that fF \f\ dp < sp*{Ff) < £, and so
I \f\dp= [\f\dp- [ \f\dp>l-e. (*)
JEf JX JFf

288
Chapter 5: NORMED SPACES AND LP-SPACES
Now, if 1 < q < oo satisfies ^ -f ^ = 1, then Holder's inequality implies
/j/i^(/j/i^)!(/£i^)!
= (J i/r^)"[M*(£/)]! < 2 *[»•(£/>]*.
ir ,1
A glance at (•) shows that 1 — e < 2* [p*(Ef)\q, or
holds for each / € £p, and the desired conclusion follows.
Problem 31.29. Let (X, <S, /x) be a measure space and let 1 < p < oo tfrtd 0 <
77 < p.
a. Show that the nonlinear function \j/:Lp{p) —> Ln(p), where V(/) =
| /1n, is norm continuous.
b. V fn _> / #^ gn -+ £ AoW //2 Lp(p), then show that
\im J \fflr^\gn\^dti = J i/i^igr dM.
Solution, (a) It should be clear that ^ maps indeed L „(p) into L z: (/z), and that
\j/ is nonlinear. Let /„ -» / in Lp(p) (i.e., let ||/„ — /||p —► 0) and assume
by way of contradiction that V(//i) "A ^(/) in Ln(p). So, by passing to a
subsequence, we can assume that there exists some e > 0 such that
U(fn) - W/)| j = (J ll/nl" - l/l" |5 <//i)? > *• (**)
Now, by passing to a subsequence again, we can assume that there exists some
function 0 < g e Lp(p) such that \f„\ < g /z-a.e. holds for each n and
fn —► / a.e.; see Lemma 31.6. Therefore, the relations \\fft\n - l/l17!" <
(l^r + l/r)" € Li(jLt) and ll/JM/l"!" —► 0 a.e., coupled with the Lebesgue
Dominated Convergence Theorem, imply J\\fn\n — |/r|" dp. —> 0, which
contradicts (••). Consequently, the nonlinear mapping V is norm continuous,
(b) Notice that the two nonlinear functions \f/\:L„(p) -> L__e_(/z) and fa:
' p-n
L-p{ix) ->• Ln(ix), defined by
^,(U) = |W|P-" and ^r2(«) = Iwl"-

Section 31: L^-SPACES
289
are—by part (a)—both norm continuous. Therefore,
fll/J^-l/rlU^O and |||g/1|»-|sr|L-+0.
p-n . n
Now, observe that (L_£_(/x))* =La(fji) holds. Consequently, from the duality
x p-n ' n
{L jl. (£i)» L c. (/x)) , we see that
p-n n
j\fn\p-"\gn\"dll = (\fn\'-\\gn\')
as claimed.
Problem 31.30. Let T: Lp(fji) —► Lp(fi) be a continuous operator, where 1 <
p < co, and let 0 < rj < p. Show that:
a. Iffe Lp(fi), then \f\P-«\Tf\" € L,(/x) and
j\f\P-*\Tf\"dn< \\T\\"(\\f\\p)p.
b. If for some f e Lp(fi) with \\f\\p < 1 we have J \f\p-n\Tf\n dfi =
\\T\\\then |r/| = ||r|||/|.
Solution. Assume T, 77, and / are as stated in the problem.
(a) If 7) = 0 or rj = p, then the desired inequality is obvious. So, assume
0 < n < p and consider the conjugate exponents
'• = ^
and , = (1 - I)"'= £.
Since \f\P~" e Lr(/A) and \Tf\n e Mm), we see that |/|p_''|r/|'' belongs to
L\(ix). Also, applying Holder's inequality with exponents r and 5, we obtain
j\f\p-*\Tf\«dn < [|(i/r")-^]¥-[/(ir/i")'^]"
^(J\f\"d^)e?.(J\Tf\"d^i
= (ii/iiP)p""(iir/nP)'' < (\\f\\Py-,'\\Tr(\\f\\py
= \\T\\"{\\f\\Py.

290
Chapter 5: NORMED SPACES AND LP-SPACES
(b) Assume that some / e Lp(/i) with ||/||p < 1 satisfies
j\f\p"l\Tf\'>dv.= \\T\\\
From Holder's inequality, we see that
II7T = f l/r^ir/i^/x < \\\fr% • \\Tf\%
= (ll/llp)p",,(lir/iii>),'<iirii1'.
Thus./l/l^-'ir/I'dAt = ||l/lp",'||r-|||3r/|,»||J. From Problem 31.4, there exists
a constant c > 0 such that (iTf^)* = c(\f\P-tl)ri or
\Tf\p = c\f\P.
Therefore, \Tf\=k\f\ holds for some A. > 0. This implies A.||/||p = ||r/||p <
lirilH/llp and so k < \\T\\. Also, from
ww = J ur^Tf^dn = j i/rvi/r^ < w,
we see that ||71 < k. Hence, k = ||7||, and so \Tf\ = \\T\\\f\.
Problem 31.31. Let (X, <S, /x) be a measure space and let f e Lp(ii)for some
1 < p < oo. Show that the function g: [0, oo) -> [0, oo] defined by
g(t) = ptp-ln*({xeX: \f{x)\>t})
is Lebesgue integrable over [0, oo) a/?d that
f\f\pdfM= f g{t)dk(t) = p (°VV({*eX: \fW\>t])dt.
J J[0,oo) JO
Solution. Let / e Lp(ii)\ we shall assume that f(x) e R holds for each
x e X. Let T = [0, oo) and consider the product measure space T x X. Also,
let
A = {(t,x)eT xX: 0<f <|/(x)|},
and note that (by Problem 26.8) the set A is \x x A-measurable. Now, consider

Section 31: LP-SPACES
291
the function h:T x X —► [0, co) defined by
u(f ^ \ PtP~l if 0 < r < |/(a-)| fP-\ ,, x
In addition, we have
J[Jh(t,x)dk(t)\dp(x) = j[j ptP-lXA(t,x)dk(t)\dn{x)
r r fl/(A)l i
-jLU **■'*]•»*»
= [\f(x)\pdp(x)<oo.
Jx
By Tonneli's Theorem (Theorem 26.7), the function h is integrable over T x X
and
f\f(x)\pdfi(x) = J[Jh(ttx)dfj.{x)\dHty (*)
Put £, = {jc 6 X: |/(a-)| > t) and note that
[h(t,x)dfj.(x) = f pt?-] dp(x) = pt"'lfJL*(Et).
JX JE,
By Fubini's Theorem (Theorem 26.6), we know that the function
t >_> ptP-lfx*(E()
is integrable over [0, 00) and from (•), we see that
f\f(x)\pdp(x) = p ft"'lfjL*(Et)dm = p fVV(£r)^.
Jx Jt Jo
That the Lebesgue integral fTtp~[p*{E()dk(t) is also an improper Riemann
integral follows from the fact that the function t 1—> p*(Et) is decreasing—and
hence continuous for all but at-most countably many /.
Problem 31.32. Let (X, 5, p) be a measure space and let f:X-+lRbea
measurable function. If p*({x e X: \f(x)\ > t}) < e~l for all t > 0, then show
that f e Lp(p) holds for each 1 < p < co.

292
Chapter 5: NORMED SPACES AND Lp-SPACES
Solution. Let g(t) = /**({* e X: \f(x)\ > /}), t > 0. In view of
Problem 31.31, we must show that f£°tp~lg(t)dX(t) < oo. Since for each t > 0 we
have 0 < g(t) < e~\ it suffices to establish that f£°tp~]e~' dt < oo.
To see this, start by observing that by L'Hopitars Rule we have
limf'"1*-* = lim ^=0.
/-*oo /-*>oo el
So, there exists some M > 0 satisfying 0 < tp~]e~* < M for all t > 0.
Hence,
/•OO /»oo
0< / tp-]e-ldt = / tp-]e-"e~2dt
Jo Jo
/»oo
< / Me~'^ dt = 2M < oo,
Jo
as desired.
Problem 31.33. Consider the vector space of functions
E — {/: R" —> IR| / is a C°°-function with compact support and I f dk = 0}.
Show that for each 1 < p < oo the vector space E is dense in Lp(Rn). Is E dense
//^L1(R,')?
Solution. We shall prove the result for the special case n = 1. The general case
(whose details can be completed as in Problem 25.3) is left for the reader. The
proof will be based upon the following property: If 1 < p < oo, e > 0, h > 0,
and a positive integer n are given, then there exists a C°°-function (j>: R —> R
such that
1. Supp$ is compact and Supp0 c [«, oo);
2. 0 < 0(a) < h for all a* g R;
3. /R0^ = l;and
4. U\\p = (fR(ppdXy <s.
To see this, assume 1 < p < oo, £ > 0, h > 0, and the positive integer n are
given. If k is an arbitrary positive integer, then (by Problem 25.3) there exists a
C°°-function /:R —► R such that:
a. Supp/ c [wf/2 + A + 2];

Section 31: L,-SPACES
293
b. 0 < f(x) < 1 for each x e IR and f(x) = 1 for each x e [n + 1, n +
If c = fRf dX > 0, then the C°°-function 0 = £/ satisfies Supp0 c [/?, /7 -f
* + 2] C [/7,oo), fR(f>dX = 1, and (in view of c = fRf dX > f"+™ldx = k)
0 < 000 < j: for each x € IR. In addition, we have
11011/
■ (I/"1)' -- if ('»'dx]
^=(¥)-•*
In view of 1 < p < oo, we see that lim;_>00(^)/' ♦ k^~x = 0, and so a
sufficiently large k will yield a function 0 with the desired properties.
To complete the proof, let / <= LP(R) and let e > 0. As in Problem 25.5(b)
(how?), there exists a C°°-function g with compact support such that ||/ —gllp <
6. If m = fRg dX = 0, then g e £, and we are done. So, assume that m ^ 0.
Pick a positive integer /? such that Suppg Pi [/z, oo) = 0, and then (by the prior
discussion) pick a C00-function 0 with compact support such that:
Supp0 C [n, oo);
0(a*) > 0 for each x e IR;
fR(/)dX = 1; and
iv. ||4>||p = (/R«K</A)'<^.
Now, consider the function j// = g — rrKp, and note that fe£ and
11/-*1, = |l</-*> + ™*l,
* 1/-*I, + M|WI,
< e + |m||0|| < e + e = 2e.
Therefore, £ is dense in Lp(R).
The vector space E is not dense in Li(IR). For instance, consider the function
/ = X[o.i] € Li(R). If 0 e Lj(R) satisfies f\f -<p\dX < ±, then from
1- f(PdX= f{f-4>)dk< [\f-4>\d\<±,
»/R */ R t/R
it follows that JR(t>dX > 1 — -1 = 1, and so 0 £ £. This shows that £ is not
dense in Li(lR).

294
Chapter 5: NORMED SPACES AND LP-SPACES
Problem 31.34. Let (0, oo) be equipped with the Lebesgue measure, and let
1 < p < oo. For each f e LP(X) let
= * l J /X(0,v
T(f)(x) = a"1 / /x(o,v)^ for x > 0.
7Vzert show that T defines a one-to-one bounded linear operator from Lp((0, oo))
P-\-
into itself such that \\T\\ = -£y.
Solution. For simplicity, we shall write Tf instead of T(f). Consider an
arbitrary function 0 < / € Cc((0too)). Choose some M > 0 so that 0 <
f{x) < M holds for all x > 0.
If / = /0°°/(r) dt, then the function
g« =
M if 0 < x < 1
^ if a- > 1
belongs to L/7((0, oo)). Since 0 < Tf < g holds, we see that Tf belongs to
Lp((0, oo)). Also, in view of the inequalities
o < a(i J'xf(t)dt)p = A-fr/CA-)]" < x[g(x)]P,
it follows that
Now, integrating by parts and using Holder's inequality, we get
/»oo
= Ay m[TmY~ldx
= A«/l-(«r<T-

Section 31: L^-SPACES
295
This easily implies that
\\Tf\\p<^\\f\\p
holds for all / e Cc((0, oo)). In other words,
r:Cc((0,cx)))—>Lp((0,oo))
defines a continuous operator such that ||7|| < -^ holds.
Since Cc((0, oc)) is norm dense in L/7((0, oo)) (Theorem 31.11), 7 has a
unique continuous (linear) extension 7* to all of Lp((Q, oo)) such that ||7*|| <
^y holds. Our next objective is to show that T*f(x) = \ f*f(t)d\(t) = 7/(a)
holds for all / 6 L/7((0, oo)) and all a > 0.
To this end, let 0 < 0 be a step function. Choose some C > 0 satisfying
0 < <t>(x) < C for all x > 0. By Theorem 31.11, there exists a sequence {/„} of
Cc((0, oo)) with lim f\f„— <P\P dk = 0. We can assume that lim /„(a) = 0(a)
holds for almost all a (see Lemma 31.6). In view of
|/„ A C - 0 | = \fn A C - 0 A C | < | fn - C I
replacing {/„} by {/„ a C}, we can assume that 0 < f„(x) < C holds for all
a > 0 and all n. Since lim \\Tfn — 7*011^ = 0, we can also assume (by passing
to a subsequence) that 7/„(a) —> 7*0(a) holds for almost all x. Next, observe
that for each fixed x > 0 we have 0 € L \ ((0, a)) and so, by the Lebesgue
Dominated Convergence Theorem, we see that
7*0(*)= lim 7/„(a)= lim i [Mt)dUf) = \ f 0(0^(0
holds for almost all a. Now, let 0 < / e Lp((0, oo)). Choose a sequence {0,,}
of step functions with 0 < 0„ t /• In view of
lim||7*0/l-7*/ll/?=O,
we can assume that 7*0„(a) —> T*f(x) holds for almost all x. Taking into
account that for each fixed a > 0, we have / e Lp((0, a)) c Lj((0, a)), the
Lebesgue Dominated Convergence Theorem implies
r/W= lim T*4>„(x)= lim i rct>n(t)dX(t) = \. f f(t)dX(t)
n-^oo n->oo JQ JQ
holds for almost all a. Thus, 7* = 7 holds.

296
Chapter 5: NORMED SPACES AND LP-SPACES
Next, we shall show that ||jT|| = -4y holds. We already know that ||r|| < -^
holds. So, it must be established that ||r|| > -£p To this end, let
fnM=\f-i)P" «°<*<l
[0 if x > 1 .
Then, (||/n||p)'' = f0x"~ -l dx = n, and moreover,
Tf„(x) =
H£_ m-(" -"p if 0<jr < 1
l+^P-i) I x-i if a- > 1 .
Consequently, we have
(\\Tf4PY = r\TMx)\pd\(x)
-[^]vy~idx+rx~'dx]
and so
T&*[n + Th]l' = \Tf'\,*[T\-\Mp=\T\-ni.
This implies
m^pfer-D + ^bj]* — A-
from which it follows that ||r|| > -f^- also holds.
Finally, we establish that T is one-to-one. Assume that Tf = 0 holds for
some / € Lp((0, oo)). Then, J*f{t)d\(t) = 0 holds for all * > 0. Now,
by Problem 22.19, we infer that / = 0 a.e. holds, and so the operator T is
one-to-one.

CHAPTER 6
HBLBERT SPACES
32. INNER PRODUCT SPACES
Problem 32.1. Let c\, C2,..., cn be n {strictly) positive real numbers. Show
that the function of two variables (•, •)• Rw x R'1 -* R» defined by (x,y) =
]C/'=i cixiyi> is an inner product on R".
Solution. Notice that for all vectors -t = (x\,..., *„), v = (vi,..., yn) and
z = (zj,..., z„) in R" we have
n n n
(ax + £y, z) = ^c/(ax/+^v/)z/=Qf ]Pc/.r/Z/+j8 ^Tc,:yizi=a(x1 z)+P(y, z),
/=i /=i /=i
U> y) = ^Q-r/ v/ = J^c/v/.r,- = (y, x), and
/=i /=i
/=i
Moreover, (*,*) = ^/Li c/*? = 0 implies Cjxf = 0 for each /, and so (since
Ci > 0 for each /) a*,- = 0 for each /, i.e., x = 0. The above show the function (•, •)
is an inner product on R".
Problem 32.2. Let (X, (•,•)) be a real inner product vector space with complex-
ification Xc. Show that the function (•,•): Xc x Xc -> C defined via the formula
{x + iy% x\ +iy\)= (*, x\) + (y, v0 + /[(y, xj) - (a, V])].
w an inner product on Xc. Also, show that the norm induced by the inner product
(•, •) on Xc is given by
li-v + iy\\ = vW) + (y,y) = (||a||2 + ||y||2 )i.
297

298
Chapter 6: HILBERT SPACES
Solution. Let x\ 4- iy\, A2 4- iy2, A3 4- iy?> € Xc. We check below the properties
of the inner product.
1. (Additivity)
((*i + iy\) + (x2 + iyi), A'3 + iy3)
= (*i + *2 4- i{y\ 4- y2), *3 4- iy3)
= (A'i + A2, *3) + 0>l + 3>2, 3>3> + ' [ (y\ + ?2* A*3) - (Ai + A'2, ^3) ]
= (*i, *3> + (yu ^3) + i[(yux3) - (xuy3)] 4- ((a2, a3) + (y2, y3)
+ l[()>2.*3)-(*2, 3>3)l)
= (ai 4- iyi, a3 + iy3) 4- (a2 + 1^2. *3 + 1^3).
2. (Homogeneity)
((a + !/5)Ui +^]),A'2-fi}'2)
= (aA! - jSyi 4- /(j8xi 4- ayi), a2 4-13^2)
= (ax\ - fiy\, a2) 4- (jSxi 4- cryi, y2) 4-1 [(£ai 4- ayi, a2)
-(qjaj -0yi,y2)]
= (a + 1 j8)[ (A! f a2) 4- (yi, y2) 4-1 [(yi, a2) - (a, , y2)] ]
= (a 4- /j8)(jri 4- iy\, a2 4- iy2).
3. (Conjugate Linearity)
(ai 4- iyi, a2 4- iy2) = (*i, a2) 4- (y\ ,y2) + i [(yi, x2) - (x\, y2)]
= (*i, a-2) 4- (yi, y2) 4- / [(a, , y2) - (yi, a2)]
= (x2n x\) 4- (y2, yi) 4- i[(y2. *i) ~ (x2, yi)]
= f^ + ^^i + '3>i)«
4. (Positivity)
(a! 4-/yi,Aj 4-iyi) = (Ai,A1)4-(yi,yi) > 0.
Moreover,
\x\ 4-/yi,Ai 4-/yi| = (Ai,A])4-(yi,yi) = 0 <=» aj = yi = 0 <=* Ai4-/yi=0.

Section 32: INNER PRODUCT SPACES
299
Problem 32.3. Let Q be a Hausdorjf compact topological space and let \x be
a regular Borel measure on Q such that Supp/x = Q. Show that the function
(-, •): C(Q) x C(Q) -» R, defined by
(/.*) = [ fgdfJ.,
Jo.
is an inner product. Also, describe the complexification ofC(Q) and the extension
of the inner product to the complexification ofC(Q,).
Solution. If f,g,he C(Q) and a, ft e R, then note that
(a/+j8g, h) = [{af+Pg)h dfi = a [ fh dfx+P [ gh dfi = «(/, A)+jB(s, A),
Jn Jo. Jn
(/.*) = [ fgdfi= fgfdn = (g,f), and
Moreover, observe that (since Supp/x = Q) a function / e C(£2) satisfies
(/,/)= IfdfJL <=> / = 0.
The complexification Cc(£2) of C(Q) consists of all complex-valued functions
/ 4- igy where /, g e C(£2). The complex inner product is given by
</.*>= f fgdfi
JQ
for all f,ge Cc(fl).
Problem 32.4. Show that equality holds in the Cauchy-Schwarz inequality (i.e.,
\(x, y)\ = ||jc|| \\y\\) if and only ifx and y are linearly dependent vectors.
Solution. Assume \(x, y)\ = \\x\\ \\y\\. If x = 0, then the conclusion is obvious.
So, assume x =fi 0. Let (*, y) = re'e. Replacing x by e~i9x, we can assume
without loss of generality that (*, y) = r > 0, and so (x, y) = ||.r|| \\y\\. Now,

300
Chapter 6: HILBERT SPACES
notice that for each real X we have
0 < (Xx + y, Xx + y) = X2||a-||2 + X[(jc, y) + 0Ty)] + lly ||2
= k2\\x\\2 + 2k(xty)+\\y\\2
= X2\\x\\2 + 2X\\x\\\\y\\ + \\y\\2
= (^llA'll + ll^ll)2.
So, if X = -jjfjj, then (Xx + J, Ajc + y) = 0 or Xx + y = 0. This implies
\\y\\x — \\*\\y = 0» which means that the vectors x and v are linearly dependent.
If a* and y are linearly dependent, then the equation |(a", v)| = ||a*|| ||y|| should
be obvious.
Problem 32.5. Jfx is a vector in an inner product space, then show that
||X||= SUp |(A,V)|.
Ib'll=i
Solution. If a- = 0, then the conclusion is obvious. So, we consider the case
a' 7^ 0. If ||v|| = 1, then the Cauchy-Schwarz inequality implies |(A\y)| <
11*11 \\y\\ < 11*11- Therefore, we have
sup |(a-,)0I<I|a'||.
For the reverse inequality, let z = a7||a'||. Then, ||z|| = 1, and so
sup |U,y)| > |(A',Z)| = \(x,x/\\x\\)\ = (a,a-)/||a-|| = ||a'||.
li.vlNl
Therefore, ||a'|| = suP||>-h=i l(A"» ^)l» w^^ the supremum being in actuality the
maximum.
Problem 32.6. Show that in a real inner product space x J_ y holds if and only
7/11* + y||2 = ||a-||2 + ||y||2. Does \\x + yf = \\x\\2 + ||y||2 in a complex inner
product space imply x 1. y?
Solution. Let x and y be two vectors in a real inner product space. If a* _L y,
then the Pythagorean Theorem gives ||a- 4- y||2 = ||a'||2 -f ||y||2. Conversely, if

Section 32: INNER PRODUCT SPACES
301
II* + yH2 = IU||2 + ||y||2, then from
\\x + y ||2 = (a + y, x + y) = (a, jc) + (a, y) + (y, a) + (y, y)
= ll*ll2 + (*,)>) + (y,*>+llyll2
= \\x\\2 + 2(x,y)+\\y\\2,
it follows that 2(a\ y) = 0, and so x JL y.
In complex inner product space the Pythagorean identity ||-v+y||2 = ll*ll2+IMI2
does not imply a J_ y. To see this, consider a non-zero vector x and let y = za.
Clearly, ||y||2 = (za, /jc) = ||jc||2. Now, note that
ll* + y||2 = ||JC + !JC||2 = ||(1+|)-V||2 = |1+!|2||JC||2
= 2||A-H2 = ||.r||2 + ||yH2,
while (a, y) = (a, ix) = -z||a||2 # 0.
Problem 32.7. Assume that a sequence [xn) in an inner product space satisfies
(a„,a) -> ||a||2 and \\xj -* ||a||. Show that xn -* a.
Solution. Observe that (jc,,,*) -> ||a||2 implies (a, a„) = (a„,a) -» ||a||2 =
||a||2. So, from
l|A-n-Ai2 = (A,, -A-,A'„ -A')
= ||JcM||2-(.r,JC,l)-U„>.r) + ||.r||2
—> lkll2-lkll2-lkn2-f-||A||2 = o,
it follows that ||a„ — a|| -* 0, i.e., a„ -» a.
Problem 32.8. Let S be an orthogonal subset of an inner product space. Show
that there exists a complete orthogonal subset C such that S CC,
Solution. Assume that S is an orthogonal subset of an inner product space X.
Let C denote the collection of all orthogonal sets that contain S. That is, an
orthogonal set A of vectors of X belongs to C if and only if S C A. If we consider
C partially ordered by the inclusion relation C, then it is easy to see that C satisfies
the hypotheses of Zom's Lemma. Now, notice that any maximal element C of C
is a complete orthogonal set satisfying S c C.
Problem 32.9. Show that the norms of the following Banach spaces cannot be
induced by inner products.

302
Chapter 6: HILBERT SPACES
a. The norm \\x\\ = max{|xi|, |jc2|, ..., |*„|} on JR.".
b. The sup norm onC[a,b].
c. The Lp-norm on any Lp(fi)-space for each 1 < p < oo with p # 2.
Solution, (a) Consider the vectors x = (1,0,..., 0) and y = (0, 1, 0,..., 0).
Clearly,
11^11 = IW = ll^ + yll = ll^-yll = i.
and so
ll* + )>ll2 + ll*--)>ll2 = 2 and 2||a'||2 + 2||j||2 = 4.
Therefore, ||x-{-;y||2+ ||jc-;y||2 ^ 2||jc||2 + 2||)>||2 and consequently the norm || • ||
does not satisfy the Parallelogram Law. This implies that the norm || • || cannot be
induced by an inner product.
(b) Again, we shall show that the sup norm || • ||oo does not satisfy the Parallelogram
Law—and this will guarantee that the sup norm is not induced by an inner product.
To see this, consider the two functions 1 (the constant function one) and/: [a, b] —>»
R defined by f(x) = f5f. Now, note that
IIHIoo = ll/lloo = l, ||1 +/Hoc = 2, and 111-/1100 = 1.
Therefore,
(111 + /Hoc)2 + (111 - /Hoc)2 = 5^4 = 2(||1||00)2 + 2(||/||oo)2,
so that the norm || • ||oo does not satisfy the Parallelogram Law.
(c) Assume that there are two disjoint measurable sets E and F such that 0 <
fi*(E) < oo and 0 < /z*(F) < oo. First, we consider the case p = oo. Then,
note that
IIXeIIoo = llxHIoo = 1 and IIxe + XfIIoo = IIXe - XfIIoo = l.
and consequently,
(llX£ + XfIIoo)2 + (IIxe - XfIIoo)2 = 2 # 4 = 2(||X£lloo)2 + 2(||XfIIoo)2.
This shows that the norm || ♦ ||oo does not satisfy the Parallelogram Law and so is
not induced by an inner product.

Section 32: INNER PRODUCT SPACES
303
Now, consider the case 1 < p < co with p ^ 2. The functions / =
|>*(£)]"X£ andg = [/x1,(F)]"Xf satisfy ll/IU = \\8\\P = 1, and hence,
2(||/||P)2 + 2(||g||A)2=:4 = 22.
Also, from |/ + g\p = 1/ - g\p = I/I77 + l£|p, we see that
(11/ + 8\\p)2 + (11/ - 2U2 = (2*)2 + (2*)2 = 2(2^)2 = 2,+^
Since /? ^ 2, we have 21+^ ^ 22, and so
(11/ + g\\p)2 + (11/ - g\\p)2 * 2{\\f\\p)2 + 2(||£||p)2.
This shows that the norm || • ||^ does not satisfy the Parallelogram Law and so it
is not induced by an inner product.
Problem 32.10. Show that a norm || • || in a complex vector space is induced by
an inner product if and only if it satisfies the Parallelogram Law, i.e., if and only if
\\x + yt + \\x-yf = *{M? + \\y\\l)
holds for all vectors x and y. Moreover, show that if || ♦ || satisfies the Parallelogram
Law, then the inner product (•, •) that induces || • || is given by
U,y) = ^(IU + yll2-lk-3'll2 + Hk + ^ll2-Hl.t-i3'll2).
Solution. If || • || is induced by the inner product (•, •), then for all vectors x and
y, we have
II* + yII2 + ||* - y\\2 = (* + y, x + y) + (x -y,x- y)
= [(x,x) + (y,x) + (x,y) + (y,y)]
+ [{x,x)-{y,x)-(x,y) + (y,y)]
= 2\\x\\2 + 2\\y\\2.
For the converse, assume that the norm || • || satisfies the Parallelogram Law.
Consider, the complex-valued function (•, •) defined by
(*. y) = ^(11* + y\\2 - II* - yll2 + * II* + iy\\2 - * II* - tyII2 )•
Clearly, (x,x) = ||*||2 holds for all vectors x. To finish the solution, we shall

304
Chapter 6: HILBERT SPACES
verify that (•, •) is a complex inner product. Start by observing that
0%AO = ^(ll}' + Al|2-||y-Al|24-/||j + /A||2--/||};-/Ai|2)
= \{ \\x + y\\2- \\x -y\\2-i||(-/)(a + iy)f +1 \\i(x - iy)\\2)
= \ (II* + y\\2 - II* - y\\2 -< II* + >y\\2 +' II* - <:vll2)
Next, note that for all vectors u, v and w, we have
4(w 4- u, w) + 4(w — i>, w)
= [||w + v 4- w||2 - ||w + u - w||2 4-1 ||u 4- v + iwf - i \\u + v -iwf]
4- [||w - v 4- w||2 - ||ii - v - iu||2 4- /||i* - v + iw\\2 — / ||w — u — /w||2]
= [||ii + w + v\\2 + \\u 4- w - v\\2] - [||ii - w 4- u||2 4- ||u - w - v||2]
4- i[||w 4- iw 4- v||2 4- ||w 4- iw - u||2] - z[||w - iw 4- v||2 4- ||u - ziu - u||2]
= 2||w 4- u;||2 4- 2|N2 - 2||k - u;||2 - 2|M|2
+ i[2||w 4- iu;||2 4- 2|M|2 - 2||h - /u;||2 - 2||u||2]
= 2[||ii 4-w||2 - ||u - w||2 4- / ||w 4- /iu||2 — / ||w — /u;||2]
= 8(w, w).
Thus, for all vectors w, v, and w we have
(w 4- v, w) 4- (u — i>, w) = 2(w, w). (•)
When u = w, (•) yields (2w, w) = 2(w, w;). Now, letting m = |(x + y), u =
\(x — y) and w = z in (•), we get
(a, z) 4- ()\ z) = (« + u, z) 4- (« - u, z) = 2(m, z) = (2m, z) = (a + y, z),
which is the additivity of (•, •) in the first variable.
For the homogeneity, note first that
(/a, y) = l-(\\ix + y\\2- ||/a - y\\2 4- / ||/a + /y||2 - / ||/a - iy\\2)
= ^(l|/A'4-y||2~-||/A-);||2 + /||A4-^||2~/||A'-y||2)

Section 32: INNER PRODUCT SPACES
305
\ (ll* + yf ~ II* - 3>ll2 " < IK* - *y)||2 + i IK* + I30II2)
\(II* + y\\2 - II* - y II2 +.1II* + iy\\2 -1II* - iy\\2)
= K*,y).
Now, as in the proof of Lemma 18.7, we can establish that (rx, y) = r(x, y) holds
for each "real" rational number /• and all jc, y € X. Since (•, •), as defined above,
is a jointly continuous function (relative to the norm || • ||), it easily follows that
(our, y) = a(x, y) holds for all a € IR and all x, y e X. Finally, for an arbitrary
complex number a + ijS and arbitrary vectors x and y, note that
({a + iftx, y) = (our + ijSjc, y) = (ax, y) + (jS(fjc). y)
= a(.rf y) + j8(z*f y) = a(.r, y) + £/(,t, y)
= (a+/£)(*, y).
This establishes that (•, •) is an inner product that induces the norm || • ||.
Problem 32.11. Let X be a complex inner product space and letT'.X -+ X be
a linear operator. Show that 7=0 if and only if (Tx, x) = 0 for each x e X.
Is this result true for real inner product spaces?
Solution. Assume that (Tx, x) = 0 holds for all x e X. From the identity
(T(x + y), x + y) = (Tx, x) + (Tx, y) + (Ty, x) + (Ty, y)
and our hypothesis, it follows that
(Tx, y) + (Ty,.r) = 0 (**)
for all x,y e X. Replacing y by ty in (••) yields (Tx,iy) -f (T(iy),x) =
i[-(Tx,y) + (Ty,x)]=0.So,
-(Tx,y) + (Ty,x) = 0 (***)
holds for all.r,y e X. Adding (*•) and (•••), we get 2(Ty,x) = 0 or (Ty,x) = 0
for all x, y e X. Letting x = Ty, we get (Ty, 7y) = 0 and so Ty = 0 for all
y 6 X, i.e. 7=0.
For real inner product spaces the preceding conclusion is false. Here is an
example. Consider the Euclidean space R2 equipped with its standard inner
product and define the linear operator T:R2 -> R2 by T(x) = (—*2>*i) for all

306
Chapter 6: HILBERT SPACES
a- = Ui, x2) G IR2. Clearly, T ^ 0, and
(7a, A') =(~A-2,A'i)-(aI,X2) = -A2A| +A*iA2 =0
holds for all x G IR2.
Problem 32.12. // {xn} is an orthonormal sequence in an inner product space,
then show that lim(An, y) = 0 for each vector y.
Solution. Let [xn] be an orthonormal sequence in an inner product space, and
let y be an arbitrary vector. Then, from Bessel's Inequality, we have
oo
X>^,)OI2<ll>>ll2<oo.
This implies !(*„, y)|2 -^ 0, and so (a„, y) -> 0.
Problem 32.13. The orthogonal complement of a nonempty subset A of an
inner product space X is defined by
A1 = [x G X: x ± y for all y e A}.
We shall denote (A1)1 by A1-1. Establish the following properties regarding
orthogonal complements:
a. A1- is a closed subspace ofX, A C A±A- and A n A1 = {0}.
b. //AC B1thenB±QA±.
c. A1- = A1- = [C(A)]1 = [C(A)]X, where C(A) denotes the vector
subspace generated by A in X.
d. IfM and N are two vector subspaces ofX, then M±± H- NJ~L C (M -f N )±A-.
e. IfM is a finite dimensional subspace, then X = M © M1.
Solution, (a) If a, v € A1- and a, ft are arbitrary scalars, then for each z G A we
have
(ax + fly, z) = a(A, z) + £(y, z) = a0 + /30 = 0,
and so ax + /?y € A1. Therefore, A1 is a vector subspace of X. Since x G A
implies x J_ v for all y e A1, it follows that x G A1-1, i.e., A c A11. Now if
a G A fl A-1, then (a, a) = 0 or a = 0, and thus A n A1 = {0}.
(b) Assume K5 and a g B1. If y e A, then 3? G B, and so y JL a. This
implies a g Ax, and so BL C A1.

Section 32: INNER PRODUCT SPACES
307
(c) From A c A and Part (b), it follows that AL c, AL. Now, let x e A1 and
let y € A. Pick a sequence [yn] c A satisfying yn -> y and note that
(y,*) = lim(y„,jc) = 0.
n—»-og
Therefore, .r gA1. Hence, A1^1, and thus i4± = A 1.
For the other equalities, note first that A c £(A) implies [ C(A) ]L Q A1. Now,
fix jc e i4x, and let y 6 £(/4). Pick y\,..., yk e A and scalars k\,..., A* such
that y = X;f=IA/y/. Then,
it *
/=! 1=1
This shows that x e [ C{A) ]±. Thus, A1 c [ C(A) ]\ and so /I1 = [ C(A) J1.
(d) From M C M + A/, it follows that M1-1 c (M + A/)11 holds. Likewise,
N c M + A/impliesA/11 C (Af + AO11. Therefore, M1J- + A/1-1 c (M + N)11.
(e) Let M be a finite dimensional subspace of dimension n. In order to establish
that X = M 0 M1, we must show that every vector can be written in the form
y + z with y e M and y e Mx. (The uniqueness of the decomposition should be
obvious.)
Start by fixing a Hamel basis {x\, xi,..., .v,,} of M. Replacing (if necessary)
{.ri, *2,..., jc,,} by the normalized set of vectors that can be obtained by applying
the Gram-Schmidt orthogonalization process (Theorem 32.11) to {x\, a'2, ..., a,, },
we can assume that the set {x\, xi,..., a,,} is also an orthonormal set.
Now, fix x e X and consider the vectors
n n
z = ]T(*, **)** and y = x - ]T(jcf **)**•
*=i *=i
Clearly, z e M and since (y, jcjt) = 0 for each k% it easily follows that y € M-1.
Now, note that ;t = y + zeM© M-1-.
Problem 32.14. Let V be a vector subspace of a real inner product space X. A
linear operator L:V -> X is said to be symmetric // (Lx, y) = (a\ Ly) holds
for all x, y e V.
a. Consider the real inner product space C[a, b] and letV = {/ e C2[a,b]:
f(a) = f(b) = 0}. Also, let p eCl[a,b] and q e C[a, b] be two fixed
functions. Show that the linear operator L:V -> C[a,b], defined by
Uf) = (pf')' + qf,
is a symmetric operator.

308
Chapter 6: HBLBERT SPACES
b. Consider Rn equipped with its standard inner product and let A: IR" —> JR"
be a linear operator. As usual, we identify the operator with the matrix
A = [ajj] representing it, where the jth column of the matrix A is the
column vector Aej. Show that A is a symmetric operator if and only if A
is a symmetric matrix. (Recall that an n x n matrix B = [bjj] is said to be
symmetric if by = fey/ holds for all i and j.)
c. Let L:V —> X be a symmetric operator. Then L extends naturally to
a linear operator L:VC = [x + ly: a, y e V) -> Xc via the formula
L(x ■+■ ly) = Lx -I- iLy. Show that L also satisfies (Lw, v) = (w, Lv) for
all u,v eVc and that the eigenvalues ofL are all real numbers.
d. Show that eigenvectors of a symmetric operator corresponding to distinct
eigenvalues are orthogonal.
Solution, (a) If /, g e V, then note that
rb
(Lf,g)= / {[p(x)f'(x)]' + q(x)f(x))g(x)dx
Ja
= / [p(x)f'(x)]'g(x)dx+ f q(x)f{x)g{x)dx
J a J a
rb rh
= pMfMg(xf
" ~ I p(x)f'(x)g'(x)dx + f q(x)f(x)g(x)dx
a Ja Ja
b
= I q(x)f(x)g(x)dx- f p(x)f(x)g'(x)dx
J a J a
(b) Recall that the transpose of a matrix B = [by] is the matrix Bx = [fey/]. In
terms of the transpose, a matrix A is symmetric if and only if A1 = A. Now, our
conclusion follows immediately from the following two identities:
(Ax, y) = (a, Aly) for all a, y € R'\ and
fl/y =(eiyAej).
(c) If u = jc + ly and v = x\ + iy\ are vectors of Vc, then note that
(Lw, v) = (L(x + ly), x\ + iy\) = (Lx + iLy, x\ 4- iy\)
= (Lx, a,) + (I?, y,) + i[(Ly, jc,) - (La, yO]
= (a, Lxfi + (y, Ly,) +1[(y, L*,) - (a, Lyx)}
= (a + /y, Lai + iLy\) = (w, Lv).

Section 32: INNER PRODUCT SPACES
309
Now, assume that X e C is an eigenvalue of L: Vc -» Xc. Fix a unit vector
u e Xc satisfying Lu = Xu, and note that
X = X(u, it) = (Aw, w) = (Lu, it) = (u, Lu) = (w, Aw) = X(u, u) = A.
This shows that A. is a real number.
(d) Assume that L: V -» X is a symmetric operator and let two nonzero vectors
w, v e Vc satisfy Lu = Xu and Lu = pv with A 7^ /z. By part (c), we know that
X and p. are real numbers. Therefore,
(X — p)(u, v) = X(u, v) — p(u, v) = (Aw, u) — (w, pv) — (Lu, v) — (w, Lv) = 0,
and so (w, v) = 0.
Problem 32.15. Let (•, •) denote the standard inner product on R", i.e., (x, )>) =
]T"=1A'/)>/ /or a// *, y e R". Recall that an n x n matrix A is said to be positive
definite if(x, Ax) > 0 holds for all nonzero vectors x e R".
Show that a function of two variables (•,•): R" x R" —► R /.y aw inner product
on R" //"flfld 0/7/>> if there exists a unique real symmetric positive definite matrix
A such that
(x,y) = (x, Ay)
holds for all x, y e R". (It is known that a symmetric matrix is positive definite if
and only if its eigenvalues are all positive.)
Solution. Let (•,•): R" xR" -> R be a function of two variables. Assume first
that there exists a real symmetric positive definite matrix A such that
{x,y) = (x,Ay)
holds for all x, y,z € R". Then, for all x, y e R" and all a, 0 e R, we have
(x,y) = (x, Ay) = (Ax, y) = (y, Ax) = (y,x)y
{ax + Py, 2) = (ax + £y, Az) = a(x, Az) + £(y, Az) = of (jc, z) + fi(y, z), and
(jc, x) = (jc, /Lr) > 0 for all x e R" and (x, x) = 0 «=» a = 0.
This shows that (•, •> is an inner product.
For the converse assume that the function of two variables (•,•): R" x Rn -> R
is a real inner product. Let e\,e2,... ,en denote the standard unit vectors, and so

310
Chapter 6: HILBERT SPACES
each vector a € R" is written as a = YH!=\ xiei- ^ f°N°ws that
/) n n n
(x> y) = (Y^wi* J2 yJeJ)= J2lL,Xiyj(ei' eJ)} = (a'* Ay)
i=l 7=1
for all a*, y e R", where A is the /? x n matrix A = [(e,-, ey)]. Clearly, i4 is a real
symmetric matrix and in view of (a, Ax') = (a, a*), we see that A is also a positive
definite matrix. The uniqueness of A should be obvious.
33. HILBERT SPACES
Problem 33.1. Let (X, S, p) be a measure space and let p: X -> (0, oo) be
a measurable function—called a weight function. Show that the collection of
measurable functions
L2(p)={feM: Jp\f\2 dp < oo)
under the inner product (•,•)• Li(p) x L2(p) -> R, defined by
is a real Hilbert space.
Solution. It should be clear that Li(p) is a vector space. Moreover, since / e
Ln(p) is equivalent to Jpf € L2(m)> it follows from Holder's inequality that
jpfgdpi\< (Jp\f\2dtiy(fp\g\2diiy
< oo,
and so (•, •) is well-defined. We leave it as an exercise for the reader to verify that
(♦, •) is indeed a real inner product. We shall prove that Liip) is a Hilbert space
by establishing that it is complete.
To this end, let {/„} c L2(p) be a Cauchy sequence. That is, for each € > 0
there exists some no such that
II/n " fmf = f P\fn ~ fm'fdfl = f \jpfn - Jp fm f d/1 < €2

Section 33: HILBERT SPACES
311
holds for all n,m > n0. This means that the sequence of functions {^/pfn} £
Z,2(a0 is a norm Cauchy sequence of the Hilbert space /^(/x)- Since L2(/x) is a
Banach space (Theorem 31.5), it follows that there exists some function g e Lidi)
such that
flStfn-gfd^^O.
Now, note that if / = g/^/p, then / € Lp{fi) and
II/„ ~ f\\2 = J P\fn - f |2 dix = y"|VP//i ~ S |2 ^ —> 0.
This shows that Ln{p) is norm complete and hence, it is a Hilbert space.
The reader should also notice that Li(p) is exactly the Hilbert space Li{v) for
the measure v: AM -> [0, oo] defined by
v(A) = f
p(x)d/i(x)
A
for each A e AM.
Problem 33.2. S/zovv r/zar the Hilbert space Li[0, oo) is separable.
Solution. Consider the countable set of functions {/*,,,: k,n = 1, 2,...}, where
, / x f jc" if
/U*)=(0 if
' a-7' if 0 < x < k
k < x.
We know that the continuous functions with compact support are dense in L 2 [0, oo)
(Theorem 31.11) and so, we need only prove that the linear span of {/a,,,} is dense
in the vector space of continuous functions with compact support. Observe that if
this is established, then the linear span of {fk,n} with rational coefficients would
be a countable dense set.
Let/ e L2[0» oo) be a continuous function with compact support, and let € > 0.
Fix an integer k such that f(x) = 0 for all x > k. By the Stone-Weierstrass
approximation theorem, there exists a polynomial P(x) = J2n=Qc"xn satisfying
| f(x) — P(x)| < €/*Jk for each x e [0, k]. Now, notice that if we consider the
function g e L2[0, oo) defined by g(x) = Y^Uo Cnfk.nW* then
~gh = (f°°\Hx)-sW\2dxy = (j \f(x)-P(x)\2dxY
<(jf *")'-«

312
Chapter 6: HBLBERT SPACES
holds. This shows that the linear span of the countable set {/*,«: /:,/? = 1, 2,...}
is dense in L2[0, oo), and so Li[Q, oo) is separable.
Problem 33.3. Let {\f/n} be an orthonormal sequence of fit fictions in the Hilbert
space Lj[a, b] which is also uniformly bounded. If {an} is a sequence ofscalars
such that an\j/n -> 0 a.e., then show that lima,, = 0.
Solution. Fix some constant C such that \i{/n(x)\ < C hold for all n and for all
x e [a, b]. Also, let {a,,} be a sequence ofscalars such that ctnT{/n(x) —► 0 holds
for almost all x.
Next, fix 6 > 0 so that eC2 < ^. Now, by Egorov's Theorem 16.7, there exists
a measurable set E C [a,b] with X(EC) < € such that the sequence of functions
{an\l/n} converges uniformly to zero on E. So, there exists an integer m such that
Itf/i^nOOl < * for all w > 7W and all x € E. Then, we have
\an\2 = f \anM)\2dt = / \an1r„0)\2dt+ f M„{t)\2dt
Ja JE JEc
< [e2dt + \an\2 f \xlf„(t)\2dt
JE JEC
< €2(b - a) + \an\2 [ C2dt
JE<
< €\b-a) + e\ctn\2C2.
This implies \\an\2 < (1 - €C2)\ctn\2 < €2(b - a) for all n > w,or
M<cy/2{b-a)
for all n > m. Since 0 < e < —r? is arbitrary, we have established that an -> 0.
Problem 33.4. Let {</>„} be an orthonormal sequence of functions in the Hilbert
space L2[— 1, 1]. Show that the sequence of functions [i/n}, where
w an orthonormal sequence in the Hilbert space L2[a,b].
Solution. Observe that the inner product satisfies
fh —
(Vr/i.Vrm)= / i/rn(x)^m(x)dx
Ja
Ja

Section 33: HILBERT SPACES
313
Making the substitution t = -^(x — ^r), we have dt = -£radx, and so
Mi, I'm) = / M)4>m(t)dt = &mn.
That is, {^rn} is an orthonormal sequence in the Hilbert space Li[a,b].
Problem 33.5. Show that the norm completion X of an inner product space X
is a Hilbert space. Moreover, if x,y e X and two sequences [xn] and {yn} ofX
satisfy xn —> x and yn —► y in X, then establish that the inner product of X is
given by
(x,y) = lim(jr/M^).
n-*oo
Solution. Assume that X is the norm completion of an inner product space X
and let x, y e X. Pick two sequences [xn] and {y,,} of X such that ||a*„ — .v|| -> 0
and \\yn — y\\ —► 0, where || • || is the norm of X (which is the unique continuous
extension of the norm of X to X). Fix some constant M > 0 such that ||.v„ || < M
and \\yn\\ < M hold for each /?. Then, using the Cauchy-Schwarz inequality, we
have
| U„, y„) - (xmt ym) | = | (x„, yn) - (*„, ym) + (a„, ym) - {xnu ym) \
= | (a'„, y„ - ym) + (a*,, - xm, ym) \
< | (xn, y,i - ym) | + | (xn - xmi ym) |
< \\Xn\\\\yn-ynA\ + \\Xn~Xm\\\\ym\\
< M{\\xn-xm\\ + \\yn-ym\\).
This shows that the sequence of scalars {(a'„, yn)} is a Cauchy sequence and hence,
convergent.
Next, assume two other sequences [x'n] and [y'n] of X satisfy ||a-,', — .v|| -> 0
and Hy,', — y|| —> 0. We can assume without loss of generality that ||a^|| < M and
||y'n || < M holds for each n. By the preceding lim^, y'n) exists, and since
| (xfl, yn) - (a-;„ y'„) | = | (a„, yn) - (a*,,, y'n) + U„, 3;;,) - Cr;„ y,',) |
= I (xnj y„ - y'n) + (a„ - a,',, )/) 1
< 1(^.^.-^)1 + 1(^-^,^)1
< iiajiii^-^ii + ik-^iiii^ii
< Mdl^-^ll + ll^-^H)—*o,

314
Chapter 6: HELBERT SPACES
it follows that lim(A'„, yn) = lim(jc^, y'n). In other words, the formula
(x,y) = lim (*„,?„) (*)
/I-*-00
gives rise to well-defined scalar-valued function onlxl
Now, it should be clear that the properties of the inner product are transferred
via (•) from the inner product of X to the function (•,-):XxX-»C. In other
words, the formula of (•) is an inner product on X. Moreover, from
||jr||2 = lim ||*„||2 = lim (*„,*„) = (*,*),
we see that the inner product given by (•) induces the norm of X.
Problem 33.6. Show that the closed unit ball of £2 is not a norm compact set.
Solution. Let U = [x e £2: 11*11 < 1} be the closed unit ball of l2. Now, for
each n let en = (0,0,..., 0, 1, 0, 0,...), the sequence with 1 in its wth coordinate
and zero elsewhere. Note that \\en || = 1 for each n and thus {en} is a sequence of
the unit ball of £2. (In fact {<?„} is an orthonormal sequence of l2.) Now, notice
that for n ^ m we have \\en —em\\ = y/l. This implies that {e,,} does not have any
Cauchy subsequences—and hence, it does not have any convergent subsequences
either. Now a glance at Theorem 7.3 guarantees that U is not a norm compact
subset of £2 •
Problem 33.7. Show that the Hilbert cube (the set of all x = (x\, x2,...) € £2
such that \xn | < £ holds for all n) is a compact subset ofln-
Solution. Let C = {(xi, jr2>...) € £2' \x„\ < £ for each n = 1,2,...}.
Clearly, C is a closed subset of £2. Thus, in order to establish the compactness of
C, it suffices to prove (by Theorem 7.8) that C is totally bounded.
To this end, let e > 0. Fix some n such that ^ju^-h f < £- Since the set
A = {(jci, ...,*„) e lRn: |jc/| < } for 1 < / < n) is closed and bounded, it must
be a compact subset of JR.". Pick xl,..., xm e A (where x' = (x\,..., x'n)) so
that A C (JJ^j B(x{, e) holds. (We consider, of course, JR." equipped with the
Euclidean distance.) Now, for each 1 < / < m let yf = (x\,..., xln, 0, 0,...).
Then, it is easy to see that C C (J^ ^(^/t 2e) holds in £2. This shows that C
is totally bounded, as required.
Problem 33.8. Show that every subspace M of a Hilbert space satisfies M =

Section 33: HILBERT SPACES
315
Solution. It should be clear that {M)L = M1; see Problem 32.13. Therefore, by
Theorem 33.7, H = ~M © M1. Also, it should be noticed that M c ~M c M±L.
Now, let x G MLL. From H = M0M1, it follows that we can write x = u + v
with// e Mandu e ML. This implies x — u = ve ML and since u e M c M11,
we have x — u e M11. Hence, x — u e M-1 O M-1-1 = {0} or x = w e M.
Therefore, M1-1 C M also holds true, and so MLL = M, as desired.
Problem 33.9. For two arbitrary vector subspaces M and Nofa Hilbert space
establish the following:
a. (M + N)^- =M-LnN±,and
b. ifM and N are both closed, then (M H N)1 = ML 4- N1.
Solution, (a) Let .rlM + iV, Then, .r JL y holds for all v e M and .r ± z holds
for all z e N. That is, x e M1- and :t € /V1. Therefore, (M 4- N)L QM^D NL.
For the reverse inclusion, let x e ML C\ NL and let y e M 4- N. Write
y = u + v with u e M and v e N and note that (x, _y) = (;c, w) 4- (*, u) = 0
holds. Hence, .v e (M + N)1, and therefore MiniV1C(M + N)1. Thus,
(M + N)1 = M-LnNJL.
(b) Now, suppose that M and N are closed subspaces. By the preceding problem
we know that M = M11 and N = N1A-. Now, use part (a) to get
(M-L + N-L)1 = Mn/v.
Therefore,
(M n yv)-1 = [{M1- 4- /v-1-)1]1 = mxT/vt.
Problem 33.10. Le/ X be an inner product space such that M = ML1- holds for
every closed subspace M. Show that X is a Hilbert space.
Solution. We need to show that X is complete in the induced norm. For this, it
suffices to establish that X = X, where X denotes the norm completion of X. (We
already know that X is a Hilbert space; see Problem 33.5.) To this end, let u e X
be a nonzero vector.
The linear functional /: X —► C defined by f(x) = (x, u) is nonzero and
continuous. So, / restricted to X is also continuous and since X is norm dense
in X, it follows that /: X -> C is a nonzero continuous linear functional. In
particular, its kernel M = {x e X: f(x) = 0} is a proper closed subspace of
X. We claim that ML ^ {0}. Indeed, if ML = {0}, then it follows from our
hypothesis that M — MLL = {0}1 = X, which is a contradiction.

316
Chapter 6: HILBERT SPACES
Next, fix a vector u e ML with ||w|| = 1 and let v = f(u)u e X. Now, taking
into account that f{x)u — f(u)x e M holds for each a* g X, it follows that
fix) = /(jc)(k, u) = /(n)(jcf w) = (a, u).
for each a € X. That is, (a, u) = (a, u) for all a g X. Since X is dense in X, we
get (a, v) = (a, u) for all a € X. That is, (a, v - u) = 0 for all a e X, and from
this we conclude that u = v e X. So, X = X, and thus X is a Hilbert space.
Problem 33.11. Consider the linear operator V: Lila, b] -* L2[a, b] defined
by
Vf{x)= ['7(f) dr.
Ja
Show that the norm of the operator satisfies \\V\\ <b — a.
Solution. By Holder's Inequality, we get
iv/(a)i < r \no\dt< f \m\dt
J a Ja
rb .,1 _ rb
s^WM1'*]'
<{b-ay-\\f\\.
Therefore, the norm of V satisfies
l|V/ll2= f Wm\2dt<(b-a) f \\ffdt
J a J a
< (b~a)2\\f\\2.
This implies \\V\\ <b-a.
Problem 33.12. Let {a,,} be a norm bounded sequence of vectors in the Hilbert
space £2, where xn = (a*{\ x\ , A3,.. .)• if for each fixed coordinate k we have
limw_oo a£ =0, then show that
lim(A-,„30 = 0
holds for each vector y e £2-

Section 33: HILBERT SPACES
317
Solution. Choose some X > 0 such that \\xn\\ < X holds for all n. Now,
fix a vector y = (y\,yi,. • •) € I2 and let € > 0. Pick some w satisfying
Since for each fixed k we have lim,,-^** = 0, there exists an integer no
satisfying \YHL\ x'l Jk\ < € for all n > no. Now, using the Cauchy-Schwarz
inequality, we see that for each n > no we have
|(-w)| = |i^r*|<|I>i,*|+ E W5ti
<+( E wi2)'(E '»'2)'
_ 00
. / \~^ . n,'>\ - (
<
k=m+\ ' Nk=m+\
< e + Xe = (1 + A.)e.
Since e > 0 is arbitrary, we have shown that lim„_*oo(-rn, y) — 0.
Problem 33.13. Let H be a Hilbert space and let [xn] be a sequence satisfying
lim (x„,y) = (x,y)
/i-*oo
for each y e H. Show that there exists a subsequence [x/,n} of [x„} such that
Solution. Start by noticing that we can assume without loss of generality that
a- = 0. Therefore, suppose
lim (A-„,y) = 0
/i-*oo
for each y e X. We claim that the sequence {xn} is norm bounded. To see
this, for each n consider the continuous linear functional /„://-> C defined by
fn(y) = (y, xn) for each y e H. By our condition, the sequence of bounded
linear functional {/„} is pointwise bounded. So, by the Principle of Uniform
Boundedness (Theorem 28.8), there exists some C > 0 such that ||/„|| < C for
each n. Now, notice that (by Theorem 33.9) \\xn || = \\fn || holds for each n.
Now, let k\ = 1 and then choose ki > k\ with Ife,, a>2)| < 1. Next, an
inductive argument shows that there exist integers k\ < ki < £3 < • • • < kn <
kn+\ < • • • satisfying
|(ax,axi+1)| < 2~" for each 1 < / < n.

318
Chapter 6: HILBERT SPACES
To finish the solution, we shall show that
Ia^+j:^ +xii+-+Mn \\2< 2+C2
n || — n
holds for each n. To see this, take inner products to get
I ELi*** |2 = EiUfa,, xki) + EiUta,. **,) + - - + EiLita.. *0
II rt II rt2
*[C2+(1 + 2-1 + 2"2 + 2~3 4- ■ - - 4- 2~n)] 2 + C2
"" A22 "" n '
This implies
72 II
as required.
Problem 33.14. Lef p: [a, b] -> (0, 00) be a measurable essentially bounded
function and for each n = 0, 1,2,... let Pn be a nonzero polynomial of degree n.
Assume that
b
p(x)Pn(x)Pm(x)dx = 0 for n # m.
Show that each Pn has n distinct real roots all lying in the open interval (a,b).
Solution. By Theorem 33.12, we know that the sequence of orthogonal
polynomials Pq, P\, P21 • • • is complete and coincides (aside of scalar factors) with
the sequence of orthogonal functions of Li{p) that is obtained by applying the
Gram-Schmidt orthogonalization process to the sequence of linearly independent
functions {1, x, x2y a*3, ...}. In particular, we have fa p{x)xmPn(x)dx = 0 for
all m = 0, 1,..., n — 1. Also, by multiplying each Pn by an appropriate scalar,
we can assume that each Pn has real coefficients and leading coefficient 1.
Now, fix one of these polynomials Pn> where n > 1. First, we shall show that Pn
cannot have any complex roots. If Pn has a complex root, then Pn has a factorization
of the form Pn(x) = [(x + a)2 + fi2]Q(x), where a and /3 are real numbers and Q
is a polynomial of degree «- 2. Hence Q(x)P„(x) = [(x+a)2+ P2][Q(x)]2 > 0,
lim
n—>-oo
/

Section 33: HELBERT SPACES
319
and from \hQ p(x)xmPn(x) dx = 0 for all m = 0, 1,..., n - 1, it follows that
Ja
b
p(x)Q(x)Pn(x)dx = 0,
which is impossible. Hence, each Pn has only real roots. This means that Pn has
a factorization of the form
pn(X) = (X - nnx - r2r- ...(*- nr,
where /*i, /*2,..., t\ are real number and mi,m2,...,mi. are natural numbers
such that m\ -\-mi-\ \-mk = n.
Next, we claim that Pn does not have any root outside of the open interval
(a, b). To see this, assume that one root lies outside of (a, b), say r\ < a. Then the
polynomial Q(x) = (a* — r2)m2 • • • (x — rk)mk has degree less than n and satisfies
Q(x)Pn(x) > 0 for each a < x < b. But then, we have
0<;
Ja
p(x)Q(x)Pn(x)dx=0,
which is a contradiction.
Finally, to see that each root appears with multiplicity one, assume by way of
contradiction that one root has multiplicity more than one, say m \ > 1. If, again
0(x) = [ (* ~ >'2)m2 ■••(*- r*)m* if m, is even
^' 1 (Jf - n)U - /-2r2 • • • (a - rkTk if wi is odd,
then <2 is a polynomial of degree strictly less than n and satisfies Q(x)Pn{x) > 0
for all a < x < b. But then, as previously,
0= I p(x)Q(x)Pn(x)dx>0%
Ja
which is absurd. Hence, each polynomial Pn has n distinct real roots all lying in
the open interval {a, b).
Problem 33.15. In Example 33 J3 we defined the sequence Po, P\, P2,... of
Legendre polynomials by the formulas
1 d" o
2nn\ dxn

320
Chapter 6: HBLBERT SPACES
We also proved that these are (aside of scalar factors) the polynomials obtained by
applying the Gram-Schmidt orthogonalization process to the sequence of linearly
independent functions {1, x, x2,...} in the Hilbert space Li([— 1, 1]). Show that
for each n we have
/>„(!)=! and \\Pn\\=y[3^v
Solution. The proof of the formula Pn{\) = 1 is by induction. Notice that for
n = 0 and n = 1 the formula is trivially true. So, for the induction argument,
assume that P„(l) = 1 holds true for some n. To complete the proof, we must
show that P„+\(l) = 1. To see this, note that
1 dn+]
Pn+lix) = 2«+'(n+l)! </*«+'^ ~ 1)n+'
2"+l(fl
fn)^-"•">']
d"-[2(n + l)x(x2 - If]
2"+](n + l)\dx'
= ^tttt[(a' -,)(A'2"1)n] + ^tit-tM2 ~ l)"
2nn\dxnL J 2nn\ dxn
where the term (a* — \)Q(x) designates the form of the expression
1 dn 1 dn
;(X - l)(X2 - 1)" = ——(* - 1)"+1(A- + D".
2nn\dx" 2"n\dx"
So,/>„+1(l) = P„(l)=l.
Next, we shall compute the norm of P„. Clearly,
p„ll2 = y P„(x)P„(x)dx
(2"n!)2 J_
1 d" d"
—(x2-l)n—(x2-l)"dx.
i dxn dx"
Next, observe that the function (x2 — 1)" = (x — 1)" (x +1)" and all of its derivatives
of order less than or equal to n — 1 vanish at the points ±1. So, integrating by

Section 33: HILBERT SPACES
321
parts n-times and using the previous observation, we obtain
(— IV f1 d2n f—IV r]
(-1)"(2/;)! /•' 2 (2n)! /•■ ,
Using integration by parts to evaluate this last integral gives
f (l-x2)ndx = / (1 -x)n(\ +x)"dx
= —^— / a - x)"-{ (i + .o"*1 d* = • •.
n + 1 7-i
= , "'' n , fd+x^dx
(/? + 1) — (2/i) ;_]
(/z!)222«+l
(2w)! (2/1+1)
Consequently, the norm of Pn is given by
(2/0! 1 r (/?!)222,,+I i 2
1 "" L(2»/2!)2J L(2/i)!(2/! + l)J
-(2»/?!)2J L(2/i)!(2/i + 1)J 2/7 + 1
as claimed.
Problem 33.16. Le/ {ra}a6/i be a family of linear continuous operators from a
complex Hilbert space X into another complex Hilbert space Y. Assume that for
each x e X and each y eY the set of complex numbers {(Ta(x), y): a € A] is
bounded. Show that the family of operators {Ta}a€A is uniformly norm bounded,
i.e., show that there exists some constant M > 0 satisfying \\Ta\\ < M for all
a e A.
Solution. Observe that if Z is a Banach space over the field of complex numbers,
then we may also consider Z as a Banach space over the field of real numbers.
Therefore, the Principle of Uniform Boundedness (Theorem 28.8) can be applied
to any Banach space over the field of complex numbers.

322
Chapter 6: HILBERT SPACES
Fix a vector x e X. For each a e A define the complex valued continuous
linear operator Ba: Y ->• C by
Ba(y) = (y.Taw).
Thus, {Ba}aeA is family of bounded linear operators from the Banach space Y to
the Banach space of complex numbers C. From Theorem 33.9, we know that
\\Ba\\ = \\Ta(x)\\.
Next, notice that for each fixed y e Y, it follows from
|fia(y)l = l(y.ra(jc))| = |(rB(jr),y)l
and our hypothesis that the family of continuous linear operators {Ba )aeA is point-
wise bounded. Hence, by the Principle of Uniform Boundedness (Theorem 28.8),
the family [Ba}aeA is norm bounded. This means that there exists a constant
Mx > 0 (that depends upon x) such that ||Ba|| < Mx for all a e A. Thus, we
have ||ra(*)|| < Mx for all a.
Therefore, the family {Ta}aeA of continuous linear operators Ta:X -> Y is
pointwise bounded. Invoking the Principle of Uniform Boundedness once more,
we conclude that there exists a constant M > 0 satisfying ||ra|| < M for all
a e A.
Problem 33.17. Let {<£„} be an orthonormal sequence in a Hilbert space H and
consider the operator T: H -> H defined by
oo
n=\
where {an} is a sequence of scalars satisfying lima„ = 0. Show that T is a
compact operator.
Solution. Let B be the open unit ball of H. We need to show that T(B) is a
compact set. For this, it suffices to show that T(B) is totally bounded. To this end,
fix € > 0 and observe that there exists an integer m such that |a„| < € holds for
all n > m.
Next, define the operator Tm ://->// by
m
Tm(x) = ^a/(x,0/)<fr.

Section 33: HDLBERT SPACES
323
Clearly, the range of Tm is a finite dimensional subspace of H and thus, Tm is a
compact operator. Therefore, Tm(B) is a totally bounded set. Thus, there exists a
finite set {y\, y2,..., yn} such that for each x e B there exists some 1 < k < n such
that \\Tm{x) -yk\\ < e. Now, Parseval's Inequality ££, |(jc, 0,)|2 < ||x||2 < 1
implies
||rW-rmW||2= J £ a/Uf0/)*/|2= £ |a/|2|U.0,)|2
00
< e2 23 |(*,&)|2<e2|W|2<e2.
Therefore, for each x e B there exists some 1 < k < n such that
\\T(x) - yk\\ < \\T(x) - Tm(x)\\ + \\Tm(x) -yk\<€ + €=2€.
This shows that 7(5) is totally bounded, and hence 7 is a compact operator.
Problem 33.18. Assume that 7, 7*: H -> H are two functions on a Hilbert
space satisfying
(Tx,y) = {x%T*y)
for allx,y e H. Show that 7 and T* are both bounded linear operators satisfying
11711 = 117*11 and \\TT*\\ = ||7||2.
Solution. By the symmetry of the situation, it suffices to show that 7 is a bounded
linear operator. We shall show first that 7 is a linear operator. To this end, fix
x, y G H and two scalars a and p. Then, for each z e H we have
(T(ccx + Py), z) = (ax + 0y, T*z) = a(x, 7*z) + P(y, T*z)
= <x{Tx, z) + 0(Ty, z) = (ctTx + PTy, z),
or (T(ax + Py) - aTx - pTy, z) = 0 for all z € H. This implies (by letting
z = T(ax + Py) - aTx - PTy) that
i.e., that 7 is a linear operator.

324
Chapter 6: HILBERT SPACES
Next, for each y € H with \\y\\ < 1 consider the linear functional fy:H-+H
defined by fy(x) = (a*, T*y). Then, using the Cauchy-Schwarz Inequality, we
see that
\fy(x)\ = |(a, r*(y)l = \<TW, y)\ < \\T(x)\\ \\y\\ < \\T(x)\\
for all y € H with ||y|| < 1. This implies that the set of linear functionals
Ify' \\y\\ < 1} is pointwise bounded and therefore, by the Principle of Uniform
Boundedness, the set of linear functionals {fy: ||y|| < 1} is norm bounded. Thus,
there exists some constant C > 0 such that || fy \\ < C for all y e H with \\y\\ < 1.
Now, assume that y e H satisfies ||y|| < 1 and T*(y) ^ 0. Letting a* =
T*(y)/\\T*(y)l we obtain
llT*m = F^Ol1^*^7^1 = l(jr'r(y))l = lfyMl " C'
This implies
|ini=sup{||r(y)||: ||y||<l}<C,
and thus, T* is a bounded linear operator. By the symmetry of the situation, T is
likewise a bounded linear operator.
Now, note that for all a, y e H with ||a|| = ||y || = 1, we have
\{Tx,y)\ = |(A,ry)| < IIaII ||:ry|| < ||a|| ||r|| \\y\\ = ||7"||,
and hence, ||ya-|| = sup{|CTA, y)\: \\y\\ < \) < \\T*\\ for all unit vectors a € H.
This implies
nni= sup hhjoii < nni.
11x11=1
Using the symmetry once more, we get ||r*|| < ||r||, and so ||r|| = ||r*||.
Finally, for each x e H with ||a|| = 1, we have
lirr*A|| < i!riiiinii|A|| = nrii2, and
lir*A||2 = (r*A, t*x) = (rr*A, a) < urni Ikll Ikii = ||rr*n,
and so by taking suprema, we get ||7T*|| = \\T||2.
Problem 33.19. Show that ifT'.H -> H is a bounded linear operator on a
Hilbert space, then there exists a unique bounded operator T*: H -> H (called

Section 34: ORTHONORMAL BASES 325
the adjoint operator ofT) satisfying
(Tx,y) = (x,T*y)
for all x, y e H. Moreover, show that \\T\\ = ||r*||.
Solution. Assume that T: H -> H is a bounded linear operator on a Hilbert
space. For each fixed y e //, the formula fy{x) = (7\t, y) defines a bounded
linear functional on H. By Theorem 33.9, there exists a unique vector T*y e H
satisfying
fy(x) = (Tx,y) = (x,T*y)
for all x e H. Now, use the preceding problem to conclude that the unique
function T*: H —> H defined above is, in fact, a bounded linear operator satisfying
lir*ii = nrn.
34. ORTHONORMAL BASES
Problem 34.1. Let {e/}/6/ and [fj}j^j be two orthonormal bases of a Hilbert
space. Show that I and J have the same cardinality.
Solution. Assume that {£/}/<=/ and {/;}y6y are two orthonormal bases of a
Hilbert space. First, suppose that / is a finite set. Then, from Theorem 34.2, it
follows that {e/}/e/ is also a Hamel basis and so H is finite dimensional. Since
the fj (as being mutually orthogonal vectors) are also linearly independent, we
conclude that J must also be a finite set. This implies that {//};<=./ is itself a
Hamel basis for //, and so / and J must have the same number of elements.
Now, suppose that / and J are infinite sets. For each / e /, we define the set
of indices // = {j € /: (e,-, fj) ^ 0}. By Theorem 34.2, we know that each 7/ is
nonempty and at-most countable. Next, we claim that
J = \JJi. (*)
To see this, let j e J. Since {£,}/<=/ is an orthonormal basis, it follows from
ParsevaFs Identity that £/€/ |(/y, e()\2 = ||/y ||2 = 1, and so (e/f /,) ^ 0 holds
true for some i e I. Thus, j € 7/ holds true for at least one / e /, and thus
Finally, to see that / and / have the same cardinality use (•) together with
the standard "cardinality" arithmetic; see, for instance, P. R. Halmos, Naive Set
Theory, Springer-Verlag, 1974, pp. 94-98.

326
Chapter 6: HILBERT SPACES
Problem 34.2. Let {e,}/€/ be an orthonormal basis in a Hilbert space H. If D
is a dense subset of H, then show that the cardinality of D is at least as large as
that of I. Use this conclusion to provide an alternate proof of Theorem 34.4 by
proving that for an infinite dimensional Hilbert space H the following statements
are equivalent:
1. H has a countable orthonormal basis.
2. H is separable.
3. H is linearly isometric to l2.
Solution. Let {e,}/G/ be an orthonormal basis in a Hilbert space H and let D be
a dense subset of H. Consider the family of open balls {D (ci, 2) }/6/ * where
B(eh±) = {xeH: lk/-Jc||<±|.
Since ||e/ - ej\\ = y/2 for i ^ j\ it follows that {B(vh i)l/e/ is a pairwise
disjoint family of open sets. Since D is dense in H for each / € /, there exists
some di € D n fl(e,-, 5). Clearly, the mapping / h* </,-, from / into D, is one-to-
one and this shows that D has cardinality greater than or equal of the cardinality
of/.
Next, we shall prove the equivalent statements. To this end, assume that H is
an infinite dimensional Hilbert space.
(1) <=» (2) Let {*i, *2. • • -} be a countable orthonormal basis for //. Then,
the finite linear combinations of the en with "rational" coefficients is a countable
dense set.
Now, assume that H is separable, and let D be a countable dense subset of H. If
[ei }/€/ is an orthonormal basis, it follows form the first part that / has cardinality at
most that of D, and hence, / is at-most countable. Since H is infinite dimensional,
/ must be countable, and so H has a countable orthonormal basis.
(2) => (3) If// has a countable orthonormal basis, then H is linearly isometric
(by Theorem 34.9) to i2(JN) = l2.
(3) ==> (1) Obvious.
Problem 34.3. Let I be an arbitraiy nonempty set and for each i e I let
ei = X{/). Show that the family of functions {^/}/€/ is an orthonormal basis for
the Hilbert space tiU)-
Solution. For each /, let ex = x\i\ a"d note that the family of functions {et-}ieI
is an orthonormal family. Now, notice that if jc = {jc/}/€/ € iiU), then
This shows that {e,-}/€/ is an orthonormal basis for the Hilbert space £2{I).

Section 34: ORTHONORMAL BASES
327
Problem 34.4. Let {e{ }/G/ be an orthonormal basis in a Hilbert space and let x be
a unit vector, i.e, \\x\\ = 1. Show that for each k e IN the set {i G I: \(x, e,)| > j }
has at most k2 elements.
Solution. From Parseval's Identity we know that
Let A = {/ G /: \(x,ef)\ >{}. If A has more than k2 elements, then by choosing
k2 + 1 indices from A, we see that
l = II^H2 = J]lCr^/)|2>(/:2 + l)^> 1,
which is impossible. Therefore, A has at most k2 elements.
Problem 34.5. Let M be a closed vector subspace of a Hilbert space H and let
{^/}i6/ be an orthonormal basis of M; where M is now considered as a Hilbert
space in its own right under the induced operations. If x G H, then show that
the unique vector of M closest to x (which is guaranteed by Theorem 33.6) is the
vector y = £/e/(.r, e,>/.
Solution. Assume that {e,-},-€/ is an orthonormal basis for a closed subspace M
of a Hilbert space H and let x G H be a fixed vector. Note first that Parseval's
Inequality guarantees that ]T/6/ |(jc, e,)|2 < ||jc||2, and so y = Yliel(xt e'\)e'\ *s a
well-defined vector of M.
We claim that x — y _L M. To see this, let z be an arbitrary vector of M, and let
z = 5Z/6/(z» ej)ej be *ts Fourier series expansion as a vector of M. Then, we have
(z, x-y) = ( ]P(z, ef)eh x - ]P(*, £?,>,]
jel /<=/
;'€/ /€/ /€/
= ]T(z, ej){ehx) - £(z, ey)(ey, .v) = 0.
76/ /€/
Now, if z is an arbitrary vector of M, then y — z g M and so x — y _L y — z.
Hence, by the Pythagorean Theorem,
||* - z||2 = ||Cr - y) + (y - z)\\2 = ||x - y||2 + \\y - z\\2 > \\x - y\\2.
This shows that y is the vector in M closest to x.

328
Chapter 6: HILBERT SPACES
Problem 34.6. Let [en} be an orthonormal basis of a separable Hilbert space.
For each n let fn = en+\ — en. Show that the vector subspace generated by the
sequence {/„} is dense.
Solution. We need to show that if a* ± f„ for all /*, then x = 0. So, let a* be a
vector satisfying x ± (en+\ — en) for each n. That is,
0 = (a, en+i - e„) = (a, en+x) - (a, e„).
This implies (a,£„+i) = (x,en) for each/?. If we let <$ = (a, ^0, then 6 = (x,en)
for all h, and by ParsevaTs Identity
oo oo
\\x\\2 = j2\(x>e»)\2 = T,s2
Therefore, <5 = 0, and hence, ||a|| = 0, or x = 0.
Problem 34.7. Show that a linear operator L\H\ —> Hj between two Hilbert
spaces is norm preserving if and only if it is inner product preserving.
Solution. Let L: H\ —► Hi be a linear operator between two Hilbert spaces. If
L is inner product preserving, then
||La||2 = (La,La) = (a,a)=||a||2
holds or || La* || = ||a|| for each a e H\, i.e., L is norm preserving. For the converse,
assume that L is norm preserving. Then, from Theorem 32.6, it follows that
{Lx,Ly)= |(||LA-fLy||2~||LA-L^||2 + /||LA-f/Lj||2-/||LA-/L^||2)
= \{\\Ux + 3OII2 - I|L(a - y)\\2 +1 ||L(a + oOll2 - / I|L(a - />0||2)
= \{l\x + ^H2 - II* - 3>H2 + '«* +/)>||2 ~ /||A - 0'||2)
= (x,y).
That is, L is inner product preserving.
Problem 34.8. Show that the vector space ti(Q) of all square summable complex-
valued functions defined on a nonempty set Q under the inner product
(x,y) = ^x(q)W)
qtQ
is a Hilbert space.

Section 34: ORTHONORMAL BASES
329
Solution. The verification of the inner product properties of the function (♦, •)
are straightforward. We shall show that i2(Q) is a Hilbert space, i.e., complete
under its induced normed.
To see this, assume that Q is an infinite set, and let (a„) C t2(Q) be a Cauchy
sequence. Since for each n we have xn(q) ^ 0 for at-most countably many q e Q,
there exists an at-most countable subset C of Q such that xn(q) = 0 for all
q e Q\C and all n. We consider only the case when C is a countable set, say
C = [q\, q2,...}. For each n, let
yn = (*/i(<7i).*„(<72), •••)•
Then, it is easy to see that we can consider {yn} as a Cauchy sequence in l2. The
completeness of l2 implies that [y,,} converges to some sequence y = (y]y y2,...)
in t2. If x: Q -* C is defined by x(q{) = y{ and x{q) = 0 whenever q e Q\C,
then x e i2(Q) and ||.v„ - jc|| = \\y„ - y\\ -► 0 holds in £2(Q). This shows that
£2(2) is a Hilbert space.
Problem 34.9. Let [ei }/€/ be an orthonormal basis of a Hilbert space H. Show
that the linear operator L: H —► t2(I), defined by
L(a)={(a-, */)}/6/f
is a surjective linear isometry.
Solution. Clearly, L is linear and by Parseval's Identity (Theorem 34.2(5)), it
is also an isometry. We shall verify next that L is also surjective. To this end,
let {A./}/6/ G InU)- From J2iei l^<l2 < °°» li follows that A.,- ^ 0 for at-most
countably many indices /. Assume that {/ e I: A.,- =£ 0} = (A.,,, A./2,...}. (We
consider only the countable case; the finite case is trivial.) Clearly, Y1T=\ l^<»> I2 <
CO.
From the Pythagorean Theorem, we have
ii m ii ^ m
and from this it follows that the series J2T=\ ^',A. ls norm convergent in H. Let
x = Y1^L\ ^ine>n = Hizi ^iei- Then, (a, e,) = A,- for each, / and so L(x) =
{^/}/g/- This shows that L is also surjective, as required.
Problem 34.10. Let [en] be an orthonormal sequence of vectors in the Hilbert
space L2[0, 2n]. Suppose that for each continuous function f in L2[0, In] we
have f = Y^L\(f* en)en. Show that [en] is an orthonormal basis.

330
Chapter 6: HELBERT SPACES
Solution. We need only show that the linear span of the set {en} is dense. Let
e > 0 and let g e L2[0, 2n]. Since the continuous functions are dense in L2K), 2n]
(see Theorem 31.10), there exists a continuous function / e LtEO, 2n] with
11/ — gll < £• By our assumption, we have / = ]C)Jlj(/, en)en and, by Bessel's
Inequality, we know that J2T=\ K/* £«)l2 < °°- Next, choose an integer m such
that [££j(/,e*)l2]'< e,and then let h = ELi (/■**)**. Then,/z is in the
linear span of the sequence {£„}, and moreover
ii* - ah = * - X>> **)** ^ ii* - /ii + £ (^' **>**
<IIS-/ll + [ £ K/,^)|2J2<e + ^ = 2^.
A=m+1
Therefore, the linear span of [en} is dense and hence, {en} is a complete orthonormal
set, i.e., it is an orthonormal basis.
Problem 34.11. Let {(/)„} be an orthonormal sequence of vectors in the Hilbert
space L2[0, 2n]. Suppose that for each continuous function f in LilO, 2n] we
have ll/H2 = Y1T=\ K/» 0n)l2- Show that [<pn] is an orthonormal basis.
Solution. It suffices to show that the linear span of the set [<pn} is dense. Let e > 0
andletg e Li[0, 2n]. Since the continuous functions are dense in L^tO, 27r], there
exists a continuous function / € Li[0, 27r] with ||/ — g|| < 6.
Now, by our hypothesis, we have ||/||2 = Y1T=\ K/» 0«)l2- Choose an integer
m such that [ YT=m K/« 0a)I2] " < e, and note that
ii m ii ii "' ii
g - £(/, foM ± "* - /ii + / - £(/> &)& I
Using once more our hypothesis, we see that
k=\
m ii r K i / m \ \i~\-
g - £(/, <t>k)4>k | < iig - /u+[ £ | (/ - £</. <t>Mi<*i) IJ:
'—' k=\ i—\
00 I
<ll*-/ll + [ £ |(/&)|2]2<e + e = 2€.
A=m+1
Therefore, the linear span of {</>„} is dense and hence, {<£„} is an orthonormal
basis.

Section 34: ORTHONORMAL BASES
331
Problem 34.12. Let (0j, 02, • • •} be an orthonormal basis of the Hilbert space
^2(a0, where \x is a finite measure. Fix a function f e L2O-O and let [ct\, «2>...}
be its sequence of Fourier coefficients relative to {0,,}, i.e., an = f f(pndfi.
Show that (although the series Y1T=\ an<l>n n.eed not converge pointwise almost
everywhere to f) the Fourier series Y1T=\ an4)n can be integrated term-by-term in
the sense that for eveiy measurable set E we have
/ fdn = y2a„ I 4>ndiL.
Je /i=1 Je
Solution. Let sn = YH=\ an<Pn, and note that ||/ — sn\\ -> 0. Now, using the
Cauchy-Schwarz inequality, we see that
\J fdn- jsndfi\2 = \j(f-sn)dfx\2<(j \f -sn\d^
< f \f-sn\2dvL. j \2dn
< \\f-stl\\2n\E)—>0.
Hence, fEfdfj. = Yl™=\ an fE4>n dfM.
Problem 34.13. Establish the following "perturbation" property of
orthonormal bases. If [el }/e/ is an orthonormal basis and {//}/€/ is another orthonormal
family satisfying
X>/-/i||2<oo,
16/
then {//}/<=/ is also an orthonormal basis.
Solution. Let {e,-}/€/ be an orthonormal basis in a Hilbert space //, and let
{//}/€/ be another orthonormal family satisfying £/€/ ||e,- — f\\2 < oo. To
establish that the orthonormal family {//}/e/ is an orthonormal basis, it suffices
to show that if a vector u satisfies u JL f for each / e /, then u = 0. So, fix a
vector u € H such that u _L f for all i e I.
From J2iei Wei "" fi\\2 < °°» we know mat the set / = {/ e I: f ^ e,} is
at-most countable. We distinguish two cases.
CASE I: / is finite, say J = [k\, k2,..., kt).
Let M = {y e H: y _L f for all i' £ J). Then, M is a closed vector subspace
of H satisfying {/*,,..., fkt] ^ M and {e*, ,...,^)CM. Moreover, we claim
that [e^,..., e*,} must be an orthonormal basis for M. Indeed, if * e M satisfies

332
Chapter 6: HELBERT SPACES
z ± e\,r forr = 1,..., £, then (in view x -L // = £/ for each/ £ 7) we have a- _L e{
for each / 6 /. Since {e/}/€/ is an orthonormal basis of //, it follows that x = 0.
Thus, {e^,..., e^} is (as being an orthonormal basis) also a Hamel basis for M,
and so M is ^-dimensional. This implies that {/^,..., /;,} is also a Hamel basis.
The latter implies that every e^r is a linear combination of the vectors /;,,..., fkr
Consequently, u _L eir for each r = 1,..., I, and hence u J_ £, for all / 6 /. This
implies u = 0, and thus, in this case, {//}/e/ is an orthonormal basis.
CASE II: / is countable, say J = {&i, &2, &3, • •.. }•
In this case, choose a natural number i such that
00
£ K.-/*yll2 = *<i.
y=*+i
and let 7i = {*i, £2,..., &*:}. Next, define the vectors
00
gr = a - £ (^' A}/^' r = 1, 2,..., £.
y=*+l
We claim the following:
• If a vector x e H satisfies x _L grforr = 1,2,..., landx _L fjfor j £ J\,
then x = 0.
To see this, assume that vector a e H is orthogonal to gr for r = 1, 2,..., I
and to each /, for j $ J\. Then, for j $ J\, we have (a, e,) = (a, Cj — /)) and
for each 1 < r < £, we have
oo
(a, ekr) = (a, ft) + J^ for. /*/><*• /*;> = °-
y=£+i
Now, from ParsevaTs Identity, we have
oo
lull2 = J2 '<*• *<)i2 = L '<*• **> - /oi2
16/ y=£+l
_ oo
< [ £ ll^-/A;l|2|lkl|2 = ^l|A'||2.
y=£+i *
This implies 0 < (1 — <5)||a||2 < 0, or a = 0, as claimed.

Section 35: FOURIER ANALYSIS
333
Next, we consider the closed vector subspace
M = [yeH: y±fi for all / £/,}.
Clearly, {g\, #2. •••»£*} ^ M. Moreover, if some vector x € M is orthogonal
to gi, g2,..., g£, then by property (•) we have x — 0. This means that the
vector space generated by g\, g2, • • •» #£ coincides with M. Since the orthogonal
vectors/^,, /*,, ..., frt belongtoM, M is ^-dimensional and so {/*£,, /*,,..., /*,}
is a Hamel basis of M. In particular, for each 1 < r < I the vector gr is a
linear combination of the vectors /*,, /;,,..., /^. This implies « _L g, for each
1 < r < I and u JL /) for j £ J\. Using (•) once more, we conclude that it = 0.
Therefore, the orthonormal family {//}/«=/ is an orthonormal basis.
35. FOURIER ANALYSIS
Problem 35.1. Show that sin" x is a linear combination of
{1, sin x, cos a*, sin 2x, cos 2x, sin 3a , cos 3a ,..., sin nx, cos nx }.
Furthermore, show that the coejficients of the cosine terms are zero when n is an
odd integer, and the coejficients of the sine terms are zero when n is an even integer.
Solution. Observe that
elx - e~lx
sin a =
2/
Then, using the binomial theorem, we get
V 2i J (2/)"£gU !(«-*)! J
1 ^r „!(-!/ „_2,),-|
(2i)" f^l k\(n - k)\ J
= (27F t[ |r^i («*« -2^+' »»c - ")*) ]
k=0
1 » r «!(-!/
■rr.
cos(/!_2,)A.]+^g[^sin(/J_2,,v].
(20" f=&U\{n-k)\
Now, observe that if// is odd, then /" = ±i and if n is even, then /" = ±1. Since

334
Chapter 6: HILBERT SPACES
sin" A' is equal to the real part of the preceding expression, we have the following
two cases:
sin" a = Y| —^ cos(/j — 2k)x for n even, and
(2/)" £gL *!(/i-*)! J
sin" x = ——- Y^ —: — sin(/z - 2k)x for n odd.
(2/)" {-zlkHn-kY. J
{2iT$^lk\{n-k)\
Problem 35.2. Show that the Dirichlet kernel Dn and the Fejer kernel Kn
satisfy
if* i r
- / Dn(t)dt = - / Kn(t)dt = \.
Solution. The Dirichlet kernel is given by
1 "
Dn(t) = - + Y^cos kt.
2 w
Integrating gives
1 fn 1 J1 fn
- Dn(t)dt = 1 + -Y^ / cosktdt = l.
* J-JT * fr{ J-7T
Likewise, the Fejer kernel is defined by
Kn(t) = -^— TDk«).
n 4- 1 ^—'
72 + 1 *=0
So, integrating and using the previous result on the Dirichlet kernel, we get
1 Cn 1 " 1 C71 1 n
- Kn(t)dt = ——J2- Dn(t)dt = —— £l = l.
Problem 35.3. Let X denote the Banach space of all continuous periodic real-
valued functions defined on [0, 27r]. Fix some x e [0, 2tt] and define the linear

Section 35: FOURIER ANALYSIS
335
functional Sn'.X —> R by the formula
W) = - rf(t)Dn{x-t)dt.
Show that the norm of the linear functional Sn satisfies
\\Sn\\ = - [~n\Dn(x-t)\dt.
x Jo
Solution. The norm of Sn is defined by
\\Sn\\ = sup \- fnf(t)Dn(x-t)dt
ll/lloo<l ,7r JO
Since \\f\\oo < 1 implies \f(t)D(x - t)\ < \D(x - t)\ for each /, we see that
»2tt
\\Sn\\<- f"\D„(x-t)\dt.
n Jo
Next, we shall establish the reverse inequality. Since the Dirichlet kernel Dn
has period 2tt, it follows that the continuous function
\Dn(x-t)\+e
also has period 2n with respect to r, and clearly |/(e, /)| < 1 for each t and all
€ > 0. This implies
* Jo |£>nU-0l + €
Taking into account Theorem 24.4 and letting € -> 0+ yields
nsnii>- r\on{x-t)\dt.
K Jo
Therefore, \\S„\\ = £ f£*\Dn(x - t)\ dt holds true.

336
Chapter 6: HILBERT SPACES
Problem 35.4. Show that the sequence of functions
{(f)-. (£)W (^coslx, (£)*cos3*, (l)*cos4*,... }
is an orthonormal basis in Li[0, n]. Also show that the preceding sequence is an
orthogonal sequence of functions in L2[0, 2n] which is not complete.
Solution. It is easy to verify that the functions
(?)*• (?)*«»*. (f)^os2x, (i)*cos3*,...
are mutually orthogonal and of norm one in L2W, n].
To show that the preceding orthonormal sequence is complete (i.e., that it is an
orthonormal basis), we need to show that if / e L2IX), 7r] is perpendicular to the
functions 1, cos a, cos2jc, cos 3a, ..., then / = 0. To this end, suppose that a
function / 6 L2W, n] satisfies
/ /(a) cos ha dx = 0
Jo
for all n = 0, 1, 2, Define the function g: [0, 27r] -» 1R by
/(a) if 0 < a < 71
gW - J jr(2jr __ x) if n < x < 27T.
Then, in Li[0, In], for each n we have
plTT
(g.COSnx) = / g(A*)C0S/2A d/A
JO
/»7T pin
= / /(■*)cosnx dx + / /(27r — a)cosnxdx.
Jo JjT
The change of variable / = 27r — a gives
(£,cos/2a)= / f (J) cos ntdt + / f(t) cos ntdt = 0.
Jo Jo
Next, observe that
/»2tT /»7T /»27T
(g,sin/7A) = / g(x)sinnxdx= I f(x)sinnxdx+ I f(2n—x)sinnxdx
J0 'Jo J ix
= / f(t)sinntdt- f(t) sin nt dt = 0.
Jo Jo

Section 35: FOURIER ANALYSIS
337
The preceding show that g is perpendicular to every vector of a complete orthogonal
sequence of Li[0, 2n]. Therefore, g = 0 and hence, / = 0. Thus, the sequence
(?)MI)^OSA-, (f)*COs2*. (^COs3a,...
is an orthonormal basis of L2[0, n].
To see that the orthogonal set
(i)Ml)-cos*, (i)*cos2*. (f)=cos3.v,...
is not complete in L2[0, 27r] notice that the nonzero function sin x is perpendicular
to each of these functions in Li[0, 2n].
Problem 35.5. Show that the sequence of functions
{(£)-«™. (f)-sin2.t, (£)*sin3*. (f)'sin4.v, ... }
is an orthonormal basis of LjW, tv]. Also prove that this set of functions is an
orthogonal set of functions in L2[0, 2n] which is not complete.
Solution. It is easy to verify that the collection of functions
{(f)Wv, (f^sinZr, (f)-sin3x, (j^sinto, ... }
is an orthonormal set of functions of L2[0, n].
Thus, in order to show that it is an orthonormal basis, we need to show that
if a function / e L2[0, n] is perpendicular to each function of the preceding
set in Lo[0,7r], then / = 0. To this end, assume that a function / e Li[0, n]
satisfies
n
f(x)sinnxdx = 0
for all n = 1, 2, 3 Define the function g: [0, 27r] —> R by
= | M if 0 < x < n
° \ -f&t - x) if n < x < In .
L

338
Chapter 6: HTJLBERT SPACES
Then, in Z,2[0, lit], for each n we have
p2n
(g.sinnx) = / g(x) sin nx dx
Jo
r«7r pin
fill nlTC
= / /(a) sin ha Ja — / /(2jt — A')sinA2A*flfjc
JO Jar
/»7r pn
= / f(t)smntdt+l f (t) sin ntdt =0 + 0 = 0.
Jo Jo
Next, observe that
p2n
(g, COS HA) = / g(x) COS /2A Ja
JO
p7T plTl
= / f(x)cosnxdx — I f (2tt — x) cos nxdx
Jo Jit
pTt p3T
= / f(t)cosntdt- f (t) cos ntdt = 0.
Jo Jo
Therefore, g is perpendicular to every vector of a complete orthogonal set of
functions in L2[0, In], Therefore, g = Oandhence, / = 0. Thus, the orthonormal
set of functions
{(?)*«»*. (f)'sin2*- (f)*™3*, (f)'sin4A-, ... }
is an orthonormal basis of L2[0, n].
To see that the orthogonal set of functions
U?)*^*. (f)^in2A-, (f)^sin3A-, (j^sin^, ... }
is not complete in L2[0, 2n], observe that cos a is perpendicular to each of these
functions.
Problem 35.6. The original Weierstrass approximation theorem showed that ev-
ery continuous function of period 2n can be uniformly approximated by
trigonometric polynomials. Establish this result.
Solution. Weierstrass originally gave a direct proof, however, the result can be
derived directly from Fejer's Theorem 35.8. Let / be a continuous function of
period 2tt defined on the entire real line. Let € > 0 be fixed. Let [sn] be the
sequence of partial sums of the Fourier series of / and let {an} be the sequence
arithmetic means.

Section 35: FOURIER ANALYSIS
339
From Theorem 35.8 we know that the sequence {an} converges uniformly to /
on [0, 2tt]. Now, notice that each an is a trigonometric polynomial, and the claim
is established.
Problem 35.7. Find the Fourier coefficients of the function
ji jfo<*<!
nx)~ \0 if \ <x <2n.
Solution. The Fourier coefficients are given by the formulas
i rln l [-- l
0o = ~ / f(x)dx = - / dx = -,
7T Jo 7t J0 2
1 fl7T l rf 1
an — — I f(x) cos nxdx = — / cosnxdx = —sin(rtf), and
x Jo 7t Jo nn
1 f2n 1 /*- 1
bn = — / f(x)s\nnx dx = — / sin nxdx = [cos(rtf) — ll.
tc Jo n Jo nn
Simplifying yields
^o = 5,
*« =
0 if n = 2, 4, 6,...
1/htt if w = 1,5,9,...
-1//Z7T if /i = 3,7, 11,...
f 1//27T if * = 1,3,5,...
2/A27T if n = 2, 6, 10,...
10 if /2=4,8, 12,... .
Problem 35.8. Find the Fourier series of the function
sin* // 0 < x < n
f(x) = t
^ y ] -sin* // it <x <2n.
Solution. The function / is continuous, even, and periodic. Its Fourier
coefficients are given by
r*7T
4
ao = I \ sinxdx = ±
Jo
an = ^ / sinjccos/ijcdjc = I ~^(/«2-D lfw 1S
n Jo 10 if /i i.
is even
is odd, and
bn = 0.

340
Chapter 6: HILBERT SPACES
So, the Fourier series of / is given by ^ — £ J2T=\ Tn--7 - Since this series
converges at every a* and the periodic function / is continuous everywhere, it
follows from Corollary 35.9 that the series converges to /(a) for each a. That is,
we have
„, x 2 4 ^cos2/2A
for each real number a.
Problem 35.9. Show that for each 0 < a* < 27T we have
oo
Solution. We consider the periodic function /: [0, 27r] -> R defined by
m = (
a if 0 < x < 2n
0 if a = 2tt .
Computing the Fourier coefficients of /, we obtain
Jo
/*2rr /»2tt
#„ = i / a' cos A7A dx = ^ I x d(sin nx)
Jo JO
= — a sin /?a — / sin /ia Ja = 0, and
nnl lo Jo -1
r2n pin
bn = £ / AsinA2Ac/A = — —■ / xd(cosnx)
Jo Jo
= —- acos/7a — I cosnxdx \ =—~\2jt —-smnx\ = — ff.
n7rL lo Jo -I L " lo J n
So, the Fourier series of the function / is tz — 2 X^i §iv£- Given that the
function / is continuous at every 0 < x < 2n and that the preceding Fourier series
converges for each 0 < x < 2n (see Example 9.7), it follows from Corollary 35.9
that
holds for each 0 < a < 27r.
00 sinnx

Section 35: FOURIER ANALYSIS
341
X" TV
Problem 35.10. Show that
o = ttjc - — + 2 >
2 3 +-i n2
holds for all 0 < x <2n. Letting x = 0 we obtain the formula Y1T=\ ~p = ^
Solution. Consider the periodic function /: [0, 27r] -> 1R defined by /(a) =
■y — 7TA'. Computing its Fourier coefficients, we get
r-n .2tt
JO
r27r
fl/f = i / (-y — 7Ta) cos/?a ^a = 4-, a;
'T Jo
/»2tt
bn — ^ I (y- — 7ta) sin/?A Ja = 0.
'T Jo
Therefore, the Fourier series of the function / is y 4- 2J2T=\ ^f1. Since
this series converges for each a and / is a continuous function, it follows from
Corollary 35.9 that
9 2 oo
A" 1XL ^v-^COS/2A
-Z-*X = -—+2 > —,
2 3 ^ /i2
and the desired identity follows.
Problem 35.11. Show that
i a ? v^ / cos nx n sin nx \
-- = !*2+4£(^ —)
holds for each 0 < a < 2jt.
Solution. Consider the periodic function /: [0, 27r] —► R defined by
f(x)=\x2 if0<x<2jt
J 1 0 if a- = 2tt .
Computing the Fourier coefficients of /, we obtain
pin
JO
|2tt
„ __ 1 / v2 jv_ .v3 8^.2
10

342
Chapter 6: HBLBERT SPACES
pl-JT
an = i / x2 cos nxdx = 4, and
Jo '
b. = 1 / -v2si
Jo
2;r
2 • j 4;r
sin n x dx = — —.
Therefore, the Fourier series of the function / is \tt2 + 4 £~ , (^ - £*ia»£).
Since this series converges for each a* (see Example 9.7) and / is continuous at
each 0 < x < 2xr, it follows from Corollary 35.9 that
x2
a 7 ,^/cosnx 7tsinnx\
= l*2 + 4U—2 —)'
/» = 1
for each 0 < x < 2n.
Problem 35.12. Consider the "integral" operator T: La[0, it] -> Z,2[0, x]
defined by
Tf(x) = [* K(x%t)fif)dt.
Jo
where the kernel K: [0, n] x [0, n] —► IR is given by
[sin(/z -f l)jt]sin/if
K{x,t) = J2
/i=l
S/wvi' r/?a/ r/ze norm of the operator T satisfies \\T\\ = n/2.
Solution. By Problem 35.5, we know that the sequence of functions
{(f)W-:« = l,2,...}
is an orthonormal basis for LjIO, tc]. Also, as usual, the norm of the operator is
given by
||71=sup{||r(/)||: /6L2[0,tt] and ||/|| = 1 }.
Now, fix a function / e Lj[0, n] with ||/|| = 1, and write
/ = £Cw(f)'sin,7A'
n=\

Section 35: FOURIER ANALYSIS
343
in its Fourier expansion relative to the above basis. By Parseval's Identity, we have
00
n/n2 = X>j2.
Next, notice that the operator satisfies
pn - oo oo
Tf{x)= [zsinu,+)r>nn,][j:c^r-^mt]dt
Jo n=\ m=l
= E[^( /""»ini./f;cm(a)*8in«irfr)]
= f£[>(f)W+ !)-*]•
Now, notice that the latter expression is the Fourier expansion of T (/) with respect
to the orthonormal basis described at the beginning of the solution. Thus, by
Parseval's Identity, we have
CO oo
F(/)H2 = tE7^tEIc«I2 = tII/II2.
from which it follows that ||7|| < n/2 holds.
Finally, if/oU) = (j)1 sinx,then ||/0|| = 1 and by Parseval's Identity we have
cx = 1 andc„ = 0 for n ^ 1, and so ||r(/0)||2 = ~. Therefore, ||T|| > tt/2, and
hence, ||r|| = n/2.

CHAPTER 7
SPECIAL TOPICS IN
INTEGRATION
36. SIGNED MEASURES
Problem 36.1. Give an example of a signed measure and two Hahn
decompositions (A, B) and (A], B\) of X with respect to the signed measure such that
A^ Ax andB ^ Bx.
Solution. Let X = R and let S be the or-algebra of all Lebesgue measurable
sets. Consider the measures p\, P2 e M(E) defined by p\(E) = X(E f) [0, 1])
and fJ,2(E) = HE H [1,2]) for each E e E (where X denotes the Lebesgue
measure on R). Now, consider the signed measure p = p\ — p2, and note that
((-co, 1), [l,oo)) and ([0, 1), (—oo, 0)U[1, oo)) are two Hahn decompositions
of X with respect to the signed measure p.
Problem 36.2. If p is a signed measure, then show that p+ a p~ = 0.
Solution. Let (A, B) be a Hahn decomposition of X with respect to p. If
£ e £, then note that
0 < p+ a p~(E) = p+ A p-(E nfi) + /i+A /z~(£ n A)
< p+(EC)B) + p-(EnA)
= p(E nBHA)- p{E n A n 5) = o.
Problem 36.3. If p is a signed measure, then show that for each A e T>we have
OO 00
\p\(A) = sup! J^ \p(An)\: [An} is a disjoint sequence o/E with [J An = A .
345

346
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Solution. Fix A e E. From Theorem 36.9, we know that
f k k 1
|At|(A) = supl^|/x(i4„)|: [Au ..., Ak] c £ is disjoint and [J A„ C A .
Also, let
OO 00
^ = sup|y^|/x(Aw)|: {A„} C D is disjoint and A = l) A„ |.
Now, let {A„} be a pairwise disjoint sequence of S such that U/^i An = A.
Clearly, ^I=1 Im(A„)| < |/x|(A) holds for each k, and so
oo A-
T |/z(A„)| = lim ]T |/x(A„)| < |/x|(A).
Therefore, s < |/x|(A). On the other hand, if {Aj,..., A*} is a finite pairwise
disjoint collection of S satisfying Ui=i ^n ^ ^»men
A = Ax U-..UA/;U(a\ (J A^) U0U0---,
n=l
and so
£>(A„)| < £|/*(An)| + |/x(A \ |J An) I + M(0) + MftZJ) + •••<*•
Consequently, |/x|(A) < s also holds. Thus, |/x|(A) = s, as claimed.
Problem 36.4. Verify that if jjl and v are two finite signed measures, then the
least upper bound /z v v and the greatest lower bound \jl a v holds in M(X) are
given by
/i v y(A) = sup{/x(£) + y(A \B): B e S and B C A}, a/u/
At a u(A) = inf{/z(£) + v(A \ B): B € S a/irf B Q A}
for each A € E.

Section 36: SIGNED MEASURES
347
Solution. The proof parallels the one of Theorem 36.1. We shall verify that if
\i, v e M(E), then the formula
co(A) = inf{/x(B) + v(A \ B): B e E and B c A }, A e X),
defines a finite signed measure (i.e., a) 6 M(Z)), and that a> is the greatest lower
bound of \x and v in M(£).
Since /Lt and v are both bounded from below (and also both bounded from
above), it follows that co(A) e JR for each AeE. Clearly, co(0) = 0 holds.
Next, we shall establish that co is cr-additive. To this end, let [An\ be a pairwise
disjoint sequence of Z and let A = IJ^Li An. If B e Y, satisfies KA, then
CO CO
fi{B) + v(A \ B) = (i(jj a„ n a) + w((J(A„ \ A„ n 5))
CO
= ^[At(^nB) + v(Alf\Awnfi)]
n=l
CO
and so co(A) > J2T=\ ^C^/i) holds. For the reverse inequality, let s > 0. Then,
for each n pick some 5„eS with Bn C A,, and
M(B«) + v(i4„\B/l)<cw(i4II)+|r,
Obviously, {5,,} is a pairwise disjoint sequence of S. Put B = U^li ^/i ^ ^>
and note that U^=i(^n \Bn) = A\B holds. Moreover,
fl>(A) < fi(B) + v(A \ B) = /x((J B„) + v(Q(An \ £„))
co
= ^2[fJL{Bn) + v{An\B„)]
n=\
CO CO
< 2[ft>(i4„) + £]= 1] w(/ln) + e.
n=l n=l
Since e > 0 is arbitrary, we infer that co(A) < Y1T=\ W(A„) also holds, and so
to e Af (E).

348
Chapter 7: SPECIAL TOPICS IN INTCGRATION
Finally, we shall establish that co is the greatest lower bound of /x and v in
M(S). Note first that co is a lower bound for both /x and v. Indeed, if A € E,
then (by letting B = A), we see that
(o(A) < fx(A) + v(0) = /x(/4) and w(A) < /x(0) + v(A) = v(A).
On the other hand, if n € M(E) satisfies 7r < /x and 7r < v and i4 € £, then
for each 5eS with B c A we have
jr(i4) = ar(fl) + jt(A \ B) < il(B) + v(i4 \ £),
from which it follows that tc(A) < co(A), i.e., jt <co. This shows that co = fiAv
holds in M(E).
Problem 36.5. Le/ A. be the Lebesgue measure on the Lebesgue measurable
subsets of JR. Iffi is the Dirac measure, defined by fi(A) = 0 ifO £ A and
fji(A) =\ifOe A, describe X v /x and X A \i.
Solution. If A = R \ {0}, B = {0}, and E is an arbitrary Lebesgue measurable
set, then
0 < X A fi(E) < X(E DB) + /x(£ H A) = 0
holds. That is, X A /x = 0. Moreover, we have
Xv/x = Av/x-f-XA/x = A-f/x.
Problem 36.6. Show that the collection of all a-finite measures forms a
distributive lattice. That is, show that if n, v, and co are three a-finite measures,
then
(H v v) A co = (/x A co) v (v a co) and (/x A v) v co = (/x v co) A (W a;).
Solution. We shall show first that every vector lattice satisfies the distributive
law. To do this, we shall use the identity (a) of Problem 9.1.
Let x, y, and z be elements in a vector lattice. Since x v y > jc, it follows
that (x v y) az > x A z, and similarly (x v y) A z > y A z. Thus,
(a- v y) a-z > (jc A z) v (y A z).
On the other hand, if u = (jc a z) v (y a z), then u > x az = jc-f-z—jcvz holds.

Section 36: SIGNED MEASURES
349
Hence, x < u — z+xvz < u — z+{xvy)vz, and similarly y < u — z+(xvy)vz.
It follows that x v y < u — z 4- (x v y) v z, and so
(x A z) v (y a z) = u > x v y + z - (x v y) v z = (x V y) a z.
Therefore, (i v j) a z = (jc A z) v (y a z) holds. The other identity can be
established in a similar manner.
Now, let {Xn} C E satisfy /x(X„) < oo, v(X„) < oo, a){Xn) < oo for all a,
and Xn "[ X. If Y,n = {AHXn: A G E }, then clearly /x, v, and a> (restricted
to E„) belong to the vector lattice M(En). Thus, if £ e E, then
(/iW)A w(£ PI X„) = (/x A a)) V (y A o>)(£ H X„)
and
(/x a v) V a;(£ H X J = (/zVw)a(w a>)(£ H X„)
hold. To finish the proof note that £HX„ t £, and then use the "order continuity"
of the measure (Theorem 15.4).
Problem 36.7. // E is a a-algebra of subsets of a set X and /x: E -> ^ is a
signed measure, then show that
AM+ O AM- = AjM|.
Solution. Assume that /x: E —> R* is an arbitrary signed measure. Let £ be
in AM+ n AM- and let A e E be an arbitrary set. Then,
\li\(A) = }i+(A) + n-(A)
= [/x+(A n £) + /x+(/4 fl £c)] + [/x"(i4 H £) + fi"(A n £c)]
= [/x+G4 H £) + /x"(A fl £)] + [fi+(A n £c) + /*-(A n £c)]
= |Ax|(An£)+|/x|(An£c),
and so £ e AjMj, i.e., AM+ fl AM- C AjM|.
For the reverse inclusion, let £ e A)M|. If A € E is arbitrary, then note that
/x+(A) + M"(A) = |m|(A) = |/x|(A H £) + |/x|(A n £c)
= [/x+(A n £) + /x+(A n £c)] + [ir{A n £) + /x"(A n £c)].
Since /z+(A) = /x+((/\ O £) U (A n £c)) < /x+(A H £) + /x+(A fl £c) and

350
Chapter 7: SPECIAL TOPICS IN INTEGRATION
\jl (A) < /x (A fl E) + /x (AD EQ) both hold, it follows from the preceding
equality that
fji+(A) = fjL+(A fl E) + /x+(A H Ec) and /x"04) = /x-(A fl E) + /x"(A fl £c),
which shows that £ € AM+ fl AM-. Thus, A|M| C AM+ fl AM-, and consequently,
A|M| = AM+ fl AM- holds, as desired.
Problem 36.8. Lef /x and v be two measures on a a-algebra X) with at least one
of them finite. Assume also that S is a semiring such that S C £, X € S, and that
the a-algebra generated by S equals E. Then show that \i = v on E //#/2d o/?/;y
Z//X = v on S.
Solution. Assume that /x is finite and that /x = v on S. If we consider the
measure space (X, 5, /x), then it is easy to see that S C E C AM holds. Now,
apply Theorem 15.10 to get that /x = /x* = v holds on D.
Problem 36.9. Ler (X, 5, \i)bea measure space, and let f e L\ (/x). Then show
that
v(A) = j fdiL
for each A e AM defines a finite signed measure on AM. Also, show that
v+(A)= [ f+dfi, v-(A)=[f-dii and \v\(A) = f \f\dfi
J A J A J A
holds for each A e AM.
Solution. If {An} is a pairwise disjoint sequence of AM satisfying A = (J^Li ^/i»
then WmY.UxfXAi = /x* and |E/=i/x>4,| < l/l holds for each a. Thus,
from the Lebesgue Dominated Convergence Theorem, it follows that
Therefore, v is a finite signed measure.
Now, note that if
A = {jc € X: f(x) > 0} and B = {* e X: f(x) < 0},

Section 36: SIGNED MEASURES
351
then it is easy to see that (A, B) is a Hahn decomposition of X with respect to
v. Since fxEnA = f+XE holds, we see that
v+(E) = v(EHA) = f 'fdfi= [ f+dfi
Jeha Je
for each E e AM. The proof for v~ is similar. The absolute value formula follows
from the identity \v\ = v+ + v~.
Problem 36.10. Let v be a signed measure on S. A function f: X -> 1R is said
to be v-integrable iff is simultaneously v+- and v~-integrable {in this case, we
write ff dv = // dv+ — ff dv~). Show that a function f is v-integrable if and
only if f e L,(|v|).
Solution. Assume that / is simultaneously v+- and v~-integrable. We can
assume that f(x) > 0 holds for all x e X. Since each set of the form (a* e X: a <
f(x) < b) belongs to AM+ n AM- = A|M| (for this identity see Problem 36.7), we
see that there exists a sequence {(pn) of simultaneously v+- and v~-step functions
such that (pn(x) t fW holds for all .v e X\ see the proof of Theorem 17.7.
Clearly, each 0„ is a |/z|-step function and from
f<j>n d\lL\ = j(pn dfJ.+ + j(t>n d[T t ff d^ + ffd[L
< CO,
we see that / is |/x| -integrable and that ffd\ti\ = // dfj,+ + ff dji~ holds.
For the converse, assume that / belongs to Lj(M). We can assume that
fix) > 0 holds for each x. Note first that if / = xa for a Iv (-measurable set
A with |v|*(i4) < oo, then there exists (by Theorem 15.11) some B e H, with
A C B and \v\*(A) = \v\*(B). It follows that \v\*(B\A) = 0, and in view of
0 < v+ < |y|, we have (v+)*(B \ A) = 0. Thus, B \ A is a y+-measurable set,
and consequently, A = B \ (B \ A) is also v+-measurable. This shows that xa
is v^-integrable.
Now, choose a sequence {0,2} of |u|-step functions with 0 < <t>n(x) t f(x)
for each .v. By the previous discussion, {<£„} is a sequence of v+-step functions.
Moreover,
f(l>ndv+ < f<t>nd\v\< ffd\v\
< CO
holds for all n. Thus, / is v+-integrable.
The v~"-integrability of / can be established in a similar manner.

352
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Problem 36.11. Show that the Jordan decomposition is unique in the following
sense. If v is a signed measure, and fi \ and //2 we two measures such that
v = fi\ — iii and fJL\ A [Li = 0, then fi\ = v+ and M2 = v~-
Solution. First, we shall establish that v+ = Mi holds. Start by observing that
v < fi\ implies v+ < fi\.
Now, let E € E. If v+(£) = 00, then v+(£) = mi(£) = 00 holds trivially.
Thus, we can suppose v+(E) < 00. Since v(E) = /jl\(E) — 112(E) < v+(£) <
00, it follows that ii\(E) < 00. Let e > 0. Then, in view of
0 = mi A /z2(£) = inf{/xi(£ \ £) + /x2(fi): # € E and B C £ },
there exists some £ € E with B C £ and Mi(£ \ #) + M2(#) < £• Thus,
v+(£) = sup{u(F): FeEand£C£}> v(fl) = Ml(B) - /x2(5)
> mi(B) - e = mi(£) - Mi(£ \ B) - b > fii(E) - 2e
holds for all e > 0. That is, v+(£) > am(£) for each £ 6 E, and therefore
v+ = Mi holds. For the other identity note that
v~ = (-v)+ = (m2 - Mi)+ = M2-
Problem 36.12. 7/2 a vector lattice xn I x means that xn+\ < x„for each n and
that x is the greatest lower bound of the sequence [xn]. A normed vector lattice is
said to have o -order continuous norm if xn I 0 implies lim ||a'„ || = 0.
a. Show that every Lp(/x) with 1 < p < 00 has a-order continuous norm.
b. Show that Loo([0, 1]) does not have o-order continuous norm.
c. Let Y*bea a-algebra of sets, and let [fin} be a sequence ofMiX) such that
\xn I fi. Show that lim fjbn(A) = fi(A) holds for all A e E.
d. Show that the Banach lattice M(E) has a-order continuous norm.
Solution, (a) Note first that /„ I f in Lp(/jl) is equivalent to fn I f a.e.
(why?). If for some 1 < p < 00 a sequence {/„} of Lp(fi) satisfies /„ I 0
a.e., then
WP = {f \fn\"^f 10
holds by virtue of the Lebesgue Dominated Convergence Theorem.
(b)If fn = x(o,i)>then fn | 0 holds in Loo([0, 1]). However, note that ||/„||oo =
1 holds for eacn n.

Section 37: COMPARING MEASURES AND THE RADON-NKODYM THEOREM 353
(c) Let pn I fi in M(£). Then 0 < p\ — pn \ p\ — p in M(E). By
Theorem 36.2, it follows that p\(A) — pn(A) f f*\(A) — p(A) holds for each
A e S. Thus, pn{A) I p(A) holds for each A e E.
(d) If pn I 0 in M(D), then from part (c) it follows that \\pn\\ = pn(X) I 0.
Problem 36.13. Prove the following additivity property of the Banach lattice
M(£)://>, v e M(S) are disjoint (i.e., \ijl\a\v\ = 0),then \\p + v\\ = HmII + IMI
Solution. If \p\ a M = 0 holds in M(£), then |/x + v\ = |/x| + M holds
(see Problems 9.2 and 9.3). Thus, \\p + v|| = \p + u|(X) = \p\(X) + \v\(X) =
IMI + HI-
Problem 36.14. Lef T, be a a-algebra of subsets of a set X and let [pn] be a
disjoint sequence of M(E). If the sequence of signed measures [pn] is order
bounded, then show that lim ||/x„ || = 0.
Solution. Let [pn] be a disjoint sequence of the Banach lattice M(S) such that
for some 0</xG M(E) we have \pn | < p for each a*. From |/x„ | a |/xm | = 0 for
n ^ w, we see that
k k
holds for each &. In particular, we have
k k k
£ ll/Xnll = £ l^l(X) ^ [V iMnljW ^ M(X) < OO
holds for each /7, and so X^Jlj HM/iII < oo. The latter easily implies lim ||/xn|| = 0.
37. COMPARING MEASURES AND THE RADON-NKODYM
THEOREM
Problem 37.1. Verify the following properties of signed measures:
a. p <£ p..
b. v <£ /x a/7j p <&a) imply v <£ co.
c. If 0 < v < p, then v <£ /x.
d. //> « 0, /ten /x = 0.

354
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Solution, (a) From Theorem 36.9, we have |/x(A)| < |/z|(A), and so if |/x|(A) =
0 holds, then /x(A) = 0 likewise holds. That is, /x <£ M-
(b) Assume v <£ /x and /a <& co and M(A) = 0. Theorem 37.2 applied twice
shows that |/x|(A) = 0 and |v|(A) = 0. Hence, v < a> holds.
(c) Let 0 < v < fi. If |/x|(A) = fji(A) = 0, then clearly v(A) = 0, and so y « /x
holds.
(d) Let /x <^0. Since the zero measure assumes the zero value at every A e E,
it follows that ijl(A) = 0 holds for every A € E. This means that fi = 0.
Problem 37.2. Vfenjfy the following statements about signed measures on a a-
algebra E of sets:
1. If fi <£co and v«w, f/zew |/x| 4- \v\ <$C w.
2. IffjL±o) and v A. co, then |/x| -f- |v| _L co.
3. // fu, <& co and \v\ < \fji\, then v <£ co.
4. If fi A. co and |v| < |//,|, //2e/? v 1. co.
5. If v <$C n and vl/z, r/ze/z v = 0.
Solution. (1) This follows immediately from Theorem 37.2.
(2) Since /ilw, there exists (by Theorem 37.5) some A\ e E with |w|(Ai) =
|/i|(A^) = 0. Similarly, there exists some A2 e E with M(/42) = M04|) = 0-
Put A = A{UA2 and B = (A]UA2)C = A\^A\. Then A, £ € E, AUfi = X,
A H B = 0, M(y4) = 0, and (|/x| + |v|)(J5) = 0. By Theorem 37.5 we infer that
co J_ \/jl\ + \v\ holds.
(3) This follows easily from Theorem 37.2.
(4) This follows immediately from Theorem 37.5.
(5) Since v ± fi, there exists some A e E such that \v\(A) = |/x|(/4c) = 0. By
v <£ fi and Theorem 37.2, \v\(Ac) = 0, and so \v\(X) = \v\(A) + \v\(Ac) = 0.
That is, |v| = 0, so that v = 0.
Problem 37.3. Le/ /x <2/2<i v be two measures on a a-algebra E. Ifv is a finite
measure, then show that the following statements are equivalent.
a. v <£ fi holds.
b. For each sequence [An] ofY, with lim ii(An) = 0, we have lim v(An) = 0.
c. For each e > 0 there exists some 8 > 0 (depending on e) such that whenever
A e E satisfies ijl(A) < 8, then v(A) < e holds.
Solution, (a) ==>• (b) If (b) is not true, then there exists some e > 0 and some
sequence [An] of E such that /i(y4„) < 2~n and v(An) > e for each n. Set

Section 37: COMPARING MEASURES AND THE RADON-NKODYM THEOREM 355
A = f)Zi IX„ Ai 6 E. From A C (X„ *,, we see that
PC 00
holds for each /z, and so /z(A) = 0. However, from Theorem 15.4(2), we see that
v(A) > £, contrary to v « /x. Hence, (a) implies (b).
(b)=> (c) If (c) is not true, then there exist some e > 0 and a sequence {An}
of E such that /z(A„) < £ and v(An) > e hold for all n. Clearly, this
contradicts (b).
(c) =» (a) Let A e E satisfy /x( A) = 0. Given e > 0, choose some <5 > 0 so
that (c) is satisfied. In view of fi(A) < 5, it follows that v(A) < e. Since e > 0
is arbitrary, v(A) = 0, and so v « /x holds.
Problem 37.4. Let /x fee a finite measure, and let [vn] be a sequence of finite
measures (all on E) sz/c/z //zar vn <& \x holds for each n. Furthermore, assume
that lim vn(A) exists in IRfor each A e E. Then, show that:
a. For each e > 0 zVzere ex/ste sowe 8 > 0 jzzc/z r/zar whenever /leE satisfies
11(A) < 8, then vn(A) < € holds for each n.
b. 77ze set function v: E -» [0, oo], defined by v(A) = lim u„(A)jfo/* earc/z
A G Y,, is a measure such that v <£ \i.
Solution, (a) From Problem 31.3, we know that E under the distance d(A, B) =
fx(AAB) is a complete metric space. From v* <£ /x and the inequality
\vk(A)-vk(B)\<vk(AAB)1
it easily follows that the function vk: E —> R is well defined (i.e., vk( A) = vk(B)
holds whenever fji(AAB) — 0) and is continuous.
Now, let s > 0. Define
Ck = {A e E: |v„(A) - vw(A)| < e for all ny m>k }.
Note that each C* is closed and that E = UjuLi ^ holds. By Baire's Category
Theorem 6.18), we have C° ^ 0 for some k. Thus, there exist Aq e Ck and
8\ > 0 such that A e E and /x(AAA0) < <$i imply A e C*.
From \>i <£ \x (1 < z < &) and the preceding problem, there exists some
0 < 8 < 8\ such that A e E and /x(A) < 5 imply v((A) < s for all I < i < k.

356
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Now, if A e E satisfies p(A) < <5, then A U (A0\i4) = A U Aq satisfies
p((A U Aq)AAq) < p(A) < 8\, and so
\v„(A) - vk(A)\ = \(vn-vk)(AVA0)-(vn-vk)(A0\A)\
< |(vw ~ v*)(i4 U A0)| + |(vn - va)(^o\A)| < 2e
holds for all n > k. Thus, A € S and /x(A) < 5 imply
|v„(A)| < 2* + v*04) < 3e
for all /z > /: (and all 1 < n < k).
(b) Let A = U^li ^« wim me sequence {A„} of £ pairwise disjoint, and let
e > 0. Choose some 5 > 0 so that statement (a) is satisfied. Next, choose some
m so that p(A \ (JJLi A) < s nolds for aI1 ri>m. Then,
/=1 /=!
holds for all k and all n > m. Thus, |u(A) - YH=\ VW/)| < * holds for all
77 > /w, and so \v(A) — Y^L\ V(A() \ < e. Since e > 0 is arbitrary, we see that
v(A) = Y1T=\ y(^n). Thus, v is a measure, and from part (a) and the preceding
problem it follows immediately that v <& p holds.
Problem 37.5. Let [vn] be a sequence of nonzero finite measures such that
lim v„(A) exists in JRfor each A e £. Show that v(A) = lim vn(A)for A € £ is
a finite measure.
Solution. Consider the set function p: £ —> [0, oo) defined by
oo
and note that p is in fact a measure. In addition, note that vn <& p holds for each
n. Now, invoke part (b) of the preceding problem to conclude that the set function
v is also a measure.
Problem 37.6. Verify the uniqueness of the Radon-Nikodym derivative by
proving the following statement: If('X,S,p) is a measure space and f e L\(p)
satisfies fA f dp, = Ofor all A e S, then / = 0 a.e.

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 357
Solution. From the given condition, it is easy to see that fAf dp = 0 must hold
for each cr-set A. Now, consider the measurable sets
A = {jc e X: f(x) > 0} and B = {x e X: f{x) < 0}.
By Problem 22.7 we know that A and B are both a -finite sets. Now, in view of
fAfdp = fBfdp = 0, it follows from Problem 22.13 that p*(A) = p*(B) = 0.
Therefore, / = 0 a.e. holds.
Problem 37.7. This problem shows that the hypothesis of o-finiteness of p. in
the Radon-Nikodym Theorem cannot be omitted. Consider X = [0, 1], E the
a-algebra of all Lebesgue measurable subsets of[0, 1], v the Lebesgue measure
on S and p the measure defined by p{0) = 0 and p(A) = oo // A ^ 0.
{Incidentally, p is the largest measure on S.) Show that:
a. v is a finite measure, p is not a-finite, and v <£ p.
b. There is no function f e L\{p) such that v(A) — fA f dp holds for all
A € £.
Solution, (a) Note that p(A) = 0 means A = 0, and so v <£ p holds,
(b) Observe that L\{p) = {0}.
Problem 37.8. Let p be a finite signed measure on S. Show that there exists a
unique function f e L\(\p\) such that
p{A) = j fd\p\
holds for all A e E.
Solution. The conclusion follows from the Radon-Nikodym Theorem by
observing that p <£ \p\ holds.
Problem 37.9. Assume that v is a finite measure and p is a a-finite measure
such that v <& p. Let g = dv/dp e L\(p) be the Radon-Nikodym derivative of
v with respect to p. Then show that:
a. If Y = [x £ X: g(x) > 0}, then Y f! A is a p-measurable set for each
v-measurable set A.
b. If f e L\(v),then fg e L\(p)and ff dv = ffgdp holds.
Solution, (a) Note first that by Theorem 37.3, S C AM C Av holds, and that
Y e AM.

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Chapter 7: SPECIAL TOPICS IN INTEGRATION
First consider the case when A e Av satisfies A c Y and v*(A) = 0. By
Theorem 15.11 there exists some B € E with A C. B and v*(B) = 0. Now, if
fM*(B (1 K) > 0, then we have the contradiction 0 = v*(B DY) — fBnY g d/i > 0
(see Problem 22.13). Consequently, fi*(B DY) = /i*(A) = 0 holds, and so
A e A„.
Now, let A e Av. Choose some B e E with A c £ and v>*04) = v*(B).
Thus, v*(£ \ A) = 0, and so (5 \ A) H y € AM. Now, note that
Any=5ny\(5\/i)ny€AM.
(b) It follows immediately from Problem 22.15.
Problem 37.10. Establish the chain rule for Radon-Nikodym derivatives: Ifco
is a a-finite measure and v and \jl are two finite measures (all on E) such that
v <£ fi and (A <£to, then v <&co and
dv dv dfi
dco dfi dco
holds.
Solution. Clearly, v «; /x and £ c Aw c AM C A„. Put / = ^ e Lj(m)
and g = 4j± e L\(co). If A e E, then by part (b) of the preceding problem, we
infer that
v(A) = / fdfi= I fXAdti= fxAgdco= / fgdco.
This combined with the Radon-Nikodym Theorem shows that
dv
— = fg, co-a.e.
dji
Problem 37.11. All measures considered here will be assumed defined on a fixed
o-algebra E.
a. Call two measures \x and v equivalent (in symbols, fi = v) if fx <£ v
and v <£ fj. both hold. Show that = is an equivalence relation among the
measures on E.
b. If[i and v are two equivalent a-finite measures, then show that AM = Au.
c. Show that if fi and v are two equivalent finite measures, then
d\i dv
— • -7— = 1 a.e. holds,
dv dfi

Section 37: COMPARING MEASURES AND THE RADON-NBKODYM THEOREM 359
d. If [i and v are two equivalent finite measures, then show that
dv
from L\(jjl) to L\(v), is an onto lattice isometry. Thus, under this
identification L\(n) = L\(v) holds.
e. Generalize (d) to equivalent a-finite measures. That is, ifn and v are two
equivalent a-finite measures, then show that the Banach lattices L \ {fi) and
L\(v) are lattice isometric.
f. Show that if [i and v are two equivalent a-finite measures, then the Banach
lattices Lp{^jl) and Lp(v) are lattice isometric for each 1 < p < oo.
Solution, (a) Straightforward.
(b) It follows immediately from Theorem 37.3.
(c) Use the relation v « // « v and the preceding problem.
(d)Let / h-► /.*£ = T{f). Since ^ e L,(v), it follows from Problem 37.9(b)
that T(f) = f- % G L,(v) and ffdfi = ff-%dv hold for each / e L,(m).
Thus, T defines a mapping from L\{^i) to L\(v) which is clearly linear. Since
^ > 0 holds, it follows that
av —
r(l/l) = l/l-^ = |/-^|=|r(/)|
and
\\nf)l = f\f-%\dv = f\f\-%dV = j\f\d* = \f\x
hold for each / e Li(/x). Thus, T: L|(/x) —> L\(v) is a lattice isometry. To see
that T is also onto, note that if g e L\(v), then g • j- e Lj(/x) and by part (c),
we see that
(e) Let {£„} be apairwise disjoint sequence of E such that (J^li £/i=^> M(£/») <
oo, and v(En) < oo for each n. Let
r„:L|(£fl1iLt) —► L\(E„,v)
be the onto lattice isometry determined by part (d) previously. Now, it is a routine
matter to verify that T:L\(/x) —> L\(v) defined by
CO
r(/) = £r«(/x£.)
for each / e Li(ju) is an onto lattice isometry.

360
Chapter 7: SPECIAL TOPICS IN INTEGRATION
(f) Suppose first that p and v are finite. Then,
/■—/•(ft)*
is a lattice isometry from Lp(p) onto Lp(v) for each 1 < p < oo. Now, if /x
and u are cr-finite, then use the arguments of part (e) to establish that Lp(p) and
Lp(v) are lattice isometric.
If p = cxi, then from part (b) it follows that Loo(p) = £oo(v) holds, and so in
this case the identity operator is a lattice isometry.
Problem 37.12. Let pbe a o-finite measure, and let AC(p) be the collection of
all finite signed measures that are absolutely continuous with respect to p; that is,
AC(p) = {ve Af (£): v « /x}.
a. Show that AC(p) is a norm closed ideal 0/M(E) (and hence AC(p), with
the norm ||v|| = |u|(X), is a Banach lattice in its own right).
b. For each f e L\ (/x), let pf be the finite signed measure defined by p/(A) ==
fA f dp. for each A e E. Then show that f h-> pf is a lattice isometiy
from L\(p) onto AC(p).
Solution, (a) Clearly, AC(p) is an ideal of M(E). If [vn] is a sequence of
AC(p) satisfying vn —> v in M(E), then vn(A) —> v(A) holds for each
A e S. Problem 37.4 shows that v e AC{p). Thus, AC(/x) is a closed vector
sublattice of M(E), and hence, a Banach lattice in its own right,
(b) Clearly, / i—> /x/ is a linear operator. By Problem 37.6 this operator is
one-to-one. From Problem 36.9, it follows that / i—> p/ is a lattice isometry,
and the Radon-Nikodym Theorem implies that it is also onto.
Problem 37.13. Let S be a a-algebra of subsets of a set X and p a measure on
E. Assume also that £* is a a-algebra of subsets of a set Y and that T:X —► Y
has the property that T~l(A) e E/or each A € E*.
a. Show that v(A) = p(T~l(A))for each A e E* is a measure on E*.
b. /// e Li(v), then show that f o T e L\(p) and
j fdv = f foTdp.
c. Ifp is finite and co is a a-finite measure onT,* such that v <& co, then show
that there exists a function g e L\(a>) such that
/ f oTdp= / fgdco
holds for each f € L\(y).

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 361
Solution, (a) Straightforward.
(b) Note first that if ^4 is a v-null set, then T~](A) is a /x-null set. Indeed,
if v*(A) = 0 holds, then there exists (by Theorem 15.11) some B e E* with
A C B and v(B) = 0. Therefore,
0 < /i*{T-\A)) < u.*{T-l(B)) = n(T-\B)) = v(B) = 0.
Now, let i4 beav-measurablesetwith v*(A) < oo. Choose some 5eS* with
KB and v*(£) = v*(A). Since u*(£ \ A) = 0, it follows from the preceding
discussion that ii*(T~\B\ A)) = 0. Thus, T~](A) = T~](B)\T-](B \A) is
/z-measurable, and moreover,
fXAdv = v*(A) = n*{T-l(A))= [ XT-HA)dl*= f XAoTdfi.
JY Jx Jx
It follows that for every u-step function 0 we have (poT e Li(/x), and fY<t>dv =
fx(p oT dp,. An easy continuity argument can complete the proof.
(c) It follows immediately from part (b) and Problem 37.9.
Problem 37.14. Let (X, <S, fx) be a o -finite measure space, and let g be a
measurable function. Show that if for some 1 < p < oo we have fg e L\{pi)for all
f e Lp(fj.), then g e Lq(fi), where ± + ^ = 1.
Also, show by a counterexample that for 1 < p < oo the a-finiteness of fi
cannot be dropped.
Solution. We can assume that g > 0 holds (why?). Then the formula F(f) =
ffgdfi for / e Lp(fji) defines a positive linear functional on Lp{p). By
Theorem 40.10, F is continuous. Now, by Theorems 37.9 and 37.10 there exists
some h e Lq(fi) such that ffgdfi = ffhdp, for each / e Lp(jjl). This
implies (how?) fA(g — h)dfji = 0 for each measurable subset A. Now, a glance
at Problem 22.13 guarantees that g = h a.e. holds.
The <7-finiteness of jjl cannot be dropped. Consider X = (0, oo) with the
measure /z defined on the a-algebra V(X) by fx(A) = oo if A ^ 0 and
/x(0) = 0. Then for 1 < p < oo we have Li(/x) = Lp(ii) = Lq(ix) = {0}, and
Loo(fji) = B(X), the bounded real-valued functions on X. On the other hand, if
g(x) = x, then fg = 0 € L\(fx) holds for all / e Lp(/x) (1 < p < oo), while
8 i Lq(fi).
Problem 37.15. Let (X, <S, /x) be a o-finite measure space, g a measurable
function, and 1 < p < oo. /toHwe f/zaf r/zere ejcwAs some rea/ number M > 0 j«c/i
f/zaf #g G Li(/z) tffld ftpgd/JL < M||0||j, holds for every step function (/>. Then,
show that:

362
Chapter 7: SPECIAL TOPICS IN INTEGRATION
a. g e Lq(fi), where ^ + ^ = 1, and
b. ffgdfi < M||/1|, holds for all f £ Lp<jl).
Solution. Let L denote the vector space of all step functions. The given
conditions show that the function F:L —> R, defined by F(<f>) = /(pgdfi, is a
continuous linear functional. Since L is dense in Lp(fi) (Theorem 31.10), it
follows that F has a continuous extension (which we shall denote by F again)
to all of Lp(fi). By Theorems 37.9 and 37.10 there exists some h € L9(/x) such
that F(f) = ffh dfi holds for all / e Lp(fi). Clearly,
F(f)\ = \ffhdn\<M\\f\\
holds for all / e Lp(fi).
To complete the proof, it suffices to show that g = h a.e. holds. To see
this, let E £ AM satisfy /x*(£) < oo. Then, consider the step function 4> =
XsSgn (g —h) e L, and note that f(p(g — h)dfi = Q implies fE \g — h\ dfi = 0.
That is, g = /z a.e. holds on E\ see Problem 22.13. Since /x is a-finite, we see
that g = h a.e. holds on X.
Problem 37.16. Let fi be a Borel measure on lR.k and suppose that there exists
a constant c > 0 such that whenever a Borel set E satisfies X(E) = c, then
/x(£) = c. Show that /x coincides with X, i.e., show that /x = X.
Solution. Assume that the Borel measure /x and the constant c > 0
satisfy the properties of the problem. Clearly, /x is a cr-finite Borel measure. By
Theorem 37.7, we can write
/x = fjL\ -f lx2, where /xj <& X and /X2 -L X.
First, we shall establish that /x2 = 0. From /x2 X X, there exist two disjoint
Borel sets A and B with A U B = IR* and /x2(4) = X(B) = 0. Since X(A) =
oo, there exists (by Problem 18.19) a Borel subset C of j4 with X(C) = c.
From X(C U B) = X(C) + X(5) = X(C) = c and our hypothesis, we see that
/x(C U £) = c. Now, note that
c < c + /X2(£) <c-f /x(£) = /x(C) + /x(B) = M(C Ufi) = c,
and so 112(B) = 0. This shows that /x2 = 0, and consequently /x = /xi is
absolutely continuous with respect to X.

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 363
Next, fix a compact set K with \{K) > c and consider both fi and X restricted
to K. By the Radon-Nikodym Theorem, there exists a non-negative function
/ eL,(ff,S,A.) such that
H(E) = jfdk
holds for each Borel subset £ of K (see Problem 12.13). We claim that / = 1 a.e.
To establish this, assume by way of contradiction that the Lebesgue measurable
set D = [x e K: f(x) < 1} satisfies X(D) > 0; we can assume (why?)
that D is a Borel set. If X(D) > c holds, then pick a Borel subset D\ of D
with \(D\) = c; if X(D) < c, then pick a Borel set D\ with D C D\ C. K
and X(D\) = c; (see Problem 18.19). Now, note that in either case, we have
lx(D\) < c, which contradicts our hypothesis. Hence, A.(D) = 0. Similarly,
\({x € K: f(x) > 1} = 0, and so / = 1 a.e. Therefore, /i(£) = X(E) holds
for each Borel subset £ of K. Now, pick a sequence {Kn} of compact subsets
of R* with k(K„) > c and Kn f R*. If £ is an arbitrary Borel subset of R*,
then note that
li(E) = lim /x(£ n K„) = lim A.(£ n Kn) = A.(£).
Problem 37.17. Lef /z tfrtd v be two a-finite measures on a a-algebra E of
subsets of a set X such that v <£ \x and v^O. Show that there exist a set E e £
and an integer n such that
a. v(£) > 0; and
b. AeS flfld i4 c £ zwp/y ^m(^) < »>(4) < nfi(A).
Solution. Pick a sequence [Xn] of Y, with X = U^li ^n» W-^/i) < °o, and
/Lt(X„) < oo for each n. Since v^O, there exists some n such that v(Xn) > 0.
From v <£ /x, it follows that /x(X„) > 0 also holds. Thus, replacing X by Xn,
we can assume from the outset that both v and \x are finite measures.
Now, by the Radon-Nikodym Theorem, there exists a function 0 < / e L \ (/x)
such that
v(A) = j fdfi
holds for each A e S. From v ^ 0, we see that / ^ 0, and so the /z-measurable
set F = {x <= X: f{x) > 0} satisfies fi*(F) > 0. Next, put
En = {xeX: £ </(*)<*}

364
Chapter 7: SPECIAL TOPICS IN INTEGRATION
and note that En f E a.e. Thus, for some n, we have /x*(E„) > 0. By
Theorem 15.11, there exists some E e E with En c E and /x(£) = n*(E„).
We claim that the set E satisfies the desired properties.
To see this, note first that
v(E) = f fdfjL= f fdfx> ±/z*(£„) > 0.
Je Je„
Now, if A e E satisfies A C £, then note that \xa < f < "Xa V> — a.e., and
consequently
i//,G4) = £ XAdfi< fdfi = v(A) < / A*/* J/x = /2/z04).
J A J A J A
Problem 37.18. Let \jl be a finite Borel measure on [1, oo) such that
a. fi <£ X, and
b. jjl(B) = afi(aB)for each a > 1 and each Borel subset B of [I, oo), where
aB = [ab: b G 5}.
7/7/ie Radon-Nikodym derivative d\ijd\ is a continuous function, then show that
there exists a constant c > 0 such that [dji/dX\(x) = -%/or each x > 1.
Solution. For simplicity, let us write ^ = /. Then, the given identity fi(B) =
afji(aB) can be written in the form
JB Ja,
fdk.
B
For B = [1, a*], we get
/A /»flA'
f(t)dt=a] f(t)dt
for each tf > 1 and each a > 1. Differentiating with respect to x (and taking
into account the Fundamental Theorem of Calculus), we see that
/(a) = a2f(ax)
holds for each a > 1 and each a > 1. Letting a = 1, we obtain
for all a > 1, and our conclusion follows.

Section 38: THE RBESZ REPRESENTATION THEOREM
365
Problem 37.19. Let /x be a finite Borel measure on (0, oo) such that
a. [A <& X, and
b. p-(aB) = jx{B)for each a > 0 and each Borel subset B o/(0, oo).
If the Radon-Nikodym derivative is a continuous function, then show that there
exists a constant c > 0 such that [dfji/dk](x) = ^for each x > 0.
Solution. Let ~ = /. Then, (by The Radon-Nikodym Theorem) the given
identity p,(B) = p*(aB) can be written in the form
J B Ja
fdX.
B
For B = [1,jc] (put B = [x, 1] if 0 < x < 1), we get
/.v pax
mdt = j nodt
for each a > 0 and each x > 0. Differentiating with respect to x (and taking
into account the Fundamental Theorem of Calculus), we see that
f(x) = af(ax)
holds for each x > 0 and each a > 0. Letting x — 1, we obtain
m = ^
for all a > 0, as desired.
38. THE RIESZ REPRESENTATION THEOREM
Problem 38.1. IfX is a compact topological space, then show that a continuous
linear functional F on C(X) is positive if and only ifF(l) — \\F\\ holds.
Solution. Let F be a continuous linear functional on C(X), where X is
compact. Note first that
{/ € C(X): H/lloo < 1} = {/ g C{X): \f\ < l}.
Thus, if F is also positive, then
|F|| = sup{F(/): / g C(X) and H/IU < 1}
= sup{F(/): / g C(X) and |/| < 1} = F(l).

366
Chapter 7: SPECIAL TOPICS IN INTEGRATION
On the other hand, assume F(l) = ||F||. Let 0 < / e C{X) be nonzero, and
Put * = mZ' Clearlv' II1 - *lloo < 1. Thus,
F(l)-F(g) = F(l-g)<||F|| = F(l)
holds, which implies F(g) > 0. Therefore, F(/) = \\f\\ooF(g) > 0 holds, and
so F is a positive linear functional.
Problem 38.2. Let X be a compact topological space, and let F and G be two
positive linear functionals on C(X). //F(l) + G(l) < \\F - G\\, then show that
F a G = 0.
Solution. Since F, G > 0, it follows that F-G <FvG and G-F <FvG,
and so \F — G\ < F v G. Thus, by the preceding problem
|F-G|| < ||FvG||=FvG(l)< \\F + G\\ < |F|| + ||G||
= F(l) + G(l)< ||F-G|,
and hence, F v G(l) = F(l) + G(l) holds. From F + G = FvG + FaG, it
follows that
||F a G|| = F A G(l) = F(l) + G(l) - F v G(l) = 0,
and so F a G = 0.
Problem 38.3. Let X be a Hausdoijf locally compact topological space and let
cq(X) = {/ e C(X): V € > 0 3 K compact with \f{x)\ <eVx <£K}.
Show that:
a. co(X) equipped with the sup norm is a Banach lattice.
b. The norm completion ofCc(X) is the Banach lattice co(X).
Solution, (a) Clearly, cq(X) with the sup norm is a normed vector lattice. For
the completeness, let {/„} be a Cauchy sequence of cq(X). Then {/„} converges
uniformly on X to some function /. By Theorem 9.2 we infer that / e C(X).
Now, if e > 0 is given, pick some n with \\f„ — /||oo < £, and then choose some
compact set K with |/„(;0| < £ for x £ K. Thus,
|/(*)| < \fn(x) - fix)\ + \fn(x)\ <e + e = 2e
holds for all a: $ K, so that / € c0(X).

Section 38: THE RIESZ REPRESENTATION THEOREM
367
(b) Obviously, CC(X) is a vector sublattice of cq(X). We have to show that CC(X)
is dense in co(X).
To this end, let / e co(X) and let e > 0. Pick some compact set K with
\f(x)\ < e for x £ K, and then choose some open set V with compact closure
such that K C V. By Theorem 10.8 there exists a function g:X —> 1R with
AT -< g < V. Then, /g e CC(X) and ||/ - fgW^ < 2e holds, proving that
CC(X) is dense in the Banach lattice co(X).
Problem 38.4. Let F be a positive linear functional on CC(X), where X is
Hausdorff and locally compact, and let p be the outer measure induced by F on X.
Show that if p* is the outer measure generated by the measure space (X, B, p),
then p*(A) = p(A) holds for evejy subset A ofX.
Solution. Let A c X. We know that
p{A) = inf{p(V): V open and A C V }.
So, if A c V holds with V open, then ju*(j4) < At*(V) = /x(V) also holds,
and thus p*(A) < p(A). On the other hand, by Theorem 15.11 there exists some
B € B with A C 5 and p.* (A) = p(B). Thus, /x(A) < p(B) = M*04), proving
that n*(A) = jLt(A) holds.
Problem 38.5. Let p and v be two regular Borel measures on a Hausdorff locally
compact topological space X. Then show that p > v holds if and only if f f dp. >
ffdv for each f eCc(X)+.
Solution. Let p and v be two regular Borel measures on a Hausdorff locally
compact topological space X.
Assume first that p > v holds (i.e., assume that p(A) > v(A) holds for each
A G B). Clearly, if 0 is a /i-step function of the form (j) = 2Z/=i aiXAj with each
di > 0 and each A\ e B, then 0 is a u-step function and f(pdp> fcfrdv holds.
Now, let 0 < / e CC(X). Since
/-l ([fl, «) = [/"' ((-co, a))]c O Z"1 ((-oo, b)) € B,
it follows from Theorem 17.7 that there exists a sequence {</>„} of /tx-step functions
of the preceding type satisfying <pn(x) t f{x) for each jc € X. This implies
/fdp — lim l<pndp> lim /0„<iv = j f dv.
n-+ooJ n-+oo J J
For the converse, assume that J f dp, > ffdv holds for each 0 < / e CC(X).
In view of the regularity of the measures, in order to establish that p > v holds
it suffices to show that p(K) > v(K) holds for each compact set K. To this end,

368
Chapter 7: SPECIAL TOPICS IN INTEGRATION
let K be a compact set. Given e > 0, choose an open set V such that K c. V
and fi(V) < n(K) + e. By Theorem 10.8 there exists a function / e CC(X) such
that K < f < V. Now note that from xk < f < Xv/tt follows that
v(K) = Jxk dv <Jfdv< jfdfi < jxv dfJL = p(V) < n(K) + e
for all £ > 0. That is, v(K) < p(K) holds, as desired.
Problem 38.6. Fix a point x in a Hausdorjf locally compact topological space
X, and define F(f) = f(x)for each f e CC(X). Show that F is a positive linear
functional on CC{X) and then describe the unique regular Borel measure /x that
satisfies F(f) = // dfifor each f e CC(X). What is the support of \i?
Solution. Clearly, F is a positive linear functional. The regular Borel measure
representing F is the Dirac measure with "base point" at x. Its support is, of
course, the set {x}.
Problem 38.7. Let X be a compact Hausdorff topological space. If fx and v
are regular Borel measures, then show that the regular Borel measures fiw and
fji a v satisfy
a. SuppOz v v) = Supp fi U Supp v, and
b. Supp(/z A v) c Supp fi fl Supp v.
Use (b) to show that z/Supp/x fl Supp v = 0, then /x J_ v holds. Also, give an
example for which Supp(/x av)^ Supp/x fl Supp v.
Solution, (a) Let A = Supp(/x v v), B = Supp/x, and C = Supp v. From
/x</xvy, v < /x v i\ and /x v v(Ac) = 0, it follows that /x(Ac) = v(Ac) = 0,
and so B C A and C Q A. That is, B\JC c A. On the other hand, the inequality
fjLV v < fi + v implies
li v v(Bc fl Cc) < (/x + v)(Bc n Cc) < fi(Bc) + y(Cc) = 0,
and so A C (5C D Cc)c = 5 U C.
(b) The inclusion follows easily from the inequalities
/x a v < fi and p, av < v.
If Supp/x fl Suppv = 0, then by part (b) Supp(/x a v) = 0 holds, and so
/x a v = 0. For an example showing that equality need not hold in (b), let X = R,
/x = the Lebesgue measure, and v = the Dirac measure with "base point" at 0.

Section 38: THE RIESZ REPRESENTATION THEOREM
369
By Problem 36.5, we have p, a v = 0. Therefore, Supp(/x A v) = 0 holds, while
Supp fM n Supp v = r n {0} = {0}.
Problem 38.8. Let X be a Hausdorff locally compact topological space X.
Characterize the positive linear junctionals F on CC(X) that are also lattice
homomorphisms; that is, F(f v g) = max{F(/), F(g)} holds for each pair
f,geCc{X).
Solution. Let F be a positive linear functional on CC(X). Then we shall show
that F is a lattice homomorphism if and only if there exist some c > 0 and some
a € X such that F(f) = cf(a) holds for all / e CC(X).
Clearly, if for some c > 0 and some oeXwe have F(f) = cf(a) for each
/ G CC(X), then F is a lattice homomorphism. For the converse, assume that
F is a non-zero lattice homomorphism. Let p, be the regular Borel measure that
represents F. If .v, y e Supp/x satisfy x ^ y, then it is not difficult to see that
there exist /, g in CC(X) with / a g = 0 and /(x) = ^(3/) = 1. Therefore,
f(/ v *) = F(f + g) = F(/) + F(s) > max{F(/), F(g)}
must hold, which is a contradiction. Thus, Supp ji consists precisely of one point;
let Supp/x = [a). Set c = /x({a}) > 0, and note that for every / e CC(X), we
have
F(f) = jfdn = f(a) • fM({a}) = cf(a).
Problem 38.9. Let X be a Hausdorff locally compact topological space such
that X is an uncountable set. Then show that
a. C*(X) is not separable, and
b. C[0, 1] (with the sup norm) is not a reflexive Banach space.
Solution. (a)Foreach x e X define the positive linear functional FX:CC(X) —>
R by Fx(f) = fM and note that \\FX-Fy\\ = 2 holds for x # y. Clearly, the set
[Fx: x g X} is an uncountable subset of C*(X). Therefore, {£(Fr, 1): x e X}
is an uncountable collection of pairwise disjoint open balls. From this, it easily
follows that no countable subset of C*(X) can be dense in C*(X).
(b) By Problem 11.12, we know that C[0, 1] is separable. If C[0, 1] is reflexive,
then its second dual is likewise separable. But then (by Problem 29.8) its first dual
must be separable, contradicting part (a). Thus, C[0, 1] is not a reflexive Banach
lattice.
Problem 38.10. Let Xbea Hausdorff locally compact topological space. For a
finite signed measure [xonB show that the following statements are equivalent:

370
Chapter 7: SPECIAL TOPICS IN INTEGRATION
a. /x belongs to M^X).
b. fi+ and pT are both finite regular Bore I measures.
c. For each A e B and € > 0, there exist a compact set K and an open
set V with K c A c K mc/z f/raf |/x(£)| < € holds for all B e B with
B cy\L
Solution. (a)=^(b) Pick two finite regular Borel measures fi\ and /X2 such
that p, = jx\ — /X2. Then, /x+ = (/xi — p,i)+ = Mi v M2 ~ M2 holds. By
Theorem 38.5, /xi v /X2 is a finite regular Borel measure, and from this it follows
that fi+ is a finite regular Borel measure. Similarly, pT is a finite regular Borel
measure.
(b) => (c) Note that |/x| = /x+ + /x~ is a finite regular Borel measure. Now, let
A € B and let e > 0 be given. Then, there exists a compact set AT and an open
set V with K c A c V and |/x|(V\/0 < e. Therefore, if £ € B satisfies
B c V \ £, then |/x(£)| < |/x|(£) < |/x|(V \ if) < e holds.
(c) =$► (a) Let A e B and let e > 0. Choose a compact set AT and an open set
V so that (c) is satisfied. Then, by Theorem 36.9, we have
0<p,+(A)-p+(K) = p+(A\K) = sup{p(B): B e B and B c A\£}
and
0</x+(V)-Ax+(>4) = /x+(V\A) = sup{/x(B): B e B and B £ V\A\.
Thus, /x+(A) - p+(K) < e and /x+(V) - /x+(A) < e both hold. Hence, /x+ is a
finite regular Borel measure. Similarly, /x" is a finite regular Borel measure, and
so /x = /x+ — /x~ € Mb(X).
Problem 38.11. A sequence [xn} in a normed space is said to converge weakly
to some vector x if lim f(xn) = f(x) holds for every continuous linear functional
/■
a. Show that a sequence in a normed space can have at most one weak limit.
b. Let X be a Hausdoiff compact topological space. Then show that a
sequence {/„} ofC(X) converges weakly to some function f if and only if
{/„} is norm bounded and lim fn(x) = f(x) holds for each x e X.
Solution, (a) Assume that a sequence [xn] in a normed vector space Y satisfies
lim f(x„) = f(x) and lim f(xn) = f(y) for every / e Y*. Then, f(x - y) = 0
holds for all / e Y*. By Theorem 29.4, we see that x — y = 0, and so [xn] can
have at most one weak limit,
(b) Assume first that the sequence {/„} of C(X) converges weakly to some

Section 38: THE RIESZ REPRESENTATION THEOREM
371
function / € C(X). By Theorem 29.8, {/„} is norm bounded. If x e X, let px
denote the Dirac measure with support {x}, and note that
fnW = jfn dflx —> J'fdflx = f(x).
Conversely, if {/„} is norm bounded and lim/„(.r) = f(x) holds for each
x e X (where, of course, / e C(X)), then the Lebesgue Dominated Convergence
Theorem implies that lim ffndp — ff dp holds for every Borel measure p.
This, coupled with the Riesz Representation Theorem, shows that {/„} converges
weakly to /.
Problem 38.12. Let p be a regular Borel measure on a Hausdorff locally
compact topological space X, and let f e L\ (p). Show that the finite signed measure
v, defined by
v(E) = f fdp
for each Borel set E, is a {finite) regular Borel signed measure. In other words,
show that v € Mb(X).
Solution. We can assume that f(x) > 0 holds for all x. By Problem 22.7, the
set A = {* e X: f(x) > 0} is a a-finite set with respect to /x. Choose a sequence
[Xn] of /z-measurable sets with p(Xn) < oo for each n and Xn f A.
Now, let E be a Borel set and let e > 0; clearly, v(E) = v(EC\A). Select some
n with v(E) — v(Xn DE) < s. Also, using the regularity of p. and Problem 22.6,
we see that there exists a compact set K c Xn n E with
v(X„ 0 E) - v(K) = f fdn- f fdp<e.
Jx„nE Jk
Thus, the compact set K C E satisfies
0 < v(E) - v(K) = [v(E) - v(X„ H E)] + [v(Xn D E) - v(K)] < 2s.
Next, use Problem 22.6 and the regularity of p to see that for each n there
exists an open set Vn satisfying Xn H E c Vn and v(Vn) — v(Xn n E) < j^.
Then, the open set V = U^li Vn satisfies E c V, and in view of V \E C
U£i(y»\*nn£)fweseethat
oo
0 < v(V) - v(E) = v(V\E)<Y^ v(V„ \ X#i H £) < e.

372
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Altogether, the preceding show that v is a regular Borel measure.
Problem 38.13. Generalize part (3) of Theorem 38.5 as follows: Iffi and v are
two regular Borel measures on a Hausdorff locally compact topological space and
one of then is a-finite, then show that /x A v is also a regular Borel measure.
Solution. Let /x and v be two regular Borel measures on a locally compact
Hausdorff topological space X and assume that /x iscr-finite. Also, let co = /xav
and note (in view of co < /x) that co is a a-finite Borel measure which is absolutely
continuous with respect to /x.
Now, let E be a Borel subset of X satisfying fi(E) < co and let e > 0.
Consider co and /x restricted to the Borel sets Be of E (from Problem 12.13 we
know that BE = [B n E: Be B}\ Now, by the Radon-Nikodym Theorem there
exists a (unique) non-negative function / e L\(E, Be, /x) satisfying
co(BnE)= f dfi% for each B € B.
Jbde
Since /x is a regular Borel measure, it follows from Problem 22.6 that there exists
a compact subset K of E such that
0 < co{E) - co(K) = [ fdfi- f fdfi= f fdfi<e.
Je Jk Je\k
Therefore, we infer that
co{E) = sup{co(K): K compact and K c E). (•)
Now, use the cr-finiteness of co to show that (•) holds true for each Borel subset of
E.
It remains to be shown that the measure of every Borel set can be approximated
from above by the measures of the open sets. To this end, let E be an arbitrary
Borel set, and recall that
co(E) = /x a v(E) = inf{/x(B) + v(E\B): B e B and B c £}.
Let c = inf{o;(0): O open and E c 0} and let e > 0. Given B e B
with B C £, choose open sets V and W such that £ C V, E\B C IV,
/x(V) < /x(£) + £, and v(W) < v{E \B) + e. Then, we have
(o(E) < c < co(V U W) < co(V) + £u(W0 < /x(V) + v(W)
< /x(B) + e + v(£ \ B) + c = /x(B) + v(£ \ B) + 2e.

Section 38: THE RBESZ REPRESENTATION THEOREM 373
Thus, co(E) < c < co(E) -f 2e holds for each e > 0, and so co(E) = c, and we
are finished.
Problem 38.14. Show that every finite Borel measure on a complete separable
metric space is a regular Borel measure. Use this conclusion to present an alternate
proof of the fact that the Lebesgue measure is a regular Borel measure.
Solution. Let X be a complete separable metric space, let B be the a-algebra
of all Borel sets of X, let {x\, X2,...} be a dense countable subset of X, and let
\x\ B —> [0, oo) be a measure.
Consider the collection A of subsets of X defined by
A=[A eB: p,(A) = inf{/x(0): A C 0 and 0 open)
= sup{iA(K): K C A and K compact}}.
The collection A has the following properties:
1. A contains the open and closed sets.
To see this, assume first that V is an open set, and let e > 0. For each n let Tn
be the collection of all open balls of the form £(*/, /*) with /• a rational number
less than or equal to £ and B(Xj,r) C V. Clearly, each Tn is at most countable
and V = UB<=jrB holds. For each n pick £J\ ..., Bj!n e Tn such that
1=1 ^
Next,put_C = PlJJl, |J/=i £", and note that C is a totally bounded set. Hence,
its closure C is also a totally bounded set (why?). Since (by Theorem 6.13) C is
a complete metric space in its_own right, it follows from Theorem 7.8 that C is a
compact set. Now, note that C C.V holds, and that
0 < ul(V) - m(C) < n(V) - fi(C) = n{V \ C)
= m(0(^\ U«r)) s £m(v\LJ a?)
n=l i=l n=l /=!
/i=l Z
Therefore,
H(V) = inf{/x(0): V c O and O open}
= sup{/z(AT): A: C V and K compact)
holds, and so V e A.

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Chapter 7: SPECIAL TOPICS IN INTEGRATION
Now, let C be a closed set, and let e > 0. By the preceding, there exists a
compact subset K with /x(X) — fi(K) < e. Then, the compact subset C n K of
C satisfies
/x(C) - ai(C 0 AT) = /x(C \C 0 JO
= /x(C \ tf) < /x(X \ AT) = /z(X) - fi(K) < e.
Also, by the previous part, there exists a compact set K\ with K\ CX\C and
/x(X \C) - /z(^i) < £• Now> the open set 0 =X\K\ satisfies C c O and
M(O) - At(C) = /x(X \ ifO - /x(C) = /x(X) - /x(ffi) - M(C)
= M(*\C)-/*(*,)<£.
Thus,
At(C) = inf{/x(0): C c O and 0 open}
= sup{/z(AT): K C C and AT compact}
also holds, and so C e A
2. If A e A, then Ac e A
From /z(A) = sup{/x(Af): ^ £ -A and K compact}, it follows that
H(AC) = Ai(Jf) - M(A)
= inf{/x(X) - fi(K): K C A and AT compact}
= inf{/z(Afc): AT c A and AT compact}
= inf{/x(0): Ac C 0 and 0 open}.
Similarly, /x(A) = inf{/x(0): A C O and 0 open} implies
/x(Ac) = sup{/x(C): C c Ac and C closed}.
Since, by part (1), fi(C) = sup{fi(K): K C C and AT compact} holds for each
closed set C, we see that
/x(Ac) = sup{/x(AT): K c Ac and AT compact}.
3. If {A,,} is a sequence of A then- (J^li An € .A.
Let {An} c A let A = LXli A„, and let £ > 0. For each h pick some open set
<9„ with An c On and /z(0„ \ A„) < e2~n. Then, the open set O = IJ^Li #*

Section 38: THE RIESZ REPRESENTATION THEOREM
375
satisfies A c 0 and from O \ A = (J~ , 0„ \ U~ i AH £ UZ\(°n \ An), we
get
MO ~ M(^) = M(0 \ A) < /x(jj(0„ \ A„)) < J2 M(0„ \ An) < 6.
n=\ n=\
On the other hand, fix some k with fi(A \ \Jki=[ At) < e, and then for each
I < i < k pick a compact set Kj C A, with /x(A/ \ AT/) < e2~l. Then, the
compact set A!" = |J-=1 AT/ satisfies K C (J^=l A,- C A and
a- a
/x(A) - fi(K) = ijl{A \K) = il(a\\J At) + /x((|J A/) \ k)
/=i /=i
< e + m((U A<) \ *) = £ + m(U A- \ U *.")
i = l /=l /=!
k
< e + YI^A' \&i) <e + e = 2e.
The validity of statement (3) has been established.
Now, from the preceding statements, we see that A is a a -algebra that contains
the open sets. Consequently, every Borel set belongs to A (i.e., A = B), and so
ix is a regular Borel measure.
Now, let us use the previous conclusion to establish that the Lebesgue measure
X on R" is a regular Borel measure. To this end, let A be an arbitrary Borel set.
Also, let Vn (resp. Cn) denote the open (resp. the closed) ball of R" with center
at zero and radius n. Since each C„ is a complete separable metric space in its
own right, it follows from the previous result that X restricted to each Cn is a
regular Borel measure. Therefore, we have
X(A n C„) = sup{A.(AT): K c A n Cn and K compact}.
From A fl C t A, it follows that X(A C\Cn) | A.(A), and an easy argument shows
that
X{A) = sup{X(A:): K c A and K compact}.
Next, note that if X(A) = oo, then
X(A) = inf{A.(0): A c O and O open)

376
Chapter 7: SPECIAL TOPICS IN INTEGRATION
is trivially true. So, assume that X(A) < oo, and let e > 0. By the regularity of
X on Cn (and the fact that Vn is an open set), we see that
X(A H Vn) = inf{A(0 H C„): Vw H A c 0 fl C„ and O open}
= inf{A(0 H Vn): Vn D A c O 0 V„ and O open}
= mf{X(0): A 0 Vn c O and 0 open}.
Therefore, for each a* there exists an open set On with A fl V„ C 0„ and
X(Ort\i4 fll/„) < el~n. Now, the set O = |J!!Li 0„ is open and satisfies
ACO. From
OO 00 oo
o\A = \Jo„\[JAnv„c \J(on \ a n v„\
we see that
00
0 < MO) - A.(A) = X(0\A)<Y^ HOn \AC\Vn) <e.
Hence, X(A) = inf{A(0): ACO and O open} also holds, and so the Lebesgue
measure X is a regular Borel measure.
Problem 38.15. Let X be a Hausdoiff compact topological space. If(p: X -» X
is a continuous function, then show that there exists a regular Borel measure on
X such that
/ f o(f)dn= / fdfi
holds for each f eC(X).
Solution. Let X be a Hausdorff compact topological space and let 0: X -> X
be a continuous function. Fix some co e X and let Cim: t^ —► £oo is a Banach-
Mazur limit (see Problem 29.7). Now, consider the positive linear functional
F:C(X)-»R defined by
F(/) = Cim(f(4>(co)), /(02M), /«>3(a,)),...),
and let p. be the regular Borel measure on X representing F, i.e., F(f) = / / dp.

Section 38: THE RIESZ REPRESENTATION THEOREM
377
holds for each / e C(X). The identity
Cim(x\, x2t • • •) = Cim{x2l x3l...)
for all Ui, JC2,...) G £qo easily implies F(f) = F(/ o <p) for each / e C{X).
Consequently, the regular Borel measure \x satisfies J f dp, = / / o <$>d\x for
each / e C(X).
Problem 38.16. This exercise gives an identification of the order dual C~(X) of
CC(X). Consider the collection M(X) of all formal expressions Mi — M2 with Mi
and \x2 regular Borel measures. That is,
M(X) = {p\ — (jl2: Mi and M2 are regular Borel measures on X).
a. Define p\ —p,2 = ^1 — v2in M(X) to mean n\(A)+v2(A) = v\(A)+p.2(A)
for all A G B. Show that = is an equivalence relation.
b. Denote the collection of all equivalence classes by M(X) again. That is,
Mi — pi2 and v\ — v2 are considered to be identical if pi\ + v2 = v\ + M2
holds. In M(X) define the algebraic operations
(Mi - M2) + (v\ - v2) = (Mi + v\) - (M2 + ^2),
of Mi — <*M2 if of > 0
«(Mi - M2) = 1 / s , v -r n
1 (-g0m2-(-o:)mi '/ « <0.
S/?0w r/zcrr these operations are well defined (i.e., show that they depend only
upon the equivalence classes) and that they make M(X) a vector space.
c. Define an ordering in M(X) by p,\ — p,2 > v\ — v2 whenever
IH(A) + v2(A) > v^A) + p2(A)
holds for each A £ B. Show that > is well defined and that it is an order
relation on M(X) under which M(X) is a vector lattice.
d. Consider the mapping m = Mi — M2 »->• F^from M(X) to C~(X) defined
by FM(/) = // d\i\ - ff d/jL2for each f 6 CC(X). Show that F^ is well
defined and that m »->■ F^ is a lattice isomorphism (Lemma 38.6 may be
helpful here) from M(X) onto C~(X). That is, show that C~(X) = M(X)
holds.
Solution, (a) Clearly, Mi — M2 = Mi — M2 and Mi — M2 = v\ — v2 implies
v\—v2 = Mi~M2- For the transitivity, let Mi~M2 = v\—vi and v\— v2 = co\— co2.
That is, assume that Ml + v2 = v\ + M2 and v\ ■+- co2 = a)\ -f v2. Adding the last
two equalities, we see that
Mi + u>2 + (vi + v2) = 0)i + M2 + (l>l + V2). (*)

378
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Since all measures involved are regular Borel measures, it follows from (•) that
H\(K) -I- o)2(K) = co\(K) -f- ijli{K) holds for each compact subset K of X. The
regularity of the measures implies
Mi (A) + co2(A) = (ox(A) + fi2(A)
for each A e B, and so Mi — M2 = &>i — a>2 holds.
(b) To see that the addition is well defined, assume that Mi — M2 = v\ — V2 and
co\ — o)2 = x\ — xi- That is, Mi + vi = vi -f M2 and cl>i -f- jto = tti -f- o>2» and so
(Mi + <wi) + (V2 + ^2) = (^1 + n\) + (/X2 + co2). That is,
(jLti - fl2) + (tt>i - 0>2) = (Ml + £Wi) - (/X2 + ft>2>
= (Vl + JTi) - (V2 + 7T2) = (Vl - V2) + (TTi ~ 7T2).
Similarly, the multiplication is well defined. Now, it is a routine matter to verify
that under these algebraic operations M{X) is a vector space.
(c) To verify that > is well defined, proceed as in part (b) above. It is a routine
matter to check that > makes M(X) a partially ordered vector space.
Next, we shall show that M(X) is a vector lattice. It suffices to verify that
(Mi — fi2)+ exists in M(X) for each mi — M2 € M(X). To this end, let /xi — M2 in
jM(X). By Theorem 38.5, Mi v M2 is a regular Borel measure, and we claim that
(/xi — /X2)4" = /A| V/X2-/X2 holds in .M(X). Clearly,
Mi — M2 5: Mi v /jl2 — M2 and 0 < Mi v M2 — M2
both hold. To see that Mi v M2 — M2 is the least upper bound of Mi — M2 and 0,
assume fi\ — fi2 < v{ — v2 = v and v > 0. Then, v 4- M2 is a regular Borel
measure such that v + M2 > Mi and v + M2 >: M2 both hold. By Theorem 38.5,
^ + M2 > Mi v M2, and hence v > mi v M2 — M2 holds in .M(X). This shows
that mi v M2 ~ M2 is the least upper bound of Mi — M2 and 0.
(d) It is a routine matter to verify that m •—► ^V from .M(X) into C~(X) is
well defined and linear. Moreover, since every F € C~(X) can be written as a
difference of two positive linear functionais, the Riesz Representation Theorem
guarantees that m '—► ^V iS onto- To see that m '—> ^m is one-to-one, assume
that FM = 0. Then, ffdfi\ = ffdfjL2 holds for each / € CC(X), and so by
(the Riesz Representation Theorem) Mi = M2- Therefore, m = Mi ~ M2 = 0 and
so m 1—► ^m is one-to-one.
Finally, observe that n > 0 holds in M(X) if and only if FM > 0 holds in
C^(X), and then invoke Lemma 38.6 to see that m '—► ?V is a lanice isomorphism
from .M(X) onto C^(X). Thus, under this lattice isomorphism, we can say that
C;{X) = M(X).

Section 39: DIFFERENTIATION AND INTEGRATION 379
Problem 38.17. This problem shows that for a noncompact space X, in
general C*(X) is a proper ideal ofC^(X). Let X be a Hausdorff locally compact
topological space having a sequence [On] of open sets such that On Q On+\ and
On # a+i for each n, and with X = |J~ i On-
a. Show that if X is a-compact but not a compact space, then X admits a
sequence {On} of open sets with the preceding properties.
b. Choose x\ e 0\ andxn e On\ On-\for n > 2. Then, show that
oo
/r(/) = E/(A'^ f°r /^ccW
defines a positive linear functional on CC(X) that is not continuous.
c. Determine the (unique) regular Borel measure p. on X that represents F.
What is the support of p.?
Solution, (a) Let {Kn\ be a sequence of compact sets with Kn\X. For each n
pick an open set Vn with compact closure such that ^„ C y„. Put On = U/Li ^/
and note that On f X. Since each On has compact closure and X is not compact,
Oni^X holds for each n. By passing to a subsequence of [On], we can assume
that On 7^ On+\ also holds for each n.
(b) Let / € CC(X). Since Supp/ C (J^i On holds and Supp/ is compact,
there exists some k with Supp/ c 0*. Thus, f(xn) = 0 for n > k, and so F
clearly defines a positive linear functional on CC(X).
Next, we shall show that F is not continuous. By Theorem 10.8 there exists
some gn e CC(X) with {*!,...,*,,}■< g„ < On. Therefore, ||F|| > F(gn) = n
holds for each n, and so ||F|| = oo.
(c) The regular Borel measure p that represents F is defined on the Borel set B
by
p(B) = The number of elements of [x\, X2,...} H B.
(If (x\,X2>...} H B is countable, then p(B) = oo, and if [x\, ;c2,...} n B = 0,
then p(B) = 0.) Also, note that
Supp/z = {*h jr2, •••}•
39. DIFFERENTIATION AND INTEGRATION
Problem 39.1. If p. is a Borel measure on R*, then show that p ± X holds if
and only if Dp{x) = Ofor almost all x.

380
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Solution. Assume p, ± X. Choose two disjoint Borel sets A and B with
AU B =JR.k and p,(A) = X(B) = 0. By Lemma 39.3, Dp,(x) = 0 holds for
almost all x in A, and so Dp,(x) = 0 holds for almost all a- in Ra.
Now, suppose that Dfi(x) = 0 holds for almost all x e R;. Use the Lebesgue
Decomposition Theorem 37.7 to write//, = fx\ -f p.2 with/xj <£ X and/x2 -L X.
By the preceding, D/Z2OO = 0 holds for almost all x in R*. Thus, from
Theorem 39.4, it follows that
<%± = dMi = Dm = 0,
and so /xj =0. Therefore, jx = /X2 JL X holds.
Problem 39.2. S/zcw rto //£ is a Lebesgue measurable subset oflRk, then
almost all points of E are density points.
Solution. For each x = (a*i, ..., a*) e JR.k and each e > 0, consider the open
interval /f = fJ/Lite ~~ e» ■*/ + £)« If £ *s a Lebesgue measurable set, then
holds for almost all a. To see this, note first that we can assume without loss of
generality that X(E) < 00 holds (why?). Now, consider the finite Borel measure
fi on R* defined by
li(A)
= X(EC\A)= fXEdX.
Clearly, \i <& X and ^ = xe> By Theorem 39.4, we have Dfi = xe a.e., and
the validity of (•) follows.
Problem 39.3. Write Br(a)for the open ball with center at a € RA and radius
r. If f is a Lebesgue integrable function on R*, then a point a 6 RA is called a
Lebesgue point for f if
?r(tf)) JBAi
r-o+ x(^r(a)) ;flr(fl)
Show that if f is a Lebesgue integrable function on R*, then almost all points
offR.k are Lebesgue points.

Section 39: DIFFERENTIATION AND INTEGRATION
381
Solution. Denote by Q the set of all rational numbers of R. Fix some a e Q,
and let Bn = B(0, n). Now, define the finite Borel measure /x by
li(E) = [ \f(x)-a\dk(x).
Since /x <£ A, it follows from Theorem 39.4 that
Dfi = ^x = 1/ — ^IX^ a-*.
Consequently,
r»™ Mik/e J/o-aI<w> = 1/w- * I w
holds for almost all x in #„, and therefore (since n is arbitrary) (•) holds for
almost all x in RA. Let Ea be a Lebesgue null set for which (•) holds for ail
x <£ Ea. Set E = \Ja(.Q Ea, and note that X(E) = 0.
Now, let y £ £ and let e > 0. Choose some rational number j e Q with
U — f(y)\ < £ (we snaM assume that / is real-valued everywhere). In view of
l/to " /00l < l/to " *l + I* - /OOI, we see that
limsup^y f |/U)-/(30|dX(.r)
+^s^/^)k-/cy)|^(x)
= |/(30-*| + l*-/(30| <2*,
and from this the desired conclusion follows.
Problem 39.4. Let f: R -> IRbe an increasing, left continuous function. Show
directly (i.e., without using Theorem 38.4) that the LebesgueStieltjes measure \Xf
is a regular Borel measure.
Solution. Let (a, b) be an open interval. Then, there exists a sequence {[an, bn]}
of closed intervals with [ani b„] t (a, b). It follows that fJLf([ant bn]) t M/((<z, 6)).
Since every open subset of R can be written as an at most countable union of
pairwise disjoint open intervals, it follows that
fif(O) = sup{fj,f(K): K C. O and K compact}
holds for all open sets O.

382
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Now, let [a, b) be a finite interval. Then, for each point c < a of continuity of
/, we have [a, b) c (c, b) and fif((c, b))-nf([a, b)) = f(a)-f(c). By the left
continuity of /, we see that fif([a, b)) = inf{/x/((c, b)): c < a}. Next, consider
aa-set A with fif(A) < oo. Choose a pairwise disjoint sequence {[an,bn)} with
A = U^Lifoi* ^«)« Given e > 0, for each ;* choose some real number cn < an
with /z/((c„, 6„) \ [fl„, /?„)) < |r, and then set V = UIJLifan **)• Clearly, V is
an open set, A C.V, and
OO 00
/i=l n=l
Thus, /x/(A) = inf[iif(V): A c V and V open}.
Now, to complete the proof, use Problem 15.2. (For a general result about regular
Borel measures, see also Problem 38.14.)
Problem 39.5 (Fubini). Let {/„} be a sequence of increasing functions defined
on [a, b] such that Yl™=\ /«0O = /OO converges in IRfor each x e [a,b]. Then,
show that f is differentiable almost eveiywhere and that /'OO = ]£«Li /w'0O
holds for almost all x.
Solution. Replacing each f„ by /„ — f„(a), we can assume that /„ > 0
holds for each n. Set sn = f\ + • • • -f /«, and note that each ^n is increasing
and s,,00 t /OO holds for each *. Clearly, / is also an increasing function.
By Theorem 39.9, / and all the /„ are differentiable almost everywhere. Since
fn+l = sn+\ — s„ is an increasing function, we see that ^+10O > s'n(x) must hold
for almost all x. Similarly, since f(x) — sn(x) = 5^S«+i ^00 IS an increasing
function, it follows that /'OO > s'n(x) holds for almost all x. Thus,
oo
n=l
exists for almost all x.
Now, for each n let
oo
tn(x) = f(x) - *•„(*) = ]T /,(*) > 0.

Section 39: DIFFERENTIATION AND INTEGRATION
383
Clearly, each tn is an increasing function. Pick a subsequence {s^} of {sn} such
that
00 oo
Yl^M<YXf(b)-skl,(b)]<oo.
The same arguments applied to {fy-J instead of [sn] show that
00 oo
converges for almost all x. In particular, s'k (x) —> /'CO holds for almost all x,
and so
holds for almost all x.
Problem 39.6. Suppose {/,,} is a sequence of increasing functions on [a, b] and
that f is an increasing function on [a,b] such that \ifn \ fi/. Establish that
f'(x) = lim f'n{x) holds for almost all x.
Solution. We shall present a solution of this problem based upon the following
general continuity property of the Differential Operator D: If {/z,,} is a sequence
of Borel measures in R* and \in t M holds for some Borel measure /z, then
Dfin f D/x a.e. also holds.
If this property is established, then using Theorem 39.8, we see that
/„'(*) = Dm/.(jc) t Dfif(x) = f'(x)
must hold for almost all x.
To establish the validity of the continuity property start by observing that if two
Borel measures \x and v satisfy \i < v, then D\i < Dv a.e. holds. Indeed, by
Theorem 39.6 both fi and v are differentiable almost everywhere. If x e R* is
a point for which Dfi(x) and Dv(x) exist and Bn = B{x, £), then
Dn(x) = lim <$} < lim gfi = Dv(x).
Now, let \in t fi. Restricting ourselves to the open balls [x e R*: \\x\\ < n],
we can assume without loss of generality that all measures are finite.

384 Chapter 7: SPECIAL TOPICS IN INTEGRATION
By the Lebesgue Decomposition Theorem 37.7, we can write \in = vn + con
with vn <C A. and &>,, J_ A.. It follows from the proof of Theorem 37.7 that
fjLn AmX fm vn. Clearly, this implies vn < v„+i for each n. From formula (c) of
Problem 9.1, we get
/i„-/i„A^ = 0v (/xn - mX) < 0 v (/zn+1 - mX) = /z„+I - /z„+1 A mX
for each m. Letting m —► oo, we obtain
0)n = Mn ~ ^ < M/i+1 - V„+i = 0)n+\
for each /z. Let vn f v and oj„ t &>• Since /xn = vn+o)„t M, it follows that
/x = v + a;. The relation vn <C A. for each /z easily implies v <£ A.. In view of
(on ± X = 0 for each /z, it follows from Lemma 37.6 that co _L X = 0. That is,
v <£ A. and a; J_ A. both hold, and so /x = v + co is the Lebesgue decomposition
of /jl with respect to A..
From Problem 39.1 (or by repeating the proof of Theorem 39.6), we see that
Dfin(x) = Dvn(x) and Dfx(x) = Dv(x) both hold for almost all a*. Let D\xn =
^ = fn for each n. In view of /xn(R*) = ffndX < £i(lR*) < oo, Levi's
Theorem 22.8 shows that there exists some / € L,(RA) with /„ t /• Now
note that vn(E) = fE fn dX implies v(E) = fE f dX for each Borel set E. This
implies / = Dv a.e., and so
Dfxn(x) = Dvn(x) = /„(*) t /OO = Dv(x) = D/x(a)
holds for almost all jc, as desired.
Problem 39.7. This problem reveals some basic properties of functions of
bounded variation on an intei'val [a,b].
a. /// is differentiate at every point and |/'(jc)| < M < oo holds for
all x e [a,b], then show that f is absolutely continuous (and hence, of
bounded variation).
b. Show that the function /: [0, 1] —> IR defined by
(0 ifx = 0
fM~ (a-2cos(^) if0<x < 1
is differentiate at each x, but is not of bounded variation (and hence, f
is continuous but not absolutely continuous).
c. // / is a function of bounded variation and \f(x)\ > M > 0 holds for
each x € [a,b], then show that g(x) = -~r is a function of bounded
variation.

Section 39: DIFFERENTIATION AND INTEGRATION
385
d. If a function f:[a,b] —> R satisfies a Lipschitz condition {i.e., if there
exists a real number M such that \f(x) — f(y)\ < M\x — y\ holds for all
x, y £ [a. b]), then show that f is absolutely continuous.
Solution, (a) If (a\, b\),..., (an, bn) are pairwise disjoint open subintervals of
[a, b], then by the Mean Value Theorem we have
n n
J2\f(bi) - /to) | < AfJ>/ - at).
/=l ;'=1
The preceding easily implies that / is an absolutely continuous function.
(b) Only the differentiability of / at zero needs verification. The inequality
4^|H,cos(^)|<,
for 0 < x < 1 yields /'(0) = 0. Now if
P - (o pL. I i IT IT A
then an easy computation shows that the variation of / with respect to the partition
P„ is
*=1
This implies V/ = oo.
(c) Note that for each a < x < y < b, we have
\sM-8iy)\ = W$s&\fW-fW\-
Therefore, Vg < jp.Vf < oo holds.
(d) Let a function f:[a,b] —► R satisfy a Lipschitz condition as stated in the
problem and let e > 0. Put 8 = ~ > 0 and note that if (a\, b\),..., (ani bn) are
pairwise disjoint open subintervals of [a, b] satisfying YH=\(°i ~~ ai) < ^» men
i=i /=i
/=i
n
/=!

386
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Problem 39.8. This problem presents an example of a continuous increasing
function (and hence, of bounded variation) that is not absolutely continuous.
Consider the Cantor set C as constructed in Example 6.15 of the text. Recall
that C was obtained from [0,1] by removing certain open intervals by steps. In
the first step we removed the open middle third interval. At the nth step there were
2n~l closed intervals, all of the same length, and we removed the open middle
third interval from each one of them. Let us denote by /",..., /"„_, (counted from
left to right) the removed open internals at the nth step. Now, define the function
/:[0, l]-> [0, 1] as follows:
i. /(0) = 0;
ii. ifx e I? for some 1 < i < 2n~l, then f(x) = (2/ - \)/2n; and
iii. ifx € C with x ^ 0, then f(x) = sup(/(r): / < a* and t e [0,1] \C}.
Part of the graph of f is shown in Figure 7.1.
a. Show that f is an increasing continuous function from [0, 1] to [0, 1].
b. Show that f'(x) = Ofor almost all x.
c. Show that f is not absolutely continuous.
d. Show that p,/ ±X holds.
Solution. Notice again that parts of the graph of the function / are shown in
Figure 7.1.
(a) Straightforward.
(b) Observe that / is constant on each /". This implies that /'(a) = 0 holds
i
y
l
3
4
1
2
1
4
!
~"~
f
_L
l
9
•a
1
2
9
J
1
3
t*
r—*
9*
1 1 1
2 7 8
.3 9 9
•
J
1
X
FIGURE 7.1.

Section 39: DIFFERENTIATION AND INTEGRATION
387
for all x € [0, 1] \ C. Since X(C) = 0, it follows that f'{x) = 0 holds for almost
all x.
(c) If / is absolutely continuous, then by Theorem 39.15 we should have
l = /(D-/(0)= f f'(x)dX(x) = 0,
Jo
which is impossible.
(d) Note that if B = [0, 1] \C, then B U C = [0, 1] and fif(B) = X(C) = 0.
Hence, \if _L X holds.
Problem 39.9. Let f:[a,b] -> 1R be an absolutely continuous function. Then
show that f is a constant function if and only if /'(jc) = 0 holds for almost all x.
Solution. Assume that /: [a, b] —> R is an absolutely continuous function
such that f'(x) = 0 holds for almost all x. By Theorem 39.15, we have
/(*)-/(*) = [Xf'(t)dX(t)=:0
Ja
for each x e [a, b]. Hence, f[x) = f(a) holds for each x e [a, b], so that / is
a constant function.
Problem 39.10. Let f and g be two left continuous functions (on R). Show that
\Xf = fjis holds if and only if f — g is a constant function.
Solution. If / — g is a constant, then it is easy to see that /x/ = fig holds. For
the converse, assume \if — fxg. If x > 0, then
fix) - /(0) = /x/([0f x)) = m5([0, x)) = g(x) - g(0)
implies f(x) - g{x) = /(0) - g(0). Similarly, if x < 0, then
/(0) - /(*) = M/(|>r, 0)) = v.8([x, 0)) = g(0) - g{x),
and so /(jc) — g(x) = /(0) — g(0) holds in this case too.
Problem 39.11. This problem presents another characterization of the norm
dual ofC[a, b]. Start by letting L denote the collection of all functions of bounded
variation on [a, b] that are left continuous and vanish at a.
a. Show that L under the usual algebraic operations is a vector space, and
that f h* ix f, from L to M\>([a, b]), is linear, one-to-one, and onto.

388
Chapter 7: SPECIAL TOPICS IN INTEGRATION
b. Define f > gto mean that f—gis an increasing function. [Note that f > g
does not imply f > g.) Show that L under >: is a partially ordered vector
space such that f > g holds in L if and only if pf > pR in Mb([<2, b]).
c. Establish that L with the norm \\f\\ = V\j\ is a Banach lattice.
d. Show, with an appropriate interpretation, that C*[a,b] = L.
Solution, (a) Clearly, L under the usual algebraic operations is a vector space.
Also, it should be clear that / i—> pf from L to M\,([a,b]) is a linear mapping.
To see that / i—> pj is one-to-one, assume pf = 0. Then,
/(*) = f(x) - f{a) = pf([a, x)) = 0
holds for all a < x < b, and so / = 0.
Next, we shall show that /x i—> /if is onto. Assume at the beginning that
0 < /x € Mb([fl, b]). Define the function
fix) =
0 if x < a
/ji([a,x)) if a < x < b
li{[a,b]) if x>b ,
and note that / is increasing, left continuous, and satisfies f(a) = 0. Thus,
/ 6 L. Now, an easy argument shows that \jl = /x/. Finally, if /x € Mb([tf, &]),
then pick two increasing functions f,geL with /x+ = /x/ and pT = /xr Note
that the function /z = /—g G L satisfies /x = xx+— /x~ = /x/~xx^ = M/-/? = M/*-
(b) Straightforward.
(c) Since f > g holds in L if and only if xxy > /x^ holds in Mt,([a,b])
and Mb ([a, ib]) is a vector lattice, it is easy to see that L must likewise be a
vector lattice. Moreover, by Lemma 38.6 the mapping / i—► /xy is a lattice
isomorphism from L onto Mb([tf,xVj).
Now, note that if / > 0 holds in L (i.e.* if / is an increasing function), then
Vf=f(b)-f(a) = ,xf({a,b]) = lnf\\
holds. Thus, for each / € L we have
«M/| = IlM/ll = I/*i/||=Vl/l.
This implies that ||/|| = V\f\ defines a lattice norm on L, and that / i—> p./
from L onto Mb([a,Z7]) is a lattice isometry. In particular, L with the norm
ll/H = Ki/i is a Banach lattice.
(d) Using the notation of Theorem 38.7, we see that the composition of the two
operators
f\—^'pf^^Fl
V-f
is a lattice isometry from L onto C*[a,b].

Section 39: DIFFERENTIATION AND INTEGRATION
389
Problem 39.12. ///: [a, b] -> IR is an increasing function, then show that f is
Lebesgue integrable and that fa f'{x) dx < f{b) — f(a) holds. Give an example
of an increasing function f for which fa f'(x)dx < f(b) — f(a) holds.
Solution. Let gn(x) — n[f(x + £) — /(*)] for each x e [a,b] (where, of
course, f(x) — f(b) for x > b.) Clearly, gn(x) —> f'{x) holds for almost all
x\ see Theorem 39.9. On the other hand, the relation gn(x) > 0 for each xy and
j g„(x)dx = n[J f(x + fidx - J f(x)dx]
= /l[/ .'f{x)dx~ / fwdx]
= n[J " f(x)dx- j° °f(x)dx]
= "[}/(«- f "fWdx]
<n[Lnnb)-^nf(a)]^f(b)-f(a\
coupled with Fatou's Lemma show that /' e L\ {[a, b]) and that
f f'{x)dx = [ lim g„{x)dx < liminf f gn{x)dx < f(b) - f(a).
Ja Ja n-*°° n-*°° Ja
Finally, an example of a function that yields strict inequality is provided by the
function described in Problem 39.8.
Problem 39.13. // f:[a,b] -> IR is an absolutely continuous function, then
show that
Vf= I \f\x)\dx
Ja
holds.
Solution. By Theorem 39.15 we have /' e L\ ([a, b]) and
tif{E) = j f'(x)dx
holds for each Borel subset E of [a,b].

390
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Start by observing that if a = to < t\ < • • • < tn = b is a partition of [a, b],
then
" i fr'
f^f'\f\x)\dx = f \f'(x)\dx.
1 = 1 A-l •'A
1 = 1 1 = 1 J/'-»
Therefore, Vf < f*\f\x)\dx holds. Now, since the continuous functions are
dense (in the Li-norm) in L\([a, b]) (Theorem 25.3) and the functions of the
form
n
* = ]Cfl/X<i,-,./,).
where a = /o < t\ < • • • < t„ = b is a partition of [a, ft], are dense (in the
Lpnorm) in C[a, b], these functions are also dense in L\([a, b]). Thus, given
s > 0, there exist a partition a = to < t\ < • • • < tn = 6 and real numbers
0i,..., an so that 0 = ^=1 ff/X(/,-_i.r,-) satisfies ||0 — Sgn /' ||i < e. In view of
|(-1 v 0) a 1 - Sgn/'| < |0 - Sgn /'|, we can assume that \4>(x)\ < 1 holds
for all x e [a,b]. Moreover, we have
f <t>(x)f'(x)dx = £> [' f'Wdx = £*/[/(*) - /(//.,)]
//
< X>foO-/('/-i)|<V>.
/=i
Next, choose a sequence {0„} of step functions of the previous type satisfying
0„ —► Sgn/' a.e. (see Lemma 31.6 of the text). In view of \<pnf'\ < l/'l> the
Lebesgue Dominated Convergence Theorem implies
/ |/'(x)|d*= / f\x)-Sgn f(x) dx
J a J a
= lim f faWf'Wdx < Vf.
Thus, Vf = fi \f'(x)\ dx holds.

Section 39: DIFFERENTIATION AND INTEGRATION
391
It is interesting to observe that Vf = \fif\([a, b]) also holds. To see this, let
a = to < t\ < • • • < tn = b be an arbitrary partition of [a, b]. Then, we have
/=! /=!
£|/('/)-/fo-i)| = £|m/(['/-i.'/))|
/=!
n
< ^|M/|([r/-1,r/)) = |M/|([fl,6]),
/=i
and so Vf < \[if\([a,b]). On the other hand, if E\,..., En are pairwise disjoint
Borel subsets of [a, b], then
Ek(£.)| = El ( /'to** h E / l/'c*>l<fc
/Vw|.
./a
< / |H-r)|d-r = Vr/
holds, which (by Theorem 36.9) implies that |/x/|([a, b]) < Vf. Consequently,
\Hf\([a,b]) = Vf holds.
Problem 39.14. For a continuously differentiable function f: [a, b] -* R e.y-
tablish the following properties:
a. 77ze signed measure [if is absolutely continuous with respect to the Lebesgue
measure and d[if/d\ — f a.e.
b. Ifg: [a, b] -> R w Riemann integrable, then gf is also Riemann integrable
and
Jgd[if = J g{x)f'(x)dx.
Solution, (a) By Problem 39.7, we know that / is absolutely continuous and
so (by Theorem 39.12) [if is absolutely continuous with respect to the Lebesgue
measure. Now, combining Theorems 39.14(2), 39.8, and 39.4, we see that
^ = Dlxf = f a.e.
(b) Since /' is a continuous function and g is Riemann integrable, it follows
that gf is also Riemann (and hence Lebesgue) integrable over [a, b]. From

392
Chapter 7: SPECIAL TOPICS IN INTEGRATION
fjLf(A) = J^/' dk for every Borel subset A of [a, fc] and Problem 22.15, we see
that
f gdnf= [ gf'dX = f g(x)f'(x)dx.
J[a,b) J[a,b] Ja
Problem 39.15. For each n consider the increasing continuous function fn: R
-> IR defined by
AW =
1 // jc> 0,
/!(*- 1)+1 Z/ 1 -£ <A" < 1,
0 jfx<l-I
///: IR -* IR w a continuous function, then show that
a. / w fifn-integrable for each n, and
b. lim//£/M/.=/(l)-
Solution. Note that Supp/x/n = [1 — £, 1]. This easily implies that / is
ji/n-integrable for each n. In addition, note that /„'(*) = n holds for each
1 — £ < jc < 1. By the preceding problem, we see that
» r1 fUfWdx
ffdlifm=J f(x)Mx)dx = J nf(x)dx =
Therefore, by the Fundamental Theorem of Calculus, we infer that lim ffdfifn=
/(D.
Problem 39.16. Let /: R -> IR be a (uniformly) bounded function and let
E = [x e R: f'(x) exists in R}.
// X(E) = 0, then show that X(f(E)) = 0.
Solution. For each natural number n , let
£„ = {aeE: \f(x)-f(a)\ < n\x - a\ for all a€R}.
Since / is bounded, it is easy to see that E = U^=i £«» an(^ so /(£) ^
USli /(£n) (see Problem 1.1(6)). Thus, in order to establish that X(f(E)) = 0,
it suffices to show that X(/(£„)) = 0 holds for each n. To this end, fix n and
e >0.

Section 39: DIFFERENTIATION AND INTEGRATION
393
From k(E) = 0, we obtain A.(£„) = 0, and so there exists a sequence of open
intervals {(^ -'*i,^+a)} satisfying
CO • CO
En C [J(bk - rk, bk + rk) and 2 ]T rk < s.
k=\ k=\
Now, note that if a e £„, then there exists some m with \bm — a\ < rmt and
hence \f{bm) — f(a)\ < n\bm — a\ < nrm holds. It follows that f(En) C
IXi (/(W ~ ^. /(**) + W). Therefore,
CO CO
£>((/(&*) - /""A, /(**) + /irj) = 2/2 £V, < 2ns.
A = l *=1
Since £ > 0 is arbitrary, we infer that A. (/(£„)) = 0, as desired. (Compare this
problem with Problem 18.9.)
Problem 39.17. This problem presents an example of a continuous function
/:R -» 1R which is nowhere differentiate; this example should be compared
with Problem 9.28. Consider the function 0: [0, 2] —► R defined by 0(a*) = x if
0 < * < 1 and 0(a) = 2 - x if \ < x < 2. Extend 0 to all ofR (periodically)
so that 0(a') = 0(a* + 2) holds for all x € R. Now, de/7/?e //ze function f:
R-» Rfry
/w = EQ)>(4"4
/»=0
S/jow r/iar / is a continuous nowhere differentiate function.
Solution. S ince the series X^n=o(4) converges and 0 < 0C-O — 1 holds for
all x, it is easy to see that the sequence of partial sums of the series f(x) =
]T^o(|)"0(4"a") converges uniformly to / on R. So, by Theorem 9.2, / is a
well-defined continuous function.
Now, fix a*o e R. The proof of the nondifferentiability of / at xq will be based
upon the following property of differentiate functions.
• Ifh: (a, b) —► R is differentiate at some a*o g (a, b) and p. = /z'(a*o)» then
for each € > 0 there exists some 6 > 0 such that whenever x, y e (a, b)
satisfy x < Xq < y andy — x < 8, then |ii2!lzj£i — p\ < €.

394
Chapter 7: SPECIAL TOPICS IN INTEGRATION
This conclusion follows easily from the inequalities
h(y)-h(x)
y-x
M
[/Ky)-/>(Ao)-M(y-Ao)]+[//(v0)-//(.v)-M(An-A)]
\—x
h(y)-h(xg)-fjL{x-x0)
y-*o
V-Vp
V—.V
Ky)-h(xo) _
y-*o
M
+
+
/i(.vo)-/;(a)
A'o-A-
//(aq)-//(v)-m(aq-a))
Ao-V
AO-A
y-.\
Now, for each natural number m, then there exists a unique integer km such that
km <4mA'0 <fc« + l. Let
sm=4~mkm and /w = 4"m(^ + 1),
and note that sm < xq < tm holds for each m. From tm — sm = 4~m, we see that
lim/m = lim.ym = x$.
The reader should keep in mind that if p and q are two integers, then 0(p) —
(j)(q) = 0 if p — g is an even integer and |0(p) — 0(<?)l = 1 if p — ^ is an odd
integer. Next, observe that if n is a non-negative integer, then 4% — 4% = 4n"w.
So, from the definition of 0, we have:
a. if n > m, then 0(4%) - 0(4%) = 0,
b. if n = m, then 0(4%) - 0(4%) = 1, and
c. if 0 < n < m, then 0(4%) - 0(4%) = 4n"w.
Therefore, for each m we have
\f(tm) ~ f(sm)\ = |£(f)"[0(4"Z„) - 4>(4°Sm)]
I m
= E(3)"[*(4"'-)-*(4"j-)l
/i=0
m-l
M!r-Ear4n-m>2(!r
n=0
This implies I Zif2iLZi£a) I > Zl for each m. Now, a glance at (•) shows that /
cannot be differentiable at ao. Since a'o is arbitrary, / is differentiable at no point
of JR.

Section 40: THE CHANGE OF VARIABLES FORMULA
395
40. THE CHANGE OF VARIABLES FORMULA
Problem 40.1. Show that an open ball in a Banach space is a connected set.
That is, show that if B is an open ball in a Banach space such that B = <D\ U On
holds with both 0\ and Oi open and disjoint, then either 0\ — 0 or
O2 = 0.
Solution. Let B be an open ball in a Banach space. Assume by way of
contradiction that there exist two nonempty open sets 0\ and Oi such that B = 0\ U On
and 0\ fl 02 = 0. Fix two elements a e 0\ and b e #2, and then define
the function /: [0, 1] —► B by /(/) = ta + (1 - t)b. Clearly, /(0) = 6 and
/(l) = a. Moreover, in view of the inequality
|/(0 - m\ = ||(r - 5)fl + (j - Oft|| < (INI + l|6||)|f - si
we see that / is a (uniformly) continuous function.
Let a = inf{f e [0, 1]: f{t) € Oj}. Choose a sequence {or,,} of [0, 1]
with an —► a and /(a„) e Oi for each n. By the continuity of / we have
f(an) -> /(a). Since 02 is open and disjoint from 01, it follows that f(a) £ #2-
In particular, a > 0 must hold. Thus, there exists a sequence [f}n) of real numbers
with 0 < f}n < a for each n and /3n —> a. By the definition of a, we see
that /(/?„) 6 #2 holds for each n, and hence, as above f(a) £ 0\. Now, note
that
f(a) i Ox U 02 = 5
holds, which is impossible.
Problem 40.2. LetT'.V -> R* be? C1 -differentiate. Show that the mapping
x h> T\x)from V into L(Rk, lRk) is a continuous function.
Solution. We know that
T\x)
So, if a = (<2i,..., ak) e R* satisfies ||a||2 = (Z!/=i ^f)1 = 1.tnen usinS tne

396
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Cauchy-Schwarz inequality, we see that
|[r'(x)-rWH2 = (E[De«-i£wH2)*
i = l 7=1
sltCEt^-ewf)-^)]1
1=1 y=i y=i
-(EB^-goof)'.
\-=i i=i
Consequently,
I7"(*) - T'(y)\\ = sup{||[r(jr) - 7"(y)]a||2: IH2 = 1}
1 = 1 7=1
for each pair x, y eV. This inequality, coupled with the fact that T: V —► RA
is C ^differentiate, implies that x i—► T\x) from V into L(R*,R*) is a
continuous function.
Problem 40.3. 5/zovv //2a/ the Lebesgue measure on 1R2 is "rotation" invariant.
Solution. A "rotation" of the plane is a linear operator T: 1R2 —► R2 whose
representing matrix A is orthogonal (i.e., it satisfies A A1 = A1 A = I). Any such
orthogonal matrix is of the form
a — T cos^ sm^ 1
"" L — sin 9 cos 6 J '
where 0 represents the angle of rotation; see Figure 7.2.
In particular, note that det A = 1. Thus, by Lemma 40.4, we see that
X(A(E)) = \dtt A\k(E) = k{E)
holds for each Lebesgue measurable subset E of R2.
Problem 40.4 (Polar Coordinates). Let
E = {(r, 6) e R2: r > 0 and0<0 < 2tt}.

Section 40: THE CHANGE OF VARIABLES FORMULA
397
V \
J
y
\
,..--<
"V
X
FIGURE 7.2. Rotation by an Angle 9
The transformation T:E -> R2 defined by T(r, 9) = (r cos0, r sin9), or as it is
usually written
x = r cos 9 and y = r sin 9,
is called the polar coordinate transformation on R2, shown graphically in
Figure 73.
a. Show that X(E \ E°) = 0.
b. If A = {(a*, 0): x > 0}, then show that A is a closed subset o/R2 whose
(2-dimensional) Lebesgue measure is zero.
c. Show that T: E° -> R2 \ A is a diffeomorphism whose Jacobian
determinant satisfies JtQ', 9) = r for each (r, 9) e E°.
d. Show that ifG is a Lebesgue measurable subset of E with X(G \ G°) = 0,
then T(G) is a Lebesgue measurable subset oflR2. Moreover, show that if
f€Li(T(G)),then
/ fdX= f(rcos9,rsin9)rdr
Jt(G) J Jg
d9
holds.
1
0
2k
e
\
■.■.■-£><)>*; ••'- ■ ■ • • • •'»..- .-P./.»..-•';Irs-:
J
r
r .
FIGURE 7.3. The Polar Coordinate Transformation

398
Chapter 7: SPECIAL TOPICS IN INTEGRATION
Solution, (a) If we consider the sets X = {(r, 0): r > 0) J = {(;*, 2n): r > 0},
and Z = {(0,0): 0 < 0 < 2tt}, then E \E° = X U Y U Z. To show that
A(£ \ E°) = 0, it suffices to establish that k(X) = A.(y) = X(Z) = 0.
Let Xn = {(r,0): 0 < r < n] and Yn = {(/-, 27r): 0 < r < n}. In view
of X„ C [0, n] x [-£, e] and /„ C [0, n] x [27r - e, 2n + e], we see that
k(Xn) = A.(rfl) = 0 holds for each n. Since Xn f X and Yn t K, it follows that
X(X) = A.(7) = 0.
Also, the inclusion Z C [—£, e] x [0, 27r] implies k(Z) < 4tt£ for each s > 0,
and thus k(Z) = 0.
(b) This is proven in part (a) previously.
(c) Clearly, T:E° —> JR2\A is one-to-one, onto, and C]-differentiable. The
Jacobian determinant is
r 3a
dx
L do
ay-\
dr
3v
'69 J
= det
/r<T.*) = det|£ rv|=det| C0S* sin*
1 dx ov ' ' -r sm0 rcos0
which implies that Jj(r, 0) = /• ^ 0 holds for each (/\ 0) € £°. The preceding
are enough to guarantee that T: E° —► 1R2 \ A is a diffeomorphism.
(d) Clearly, G° c E°. Thus, by part (c), T(G°) is an open subset of R2 \ A and
T:G° —> 7(G°) is a diffeomorphism.
Since T:WL2 —> IR2 (defined by T(i\0) = (rcos0f r sin0)) is a
C1-diffeomorphism, it follows from Lemma 40.1 that k(T(G \G0)) = 0. Now, if we
consider the sets A = G, B = 7(G), V = G°, and IV = T(G°), then
X(A\V) = A.(G\G°) = 0 and k(T(G)\T(G0)) < X(T(G\G0)) = 0 both
hold. Thus, Theorem 40.8 applies and gives us the desired formula.
Problem 40.5. This problem uses polar coordinates (introduced in the preceding
problem) to present an alternate proof of Euler's formula f™e~x dx = y/n/2.
a. For each r > 0, let C, = {(a*, y) e R2: x2 + y2 < r2, x > 0, y > 0} a/z<i
5r = [0, r] x [0, /•]. Show that Cr C 5,- C Cr%/2-
b. ///(*, y) = e-^'+A rfo/2 .s/iow that
J fdX< j fdX < f fdX,
JCr JSr JCrj2
where X is the two-dimensional Lebesgue measure.
c. Use the change of variables to polar coordinates and FubinVs Theorem to
show that
f f,dk = f2 f e~'2t dt d0 = j(l - e'r\

Section 40: THE CHANGE OF VARIABLES FORMULA
399
FIGURE 7.4.
d. Use (b) to establish that
{jy**)2<->
e~2r\
and then let r —► oo to obtain the desired formula.
Solution, (a) Geometrically the three sets are as shown in Figure 7.4.
(b) Since f(x, y) - e~{x2+y2) > 0 holds for all (jc, y), we see that
fXCr < fXsr < fXcr^
and the desired inequality follows.
(c) Consider the polar coordinates transformation described in the preceding
problem. For the set G = {(/, 0): 0 < f < r and 0 < 0 < \}, we have
I fdX= I fdX = J I f(tcosQ,ts\nO)tdtdO
JCr JT{G) J JG
= f1 fe-1tdtd9 = ^{\-e-r2).
Jo Jo
(d) Note that

400
Chapter 7: SPECIAL TOPICS IN IN1EGRATION
Thus, using (b) and (c), we see that
f(l-e-r:)<(£^2^)2<|(l-e-^),
and by letting /• —► oo we get (f£°e~x dx) = j.
Problem 40.6. In R4, "double" polar coordinates are defined by
jt = rcos0, y = rsin0, z = pcos0, iu = psin0.
State the change of variables formula for this transformation, and use it to show
that the "volume" of the open ball in R4 with center at zero and radius a is
[it2a\
Solution. The transformation T: R4 —> R4 is given by
T{r, p,0,<p) = (r cos0,r sin#, pcos0, psin0)
for each (r, p, 0, 0) 6 R4. Its Jacobian determinant is
JtQ\p,0,(/)) = det
cos0 sin0 0 0 1
0 0 cos0 sin0
—rsin0 rcosO 0 0
0 0 — psin# pcos0 J
-rp.
Write R4 = R2xR2, and consider the Lebesgue measure on R4 as the product
measure of the corresponding Lebesgue measures on the two factors. Fix a > 0,
and let
E = {(r, p): r > 0, p > 0, and r2 + p2 < a2}
and
F = [0,2tt] x [0,2tt]CR2.
Put G = £ x F c R2 x R2 = R4, and note that T(G) = £, the open ball
of R4 with center at zero and radius a. Now, if C = {(/% p): rp = 0} C R2
and D = {(r, p, 0,0): r > 0, p > 0} C R4, then both sets are closed in their

Section 40: THE CHANGE OF VARIABLES FORMULA
401
corresponding spaces and their corresponding Lebesgue measures are zero. Thus,
if
V = (E\C)x [(0,27r)x(0,27r)] and W = £\D,
then both V and W are open subsets of R4 and T: V —> W is a diffeomor-
phism (onto). Since X(G \ V) = X{B \ W) = 0, Theorem 40.8 combined with
Fubini's Theorem shows that
Volumeof B = X(B) = J dX = / dX = f f J frpdr dpdO dcp
= f((rp dx\ dX = 4n2 f rp dr dp
= 47T • -g- = ^7T a .
Problem 40.7 (Cylindrical Coordinates). Let
E = {(r, 0, z) e R3: /• > 0, 0 < 6 < lit, z e R).
The transformation T: E -> R3 defined by T{r, 0, z) = (r cos0, r sm9, z) or as
it is usually written
x = /• cos c
rsin0, z — z,
is called the cylindrical coordinate transformation, shown graphically in
Figure 7.5.
a. Show that X(E \ E°) = 0.
b. If A = {(a-, 0, z) g R3: x > 0, z e R}, //zew .stow f/wrt A is a closed subset
0/R3 whose (three-dimensional) Lebesgue measure is zero.
FIGURE 7.5. The Cylindrical Coordinate Transformation

402
Chapter 7: SPECIAL TOPICS IN INTEGRATION
c. Show that T:E° -+ 1R?\A is a diffeomorphism whose Jacobian
determinant satisfies jy (;\ 0, z) = r for each (/•, 0, z) e E°.
d. Show that ifG is a Lebesgue measurable subset of E with X(G \ G°) = 0,
then T(G) is a Lebesgue measurable subset qfR3. Moreover, show that if
f eLx{T{G)),then
I fdk= I I I f{r cos 0,r sin 6,z)rdrd0dz
Jt{G) . J J Jg
holds.
Solution. Repeat the solution of Problem 40.4.
Problem 40.8 (Spherical Coordinates). Let
E = {(r, 6, (p) e IR3: r > 0, 0 < 0 < 2tt, 0 < 4> < n}.
The transformation T: E -» R3 defined by
7(r, Q,(p) = (rcos0sin0,rsin0sin0,/*cos0),
or as it is usually written
x = rcos0sin0, y = rsin0 sin$, z = /*cos0,
z\y ca//ed f/ze spherical coordinate transformation, shown graphically in
Figure 7.6.
a. Show that k(E \ E°) = 0.
b. // A = ((x, 0, z): a* > 0 <z/id z £ R}, //ze/z s/ztfw r/zaf A is a closed subset
ofTR? whose ^-dimensional) Lebesgue measure is zero.
FIGURE 7.6. The Spherical Coordinate Transformation

Section 40: THE CHANGE OF VARIABLES FORMULA
403
Show that T: E° —> 1R3 \ A is a diffeomorphism whose Jacobian
determinant satisfies JtO\ 0,$) = —r2 sin 0.
Show that ifG is a Lebesgue measurable subset of E with \{G \ G°) = 0,
then T(G) is a measurable subset vflR?. In addition, show that if f e
L\{T(G)),then
L
T(G)
holds.
f dk— I I I f(r cos 6 sincp, r sin 9 sirup, r cos (j))r2 sin(pdr d9 d$
Solution. Repeat the solution of Problem 40.4.

PROBLEMS IN
REAL ANALYSIS
2nd Edition
A Workbook With Solutions
Charalambos D. Aliprantis
Owen Burkinshaw
Professors Aliprantis and Burkinshaw's Problems in Real Analysis,
2nd edition, is designed to equip the reader with the tools to suc-
ceed in the real analysis course. Published as a companion to their
successful Principles of Real Analysis, 3rd edition, this book teaches
the basic methods of proof and problem-solving by presenting the
complete solutions to over 600 problems that appear in Principles
of Real Analysis. The problem sets cover the entire spectrum of
difficulty: some are routine, some require a good grasp of the
material involved, and some are exceptionally challenging.
This is the first book to offer complete solutions to graduate level
problems in real analysis. It is ideal for all undergraduate and first
year graduate analysis courses. Students and scholars from all
branches of science and engineering will also find this collection
of problems an invaluable reference source.
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