Problems and Solutions for Undergraduate Real Analysis II 9789887879763, 9789887879770, 9887879770

Citation preview

Problems and Solutions for Undergraduate Real Analysis II

by Kit-Wing Yu, PhD [email protected]

c 2019 by Kit-Wing Yu. All rights reserved. No part of this publication may be Copyright reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. ISBN: 978-988-78797-6-3 (eBook) ISBN: 978-988-78797-7-0 (Paperback)

ii

About the author

Dr. Kit-Wing Yu received his B.Sc. (1st Hons), M.Phil. and Ph.D. degrees in Math. at the HKUST, PGDE (Mathematics) at the CUHK. After his graduation, he has joined United Christian College to serve as a mathematics teacher for at least nineteen years. He has also taken the responsibility of the mathematics panel since 2002. Furthermore, he was appointed as a part-time tutor (2002 – 2005) and then a part-time course coordinator (2006 – 2010) of the Department of Mathematics at the OUHK. Apart from teaching, Dr. Yu has been appointed to be a marker of the HKAL Pure Mathematics and HKDSE Mathematics (Core Part) for over thirteen years. Between 2012 and 2014, Dr. Yu was invited to be a Judge Member by the World Olympic Mathematics Competition (China). In the area of academic publication, he is the author of four books • A Complete Solution Guide to Real and Complex Analysis I. • Problems and Solutions for Undergraduate Real Analysis I. • Mock Tests for the ACT Mathematics. • A Complete Solution Guide to Principles of Mathematical Analysis. Besides, he has published over twelve research papers in international mathematical journals, including some well-known journals such as J. Reine Angew. Math., Proc. Roy. Soc. Edinburgh Sect. A and Kodai Math. J.. His research interests are inequalities, special functions and Nevanlinna’s value distribution theory.

iii

iv

Preface

This book “Problems and Solutions for Undergraduate Real Analysis II ” is the continuum of the first book “Problems and Solutions for Undergraduate Real Analysis I ”. Its aim is the same as its first book: We want to assist undergraduate students or first-year students who study mathematics in learning their first rigorous real analysis course. “The only way to learn mathematics is to do mathematics.” – Paul Halmos. My learning and teaching experience has convinced me that this assertion is definitely true. In fact, I believe that “doing mathematics” means a lot to everyone who studies or teaches mathematics. It is not only a way of writing a solution to a mathematical problem, but also a mean of reflecting mathematics deeply, exercising mathematical techniques expertly, exchanging mathematical thoughts with others effectively and searching new mathematical ideas unexpectedly. Thus I hope everyone who is reading this book can experience and acquire the above benefits eventually. The wide variety of problems, which are of varying difficulty, include the following topics: Sequences and Series of Functions, Improper Integrals, Lebesgue Measure, Lebesgue Measurable Functions, Lebesgue Integration, Differential Calculus of Functions of Several Variables and Integral Calculus of Functions of Several Variables. Furthermore, the main features of this book are listed as follows: • The book contains 226 problems, which cover the topics mentioned above, with detailed and complete solutions. Particularly, we include over 100 problems for the Lebesgue integration theory which, I believe, is totally new to all undergraduate students. • Each chapter starts with a brief and concise note of introducing the notations, terminologies, basic mathematical concepts or important/famous/frequently used theorems (without proofs) relevant to the topic. • Three levels of difficulty have been assigned to problems: Symbol ⋆ ⋆

⋆

Intermediate

⋆

⋆

Level of difficulty Introductory

⋆

Advanced

Meaning These problems are basic and every student must be familiar with them. The depth and the complexity of the problems increase. Students who target for higher grades must study them. These problems are very difficult and they may need some specific skills.

v

vi • Colors are used frequently in order to highlight or explain problems, examples, remarks, main points/formulas involved, or show the steps of manipulation in some complicated proofs. (ebook only) If you find any typos or mistakes, please feel free to send your valuable comments or opinions to [email protected] Any updated errata of this book or news about my new book will be posted on my new website: https://sites.google.com/view/yukitwing/ Kit Wing Yu July 2019

List of Figures

10.1 An example of pointwise convergence. . . . . . . . . . . . . . . . . . . . . . . . .

2

10.2 An example of uniform convergence. . . . . . . . . . . . . . . . . . . . . . . . . .

2

15.1 The Inverse Function Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 16.1 The subinterval in R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 16.2 The outer and the inner Jordan measures. . . . . . . . . . . . . . . . . . . . . . . 171 16.3 The mapping φ : E → D. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 16.4 The mapping φ : E → R.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

vii

List of Figures

viii

Contents

Preface

v

List of Figures

vii

10 Sequences and Series of Functions

1

10.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

10.2 Uniform Convergence for Sequences of Functions . . . . . . . . . . . . . . . . . .

6

10.3 Uniform Convergence for Series of Functions . . . . . . . . . . . . . . . . . . . . .

15

10.4 Equicontinuous Families of Functions . . . . . . . . . . . . . . . . . . . . . . . . .

23

10.5 Approximation by Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

11 Improper Integrals

33

11.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

11.2 Evaluations of Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . .

37

11.3 Convergence of Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . .

42

11.4 Miscellaneous Problems on Improper Integrals . . . . . . . . . . . . . . . . . . . .

50

12 Lebesgue Measure

57

12.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

12.2 Lebesgue Outer Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61

12.3 Lebesgue Measurable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65

12.4 Necessary and Sufficient Conditions for Measurable Sets . . . . . . . . . . . . . .

78

13 Lebesgue Measurable Functions

83

13.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83

13.2 Lebesgue Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85

13.3 Applications of Littlewood’s Three Principles . . . . . . . . . . . . . . . . . . . .

96

14 Lebesgue Integration

105

14.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 ix

Contents 14.2 Properties of Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 109 14.3 Applications of Fatou’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 14.4 Applications of Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . 129 15 Differential Calculus of Functions of Several Variables

141

15.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 15.2 Differentiation of Functions of Several Variables . . . . . . . . . . . . . . . . . . . 147 15.3 The Mean Value Theorem for Differentiable Functions . . . . . . . . . . . . . . . 155 15.4 The Inverse Function Theorem and the Implicit Function Theorem . . . . . . . . 157 15.5 Higher Order Derivatives

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

16 Integral Calculus of Functions of Several Variables

167

16.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 16.2 Jordan Measurable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 16.3 Integration on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 16.4 Applications of the Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . 186 16.5 Applications of the Change of Variables Theorem . . . . . . . . . . . . . . . . . . 189 Index

195

Bibliography

197

CHAPTER

10

Sequences and Series of Functions

10.1 Fundamental Concepts In Chapters 5 and 6, we consider sequences and series of real numbers. In this chapter, we study sequences and series whose terms are functions. In fact, the representation of a function as the limit of a sequence or an infinite series of functions arises naturally in advanced analysis. The main references in this chapter are [2, Chap. 9], [4, Chap. 8], [22, Chap. 3] and [28, Chap. 16].

10.1.1 Pointwise and Uniform Convergence Definition 10.1 (Pointwise Convergence). Suppose that {fn } is a sequence of real-valued or complex-valued functions on a set E ⊆ R. For every x ∈ E, if the sequence {fn (x)} converges, then the function f defined by the equation f (x) = lim fn (x) n→∞

(10.1)

is called the limit function P of {fn } and we can say that {fn } converges pointwise to f on E. Similarly, if the series fn (x) converges pointwise for every x ∈ E, then we can define the function ∞ X fn (x) (10.2) f (x) = n=1

on E and it is called the sum of the series.

Definition 10.2 (Uniform Convergence). A sequence {fn } of functions defined on E ⊆ R is said to converge uniformly to f on E if for every ǫ > 0, there exists an N (ǫ) ∈ N such that n ≥ N (ǫ) implies that |fn (x) − f (x)| < ǫ (10.3)

for all x ∈P E. In this case, f is called the uniform limit of {fn } on E. Similarly, we say that the series fn (x) converges uniformly to f on E if the sequence {sn } of its partial sums, where n X sn (x) = fk (x), k=1

converges uniformly to f on E.

1

Chapter 10. Sequences and Series of Functions

2

Figure 10.1 shows the sequence of functions {fn } on [0, 1), where fn (x) = xn . This sequence of functions converges pointwise but not uniformly to f = 0 because if we take ǫ = 21 , then for each n ∈ N, one can find a xn such that fn (xn ) > 21 , i.e., points above the red dotted line. See also Problem 10.4 below.

Figure 10.1: An example of pointwise convergence.

1 Now Figure 10.2 gives the sequence of functions {fn } on [0, 1], where fn (x) = n2 +x 2 . In fact, the idea of the inequality (10.3) can be “seen” in the figure that the graphs of all fn for n ≥ N lie “below” the line y = ǫ.

Figure 10.2: An example of uniform convergence.

3

10.1. Fundamental Concepts Remark 10.1 (a) It is obvious that the uniform convergence of {fn } to f implies its pointwise convergence to f , but the converse is false. (b) Some books use fn → f and fn ⇒ f on E to denote pointwise convergence and uniform convergence respectively.

The core interest for the convergence problem here is that we want to determine what kinds of properties of functions fn that will be “preserved” under the limiting processes (10.1) or (10.2). In fact, we discover that uniform convergence preserves continuity, differentiation and integrability of the functions fn .

10.1.2 Criteria for Uniform Convergence In the following, we state some methods of testing whether a sequence {fn } converges to its pointwise limit f uniformly. Theorem 10.3 (Cauchy Criterion for Uniform Convergence). Suppose that {fn } is a sequence of functions defined on E ⊆ R. (a) The sequence {fn } converges uniformly on E if and only if for every ǫ > 0, there is an N (ǫ) ∈ N such that m, n ≥ N (ǫ) imply that |fn (x) − fm (x)| < ǫ for all x ∈ E. P (b) The series fn converges uniformly on E if and only if for every ǫ > 0, there is an N (ǫ) ∈ N such that m > n ≥ N (ǫ) imply that m X fk (x) < ǫ k=n

for all x ∈ E.

Theorem 10.4. The sequence {fn } converges uniformly to f on E ⊆ R if and only if Mn = sup |fn (x) − f (x)| → 0 x∈E

as n → ∞. Theorem 10.5 (Weierstrass M -test). Suppose that {fn } is a sequence of functions defined on E ⊆ R and {M nonnegative numbers such that |fn (x)| ≤ Mn for all Pn } is a sequence of P n = 1, 2, 3, . . .. If Mn converges, then fn converges uniformly on E.

10.1.3 Preservation Theorems Theorem 10.6 (Uniform Convergence and Continuity). Suppose that fn → f uniformly on E ⊆ R. If every fn is continuous at p ∈ E, then the uniform limit f is also continuous at p.

Chapter 10. Sequences and Series of Functions

4

Theorem 10.7 (Uniform Convergence and Riemann-Stieltjes Integration). Let a < b. Suppose that α is monotonically increasing on [a, b], all fn ∈ R(α) on [a, b] and fn → f uniformly on [a, b]. Then we have f ∈ R(α) on [a, b] and Z b Z b Z b lim fn dα = lim f dα = fn dα. a n→∞

a

In particular, if the series

∞ X

n=1

uniformly to f on [a, b], then we have Z b Z f dα = a

∞ bX

n→∞ a

fn → f

fn dα =

a n=1

∞ Z X

b

fn dα.

n=1 a

Theorem 10.8 (Uniform Convergence and Differentiability). Suppose that {fn } is a sequence of differentiable functions on [a, b], where a < b. We suppose that (a) {fn (p)} converges for at least one point p ∈ [a, b], and (b) {fn′ } converges uniformly on [a, b]. Then there exists a function f defined on [a, b] such that fn → f uniformly on [a, b] and f ′ (x) = lim fn′ (x) n→∞

(10.4)

on [a, b]. ′ P Furthermore, P if′ the sequences {fn (p)} and {fn } in parts (a) and (b) are replaced by the series fn (p) and fn respectively, then we have the same conclusion and the equation (10.4) is replaced by ∞ X fn′ (x). f ′ (x) = n=1

10.1.4 Uniformly Boundedness and Equicontinuity Now we want to find any similarity between sequences of numbers {xn } and sequences of functions {fn }. In fact, we have two important and basic questions. The first question origins from the Bolzano-Weierstrass Theorem ([25, Problem 5.25]): Every bounded sequence of real (or complex in fact) sequence has a convergent subsequence. Is there any similar result for sequences of functions? To answer this question, we have to define two kinds of boundedness first. Definition 10.9 (Pointwise Boundedness and Uniformly Boundedness). Suppose that {fn } is a sequence of functions defined on E ⊆ R. (a) It is said that {fn } is pointwise bounded on E if for every x ∈ E, there is a finite-valued function g : E → R such that |fn (x)| < g(x) for all x ∈ E and n = 1, 2, . . ..

5

10.1. Fundamental Concepts (b) It is said that {fn } is uniformly bounded on E if there is a positive constant M such that |fn (x)| < M for all x ∈ E and n = 1, 2, . . .. Remark 10.2 It is clear from the definitions that uniformly boundedness certainly implies pointwise boundedness.

It is well-known that if {fn } is pointwise bounded on a countable set E, then it has a subsequence {fnk } such that {fnk (x)} converges for every x ∈ E.a This answers the first question partially. However, the general situation does not hold even if E is compact and {fn } is a uniformly bounded sequence of continuous functions. Next, the second question origins from the fact that a sequence {xn } converges if and only if every subsequence {xnk converges (see [25, Theorem 5.3, p. 50]). Is every convergent sequence of functions contains a uniformly convergent subsequence? The answer is negative even if we assume stronger conditions that {fn } is uniformly bounded on a compact set E. Both failures of the above two questions are due to the lack of the concept of equicontinuity which is stated as follows:

Definition 10.10 (Equicontinuity). Let X be a metric space with metric d and F a family of functions defined on E ⊆ X. The family F is equicontinuous on E if for every ǫ > 0, there exists a δ > 0 such that |f (x) − f (y)| < ǫ for all f ∈ F and all x, y ∈ X with d(x, y) < δ. Theorem 10.11. Suppose that K is a compact metric space. If {fn } is a sequence of continuous functions on K and it converges uniformly on K, then {fn } is equicontinuous on K. The Arzel` a-Ascoli Theorem. Suppose that K is a compact metric space and {fn } is a sequence of continuous functions on K. If {fn } is pointwise bounded and equicontinuous on K, then it is uniformly bounded on K. Furthermore, {fn } contains a uniformly convergent subsequence {fnk } on K.

10.1.5 The Space C (X) and the Approximation by Polynomials Definition 10.12. Suppose that X is a metric space. Then C (X) denotes the set of all complexvalued bounded continuous functions on X. For each f ∈ C (X), we define kf k = sup |f (x)| x∈X

which is called the supremum norm. Then C (X) is a complete metric space with this metric k · k. a

See, for example, [18, Theorem 7.23, p. 156].

Chapter 10. Sequences and Series of Functions

6

Remark 10.3 If A is closed in C (X), then A is call uniformly closed and its closure A is called uniform closure. By Theorem 10.6 (Uniform Convergence and Continuity) or Problem 10.15, we know that the uniform limit of a sequence of polynomials is a continuous function. It is natural to ask its converse: Can a continuous function be approximated uniformly by polynomials? The answer to this question is affirmative and in fact, we state it as follows: The Weierstrass Approximation Theorem. If f : [a, b] → C is continuous, then there exists a sequence of polynomials {Pn } such that Pn → f uniformly on [a, b]. If f is real, then we may take every Pn to be real.

10.2 Uniform Convergence for Sequences of Functions Problem 10.1 ⋆ For each n ∈ N, we define fn (x) =

  

x n,

if n is odd;

1 n,

otherwise.

Prove that {fn } converges pointwise but not uniformly on R. Proof. For every x ∈ R, since lim f2k+1 (x) = lim

k→∞

k→∞

x =0 2k + 1

and

lim f2k (x) = lim

k→∞

k→∞

1 = 0, 2k

we conclude that {fn } converges pointwise to f = 0 on R. However, since x M2k+1 = sup |f2k+1 (x) − 0| = sup = +∞, x∈R x∈R 2k

Theorem 10.4 says that {fn } does not converge uniformly on R. This completes the proof of  the problem. Problem 10.2 ⋆ Verify that the sequence of functions fn (x) = not on R.

x2 +nx n

converges uniformly on [0, 1], but

7

10.2. Uniform Convergence for Sequences of Functions

Proof. Obviously, we have fn (x) → x as n → ∞ for every x ∈ R, so its pointwise limit is f (x) = x. Now for each fixed n ∈ N, we have x2 = ∞, x∈R n

Mn = sup |fn (x) − f (x)| = sup x∈R

so Theorem 10.4 implies that {fn } does not converge uniformly to f (x) = x on R. However, on the interval [0, 1], we note that 1 Mn = n for every n ∈ N so that Mn → 0 as n → ∞. Hence it deduces from Theorem 10.4 again that {fn } converges uniformly to f (x) = x on [0, 1]. We have completed the proof of the problem.  Problem 10.3 (Dini’s Theorem) ⋆ Suppose that {fn } is a sequence of continuous functions defined on [a, b] such that fn (x) ≥ fn+1 (x) for all x ∈ [a, b] and n ∈ N. If {fn } converges pointwise to a continuous function f on [a, b], prove that fn → f uniformly on [a, b]. Proof. Without loss of generality, we assume that f = 0. Since each fn is uniformly continuous on [a, b], the number Mn = sup |fn (x)| x∈[a,b]

is well-defined. Since fn (x) ≥ fn+1 (x) on [a, b], {Mn } is decreasing. If Mn does not converge to 0, then there exists a ǫ > 0 such that Mn > ǫ for all n ∈ N. This means that for each n, we can find a xn ∈ [a, b] with fn (xn ) > ǫ. (10.5) Since {xn } ⊆ [a, b], it follows from the Bolzano-Weierstrass Theorem [25, Problem 5.25, p. 68] that {xn } has a convergent subsequence. Let the limit of the subsequence be p. By the hypothesis, we have fn (p) → 0 as n → ∞. In other words, there is an N ∈ N such that fN (p) < ǫ. Since fN is continuous at p, the Sign-preserving Property [25, Problem 7.15, p. 112] ensures that there is a δ > 0 such that fN (x) < ǫ (10.6) for all x ∈ [a, b] with |x − p| < δ. Now the property fn (x) ≥ fn+1 (x) on [a, b] implies that the inequality (10.6) also holds for all fn with n ≥ N . By the definition of p, we can choose xn with n ≥ N and |xn − p| < δ so that the inequality (10.6) gives fn (xn ) < ǫ

(10.7)

for all n ≥ N . Now it is obvious that the inequalities (10.5) and (10.7) are contrary. Hence we  see that Mn → 0 as n → ∞, completing the proof of the problem. Problem 10.4 ⋆ Show that the hypothesis “f is continuous” in Problem 10.3 (Dini’s Theorem) cannot be dropped.

Chapter 10. Sequences and Series of Functions

8

Proof. Let fn (x) = xn on [0, 1]. Then it is easy to see that fn (1) = 1 for all n ∈ N and fn (x) → 0 as n → ∞ for all x ∈ [0, 1). Thus the limit function f of this sequence {fn } is given by   1, if x = 1; f (x) = (10.8)  0, otherwise.

Assume the convergence was uniform. Then Theorem 10.6 (Uniform Convergence and Continuity) shows that the function f must also be continuous on [0, 1], but it contradicts the definition (10.8). Thus the condition “f is continuous” in Problem 10.3 (Dini’s Theorem) cannot be  dropped, completing the proof of the problem. Problem 10.5 ⋆ Construct a sequence of continuous functions {fn } defined on a compact set K with pointwise continuous limit function f but it does not converge uniformly to f on K.

Proof. Suppose that K = [0, 1] and each fn : [0, 1] → R is defined by  nx, if x ∈ [0, n1 ];      2 − nx, if x ∈ ( n1 , n2 ]; fn (x) =      0, otherwise.

Then it is clear that every fn is continuous on [0, 1]. If x = 0, then fn (0) = 0 for every n ∈ N. / (0, N2 ] and thus fn (x) = 0 If x ∈ (0, 1], then there exists an N ∈ N such that x > N2 so that x ∈ for every n ≥ N . In other words, its limit function is f =0 which is continuous on [0, 1]. However, we note that for each n ∈ N, we have 1 Mn = sup |fn (x) − 0| = fn =1 n x∈[0,1] and we follow from Theorem 10.4 that {fn } does not converge to f = 0 uniformly. This is the end of the proof.  Problem 10.6 ⋆ Suppose fn → f uniformly on E ⊆ R and there is a positive constant M such that |fn (x)| ≤ M on E and all n ∈ N. Let g be continuous on [−M, M ]. Define h = g ◦ f and hn = g ◦ fn for every n ∈ N. Show that hn → h uniformly on E. Proof. Since g is continuous on [−M, M ], we know from [25, Theorem 7.10, p. 100] that g is uniformly continuous on [−M, M ]. Given ǫ > 0, there is a δ > 0 such that |g(x) − g(y)| < ǫ

(10.9)

9

10.2. Uniform Convergence for Sequences of Functions

for all x, y ∈ [−M, M ] with |x − y| < δ. Since fn → f uniformly on E, there is an N ∈ N such that n ≥ N implies that |fn (x) − f (x)| < δ (10.10) for every x ∈ E. Combining the inequalities (10.9) and (10.10), it can be shown that n ≥ N implies |hn (x) − h(x)| = |g(fn (x)) − g(f (x))| < ǫ for every x ∈ E. By Definition 10.2 (Uniform Convergence), {hn } converges uniformly to h on E. We complete the proof of the problem.  Problem 10.7 ⋆ Define fn : [0, 1] → R by fn (x) =

1 . 1 + xn

Prove that {fn } converges uniformly to f on [0, α], but not on [0, 1], where 0 < α < 1.

Proof. Let f be the pointwise limit function of {fn }. If x = 1, then fn (1) = Thus we have f (1) = 12 . Let x ∈ [0, 1). Since lim fn (x) = lim

n→∞

n→∞

1 2

for every n ∈ N.

1 = 1, 1 + xn

we obtain f (x) = 1 on x ∈ [0, 1). In other words, we have f (x) =

  

1 2,

if x = 1;

1,

otherwise.

(10.11)

Next, we fix α ∈ (0, 1). For every x ∈ [0, α], we have xn ≤ αn so that xn + xn αn ≤ αn + xn αn . By this and the definition (10.11), we know that |fn (x) − f (x)| =

Since 0 < α < 1, we have

1 xn αn − 1 = ≤ . 1 + xn 1 + xn 1 + αn

(10.12)

αn = 0. n→∞ 1 + αn lim

Therefore, we conclude from this and the inequality (10.12) that fn → f uniformly on [0, α]. q However, given a positive integer n, we let x ∈ ( n 12 , 1) so that 21 < xn < 1. By this and the definition (10.11) again, it yields that |fn (x) − f (x)| =

1 xn 1 1 2 = = − 1 > n n 1+x 1+x 1+1 4

and then {fn } does not converge to f on [0, 1] by Theorem 10.4. Hence we complete the proof  of the problem.

Chapter 10. Sequences and Series of Functions

10

Problem 10.8 ⋆ For n ∈ N, we suppose that fn = n1 exp(−n2 x2 ) on R. Prove that fn → 0 uniformly on R, fn′ → 0 pointwise on R, but not uniformly on (−M, M ) for every M > 0.

2

Proof. For every x ∈ R, note that ex ≥ 1 and so |fn (x)| =

1 nen2 x2

1 . n

≤

By Theorem 10.4, fn → 0 uniformly on R. By direct differentiation, we have fn′ (x) = −2nxe−n so it is easy to see that fn′ → 0 pointwise on R.

2 x2

,

Assume that there was a M > 0 such that fn′ → 0 uniformly on (−M, M ). Pick ǫ = e−1 . By assumption, there exists an N ∈ N such that n ≥ N implies that |fn′ (x)| =

2n|x| < e−1 en2 x2

(10.13)

1 on (−M, M ). Now we may take N to be large enough so that N > M , i.e., N1 ∈ (0, M ). Then 1 it is legal to put x = N into the inequality (10.13) to get the contradiction that

2e−1 < e−1 . Hence {fn′ } does not converge uniformly to f ′ = 0 on (−M, M ), completing the proof of the  problem. Problem 10.9 ⋆ Suppose that {fn } is a sequence of uniformly continuous functions on R. If fn → f uniformly on R, prove that f is uniformly continuous on R.

Proof. Given ǫ > 0. By the hypotheses, there exists an N ∈ N such that n ≥ N implies that |fn (x) − f (x)|
0 such that |fN (x) − fN (y)|
0. Then one can find an N ∈ N such that n ≥ N implies that |fn (x) − f (x)| < ǫ. For this N , since fN is bounded on E, there is a M > 0 such that |fN (x)| ≤ M on E. Thus we obtain |f (x)| ≤ |f (x) − fN (x)| + |fN (x)| < ǫ + M on E, i.e., f is bounded on E. The second assertion is negative. For example, we consider E = (0, 1) and fn (x) = min(x−2 , n). Then it is trivial to check that fn (x) → x−2 pointwise on (0, 1). Since x−2 is unbounded on the  bounded set (0, 1), this counterexample completes the proof of the problem. Problem 10.11 ⋆ Prove that lim

Z

n→∞ 1

π

2

e−nx dx = 0.

2

Proof. Consider fn (x) = e−nx defined on [1, π]. Clearly, we have fn ∈ R on [1, π] because fn is continuous on [1, π] and 1 Mn = sup |fn (x)| ≤ n → 0 e x∈[1,π] as n → ∞. Thus {fn } converges uniformly to 0 on [1, π] by Theorem 10.4. By Theorem 10.7 (Uniform Convergence and Riemann-Stieltjes Integration), we see easily that Z π Z π 2 −nx2 lim e−nx dx = 0 dx = e lim n→∞ 1

1

n→∞

which completes the proof of the problem.



Problem 10.12 ⋆ Let fn : R → R. Prove that if {fn } converges pointwise to f on R, then it converges uniformly to f on any finite subset of R.

Proof. Let {x1 , x2 , . . . , xm } be a finite subset of R. Given ǫ > 0. Since fn → f pointwise on R, there exists an N (ǫ, xk ) ∈ N such that n ≥ N (ǫ, xk ) implies that |fn (xk ) − f (xk )| < ǫ,

(10.16)

Chapter 10. Sequences and Series of Functions

12

where k = 1, 2, . . . , m. Take N (ǫ) = max(N (ǫ, x1 ), N (ǫ, x2 ), . . . , N (ǫ, xm )). Then it is obvious that the inequality (10.16) holds for all k = 1, 2, . . . , m when n ≥ N (ǫ). In other words, {fn } converges uniformly to f on {x1 , x2 , . . . , xm } which ends the proof of the  problem. Problem 10.13 ⋆ ⋆ Let fn : R → R. Prove that if {fn } converges uniformly to f on all countable subsets of R, then {fn } converges uniformly to f on R. Can the hypothesis “all countable subsets” be replaced by the condition “all Cauchy sequences”?

Proof. Assume that {fn } did not converge uniformly to f on R. In other words, for some ǫ > 0 and for every N ∈ N, there exists a x ∈ R such that |fn (x) − f (x)| ≥ ǫ for some n ≥ N . Particularly, pick N = 1, then one can find a x1 ∈ R such that |fn (x1 ) − f (x1 )| ≥ ǫ hold for some n ≥ 1. In fact, we can find a countable subset E = {xk } of R such that |fn (xk ) − f (xk )| ≥ ǫ

(10.17)

hold for some n ≥ k, where k = 1, 2, . . .. However, the inequality (10.17) means that {fn } does not converge uniformly on the countable set E, a contradiction. The second assertion is false. For example, we  0,      x − n + 12 , fn (x) =      1 2,

consider if x ≤ n − 12 ; if n −

1 2

≤ x ≤ n;

(10.18)

otherwise.

Let {pk } be a Cauchy sequence in R. We claim that fn → 0 uniformly on {pk }. We know from [25, Theorem 5.2, p. 49] that {pk } is bounded by a positive constant M . On the one hand, if n > M + 12 , then pk < n − 21 for every k = 1, 2, . . . and we obtain from the definition (10.18) that fn (pk ) = 0 for every k = 1, 2, . . .. Consequently, fn → 0 uniformly on {pk }. On the other hand, for every n ∈ N, since 1 Mn = sup |fn (x) − 0| = , 2 x∈R Theorem 10.4 implies that {fn } does not converge uniformly on R. This completes the proof of the problem. 

13

10.2. Uniform Convergence for Sequences of Functions Problem 10.14

⋆ ⋆ Suppose that {fn } is a sequence of continuous functions defined on a compact set K ⊂ R, fn → f pointwise on K and f is continuous. Prove that fn → f uniformly on K if and only if for every ǫ > 0, there is a m ∈ N and a δ > 0 such that n > m and |fk (x) − f (x)| < δ imply that |fk+n (x) − f (x)| < ǫ (10.19) for all x ∈ K and k ∈ N. Proof. Suppose that fn → f uniformly on K. Thus given ǫ > 0, there exists an N ∈ N such that n > N implies that |fn (x) − f (x)| < ǫ for all x ∈ K. If we take m = N and δ = ǫ, then the inequality (10.19) holds trivially.

Conversely, given ǫ > 0. By the hypotheses, there exists a m ∈ N and a δ > 0 such that n > m and |fk (x) − f (x)| < δ imply the inequality (10.19) on K and all k ∈ N. Fix p ∈ K. Since fn (p) → f (p) as n → ∞, there exists a k ∈ N such that δ |fk (p) − f (p)| < . 3

(10.20)

Recall that both fk and f are continuous at p, so there exists a neighborhood B(p, rp ) of radius rp > 0 such that x ∈ B(p, rp ) ∩ K implies |fk (x) − fk (p)|
m, then we have |fki +n (x) − f (x)| < ǫ

(10.23)

for all x ∈ B(pi , rpi ) ∩ K. Put N = max(k1 , k2 , . . . , ks ) + m. If n′ > N ≥ ki + m for every i = 1, 2, . . . , s, then we deduce from the inequality (10.23) that |fn′ (x) − f (x)| < ǫ.

(10.24)

Since x is arbitrary in K, we conclude from the inequality (10.24) that fn → f uniformly on K  and we complete the proof of the problem.

Chapter 10. Sequences and Series of Functions

14

Problem 10.15 ⋆ ⋆ Suppose that {Pn } is a sequence of polynomials on R with complex coefficients and Pn → P uniformly on R. Is P again a polynomial on R?

Proof. We claim that there exists an N ∈ N such that deg Pn = deg PN for all n ≥ N . Assume that it was not the case. Since Pn → P uniformly on R, Theorem 10.3 (Cauchy Criterion for Uniform Convergence) ensures that there exists a k ∈ N such that m, n ≥ k imply that |Pm (x) − Pn (x)| < 1 (10.25) for all x ∈ R. For this particular k, our assumption shows that there is an nk > k with the property deg Pnk 6= deg Pk and this definitely shows that sup |Pnk (x) − Pk (x)| = ∞. x∈R

which contradicts the inequality (10.25). Hence this proves the claim. Now we let deg PN = K and Pn (x) = an,K xK + an,K−1 xK−1 + · · · + an,0 ,

(10.26)

for all n ≥ N , where an,K , an,K−1 , . . . , an,0 ∈ C. If an,i 6= aN,i for some i ∈ {1, 2, . . . , K}, then we have sup |Pn (x) − PN (x)| = ∞ x∈R

which contradicts the inequality (10.25) again. In other words, we have an,i = aN,i for all n ≥ N and i = 1, 2, . . . , K. For simplicity, we write aN,i = ai and the polynomial (10.26) can be expressed as Pn (x) = aK xK + aK−1 xK−1 + · · · + a1 x + an,0 , for all n ≥ N . If an,0 → a0 as n → ∞, then the function P (x) must be in the form P (x) = aK xK + aK−1 xK−1 + · · · + a1 x + a0 which is a polynomial on R with complex coefficients. This completes the proof of the problem.  Problem 10.16 ⋆ ⋆ Suppose that fn : R → R and all fn satisfy Lipschitz condition with the same Lipschitz constant K. If fn → f uniformly on R, prove that f also satisfies Lipschitz condition with Lipschitz constant K.

15

10.3. Uniform Convergence for Series of Functions

Proof. For the definitions of Lipschitz condition and Lipschitz constant, you are suggested to read [25, Remark 8.7, p. 142]. Given ǫ > 0, there exists an N ∈ N such that n ≥ N implies that |fn (x) − f (x)|
1. Show that the series ∞ h X π

n=1

2

i − arctan[np (1 + x2 )]

converges uniformly on R. Proof. Recall the fact that arctan x + arctan x1 = 0
0, so

1 π − arctan[np (1 + x2 )] = arctan p 2 n (1 + x2 )

for all x ∈ R. Let f (x) = x − arctan x for x > 0. Since f ′ (x) = on (0, ∞) by [25, Remark 8.3, p. 129] and thus x > arctan x

x2 1+x2

(10.30)

> 0, f is strictly increasing

Chapter 10. Sequences and Series of Functions

16

if x > 0. Therefore, we further reduce the inequality (10.30) to 0
12 . Prove that the series ∞ X

n=1

np (1

x + nx2 )

converges uniformly on R.

Proof. For each n ∈ N, let fn (x) =

np (1

x + nx2 )

be defined on R. Using the A.M. ≥ G.M., we know that 1 np + np+1 x2 √ 2p+1 2 np (1 + nx2 ) x = np+ 2 |x| = ≥ n 2 2

which implies



1 x ≤ p+ 1 p 2 n (1 + nx ) 2n 2

for all x ∈ R. Since p > 21 , it yields from [25, Theorem 6.10, p. 77] that Theorem 10.5 (Weierstrass M -test) shows that the series ∞ X

n=1

np (1

∞ X

n=1

1 n

p+ 21

converges and

x + nx2 )

converges uniformly on R. We have completed the proof of the problem.



17

10.3. Uniform Convergence for Series of Functions Problem 10.19 ⋆ Suppose that each fn : [0, 1] → (0, ∞) is continuous and f (x) =

∞ X

fn (x)

n=1

is also continuous on [0, 1]. Prove that the series

∞ X

fn (x) converges uniformly on [0, 1].

n=1

Proof. Let Sn (x) =

n X

fk (x) and

k=1

Rn (x) =

∞ X

k=n+1

fk (x) = f (x) − Sn (x),

where n = 1, 2, . . . and x ∈ [0, 1]. Since f and Sn are continuous on [0, 1], each Rn is continuous on [0, 1]. By the definition, we also have Rn (x) ≥ Rn+1 (x) for all x ∈ [0, 1] and all n ∈ N. Furthermore, Rn (x) → 0 pointwise on [0, 1]. Hence it follows from Problem 10.3 (Dini’s Theorem) that Rn (x) → 0 uniformly on [0, 1], i.e., n X k=1

fk (x) → f (x)

uniformly on [0, 1]. This ends the proof of the problem. Problem 10.20 ⋆ Prove that f (x) =

∞ X x3 sin nx n=1

is continuous on R.

n2

Proof. Let N ∈ N. On (−N, N ), we have

Since

∞ X N3

n=1

n2

x3 sin nx N 3 ≤ 2. 2 n n

< ∞, Theorem 10.5 (Weierstrass M -test) implies that the series f (x) =

∞ X x3 sin nx

n=1

n2



Chapter 10. Sequences and Series of Functions

18

converges uniformly on (−N, N ). Suppose that fn (x) =

n X x3 sin kx

k2

k=1

on (−N, N ). Then the previous paragraph verifies immediately that fn → f uniformly on (−N, N ). Since each fn is continuous on (−N, N ), Theorem 10.6 (Uniform Convergence and Continuity) implies that f is continuous on (−N, N ). Since N is arbitrary, f is then continuous  on R. We have completed the proof of the problem. Problem 10.21 ⋆

⋆ Suppose that {an } is a sequence of nonzero real numbers. Prove that ∞ ian x X e

n=1

n2

converges uniformly on R to a continuous function f : R → C. Evaluate the limit Z T 1 lim f (x) dx. T →∞ 2T −T

Proof. Clearly, we have

eian x 1 2 ≤ 2 n n for all x ∈ R so that Theorem 10.5 (Weierstrass M -test) implies that the series converges uniformly on R to a function f : R → C. Since the partial sum of the series is obviously continuous on R, Theorem 10.6 (Uniform Convergence and Continuity) ensures that the function f must be continuous on R. Furthermore, since the convergence is uniform, Theorem 10.7 (Uniform Convergence and Riemann-Stieltjes Integration) gives Z T Z T X ∞ ∞ Z ∞ ian x X sin an T 1 X T eian x 1 e 1 dx = dx = . (10.31) f (x) dx = 2 2·a T 2T −T 2T −T n2 2T n n n −T n=1

n=1

n=1

By the Mean Value Theorem for Derivatives [25, p. 129], it is true that | sin x| ≤ |x| for all x ∈ R so that sin a T 1 n 2 ≤ 2 n · an T n for all n ∈ N and all T ∈ R. Next, we follow from Theorem 10.5 (Weierstrass M -test) that the infinite series ∞ X sin an T n2 · an T n=1

converges uniformly on R (with respect to T ). Therefore, it deduces from the equation (10.31) and then using [18, Exercise 13, p. 198] to conclude that Z T ∞ X sin an T 1 f (x) dx = lim lim T →∞ T →∞ 2T −T n 2 · an T n=1

19

10.3. Uniform Convergence for Series of Functions = = =

∞ X 1 sin an T  lim n2 T →∞ an T n=1 ∞ X 1 n2

n=1 π2

6

,

completing the proof of the problem.



Problem 10.22 ⋆ Suppose that f : R → R is continuous and −∞ < a < b < ∞. For each n ∈ N, we define fn : R → R by n−1 k 1X  . f x+ fn (x) = n n k=0

Prove that fn converges uniformly on [a, b].

Proof. Given ǫ > 0. Since f is continuous on R, it is uniformly continuous on [a, b + 1]. Then there exists a δ > 0 such that |f (x) − f (y)| < ǫ (10.32) for every x, y ∈ [a, b + 1] and |x − y| < δ. Particularly, we may choose an n ∈ N such that Fix x ∈ [a, b]. Then we have Z

x

x+1

Z n−1 X f (t) dt − fn (x) = k=0

x+ k+1 n k x+ n

n−1 k  1X  f x+ f (t) dt − . n n

1 n

< δ.

(10.33)

k=0

By the First Mean Value Theorem for Integrals [25, p. 162], for each k = 1, 2, . . . , n − 1, there is a yk ∈ (x + nk , x + k+1 n ) ⊆ [a, b + 1] such that Z

x+ k+1 n k x+ n

f (t) dt =

f (yk ) . n

(10.34)

Substituting the value (10.34) into the equation (10.33), we obtain Z

x+1 x

n−1  n−1 k  X f (yk ) 1 X − f x+ f (t) dt − fn (x) = n n n

≤

k=0 n−1 X

1 n

k=0

k=0

 k  f (y ) − f x + . k n

Finally, by using the uniform continuity (10.32) to the inequality (10.35), we see that Z

x+1 x

f (t) dt − fn (x) < ǫ

(10.35)

Chapter 10. Sequences and Series of Functions

20

for all x ∈ [a, b] and all positive integers n > 1δ . Now if we define F : [a, b] → R by F (x) =

Z

x+1

f (t) dt, x

then we conclude immediately from Definition 10.2 (Uniform Convergence) that {fn } converges  uniformly to F on [a, b]. This completes the proof of the problem. Problem 10.23 ⋆

⋆ Show that the infinite series ∞ X sin nx

n=1

n

converges uniformly on [δ, 2π − δ], where δ ∈ (0, π).

Proof. Given ǫ > 0. Fix δ ∈ (0, π). We claim that the series ∞ ikx X e k=1

(10.36)

k

converges uniformly on [δ, 2π − δ] first. To this end, we note that if x ∈ [δ, 2π − δ], then p p |1 − eix | = 2(1 − cos x) ≥ 2(1 − cos δ) > 0. Therefore, we have

inx

|e

i(n+1)x

+e

imx

+ ··· + e

einx − ei(m+1)x r 2 , |= ≤ ix 1−e 1 − cos δ

(10.37)

where m ≥ n ≥ 0. Recall the summation by parts [18, Theorem 3.41, p. 70] that m X

k=n

ak bk = Am bm − An−1 bn +

m−1 X k=n

Ak (bk − bk+1 ),

(10.38)

where Ak = a0 + a1 + a2 + · · · + ak . Hence we deduce from the inequality (10.37) and the formula (10.38) with ak = eikx that m−1 m X 1 1  eikx Am An−1 X − + − Ak ≤ k m n k k+1 k=n k=n r r m−1  X 1 2 2 1 1 1 ≤ + · + · − 1 − cos δ m n 1 − cos δ k k+1 k=n r r 1 2 2 1 2 + · − ≤ · n 1 − cos δ 1 − cos δ n m r 2 4 . ≤ · n 1 − cos δ

(10.39)

21

10.3. Uniform Convergence for Series of Functions

Consequently, if n is large enough, then we get immediately from the inequality (10.39) that m X eikx < ǫ. k k=n

Thus we conclude from Theorem 10.3 (Cauchy Criterion for Uniform Convergence) that the series (10.36) converges uniformly on [δ, 2π − δ] as desired. Finally, we get from the definition of the modulus easily that m m X eikx X sin kx ≥ , k k k=n

k=n

so we may obtain the same conclusion and thus we complete the analysis of the problem.



Problem 10.24 ⋆

⋆ Suppose that f : R → R is given by f (x) =

∞ X

n=1

n2

1 . + x2

Prove that f is differentiable in R and find f ′ (x).

Proof. Let fn (x) =

1 n2 +x2

and M be a positive constant. Recall that ∞ X 1 π2 fn (0) = . = n2 6 n=1 n=1 ∞ X

In addition, if |x| ≤ M , then we have

so that

|fn′ (x)| ≤

−2x 2M ≤ 4 (n2 + x2 )2 n

∞ ∞ ∞ X X X 1 ′ ′ |fn (x)| ≤ 2M · < ∞. fn (x) ≤ 4 n n=1 n=1 n=1

By Theorem 10.5 (Weierstrass M -test), the series ∞ X

fn′ (x)

n=1

converges uniformly on [−M, M ]. Now it deduces from Theorem 10.8 (Uniform Convergence and Differentiability) that f is differentiable in [−M, M ] and ′

f (x) =

∞ X

n=1

fn′ (x)

=

∞ X

n=1

−2x + x2 )2

(n2

(10.40)

Since M is arbitrary, the formula (10.40) holds for all x ∈ R and we complete the proof of the  problem.

Chapter 10. Sequences and Series of Functions

22

Problem 10.25 ⋆

⋆ Let f (x) =

Z

x 0

sin t dt, where x ∈ R. Prove that t f (x) =

∞ X

n=0

(−1)n x2n+1 (2n + 1)!(2n + 1)

for x ∈ R.

Proof. Since the power series representation of sin t is given by ∞ X (−1)n 2n+1 t , (2n + 1)!

sin t =

n=0

we have ∞

sin t X (−1)n 2n = t . t (2n + 1)! n=0 Let M > 0. Since

for t ∈ [−M, M ] and the series

(10.41)

(−1)n t2n M 2n ≤ (2n + 1)! (2n + 1)! ∞ X

n=0

M 2n (2n + 1)!

converges, it yields from Theorem 10.5 (Weierstrass M -test) that the series on the right-hand side of the representation (10.41) converges uniformly on [−M, M ]. It is definitely that each (−1)n 2n ∈ R on [−M, M ], it follows from Theorem 10.7 (Uniform Convergence and Riemann(2n+1)! t Stieltjes Integration) that f (x) =

Z

x

sin t dt t 0 Z xX ∞ (−1)n

t2n dt (2n + 1)! 0 n=0 ∞ X Z x (−1)n = t2n dt (2n + 1)! 0

=

=

n=0 ∞ X n=0

(−1)n x2n+1 (2n + 1)!(2n + 1)

for all x ∈ [−M, M ]. Since M is arbitrary, our desired result follows. This completes the proof of the problem. 

23

10.4. Equicontinuous Families of Functions

10.4 Equicontinuous Families of Functions Problem 10.26 ⋆ ⋆ Let −∞ < a < b < ∞. Suppose that {fn } is pointwise convergent and equicontinuous on [a, b]. Prove that {fn } converges uniformly on [a, b]. Proof. Given ǫ > 0. Let f be the limit function of {fn } on [a, b]. By the δ > 0 such that ǫ |fn (x) − fn (y)| < 3 for all x, y ∈ [a, b] with |x − y| < δ and all n ∈ N. Taking n → ∞ in the get ǫ |f (x) − f (y)| ≤ 3 for all x, y ∈ [a, b] with |x − y| < δ. By the definition, the inequality is uniformly continuous on [a, b]. Since [a, b] is compact, there exists {x1 , x2 , . . . , xk } ⊆ [a, b] such that [a, b] ⊆

k [

hypotheses, there is a (10.42) inequality (10.42), we (10.43) (10.43) shows that f a set of finite points

(xi − δ, xi + δ).

(10.44)

i=1

Since fn → f pointwise on [a, b], there exists an Ni ∈ N such that n ≥ Ni implies ǫ |fn (xi ) − f (xi )| < , 3

(10.45)

where i = 1, 2, . . . , k. Put N = max(N1 , N2 , . . . , Nk ). If x ∈ [a, b], then the finite open covering property (10.44) ensures that there exists an i ∈ {1, 2, . . . , k} such that |x − xi | < δ. Therefore, for all x ∈ [a, b], it follows from the inequalities (10.42), (10.43) and (10.45) that n ≥ N implies |fn (x) − f (x)| ≤ |fn (x) − fn (xi )| + |fn (xi ) − f (xi )| + |f (xi ) − f (x)| < ǫ. By the definition, {fn } converges uniformly on [a, b]. This completes the proof of the problem.  Problem 10.27 ⋆ Let −∞ < a < b < ∞. Suppose that {fn } is a sequence of continuous functions on [a, b] and each fn is differentiable in (a, b). If {fn } is pointwise convergent on [a, b] and {fn′ } is uniformly bounded on (a, b), prove that {fn } is uniformly convergent on [a, b]. Proof. For all x, y ∈ [a, b] and all n ∈ N, we get from the Mean Value Theorem for Derivatives that |fn (x) − fn (y)| = |fn′ (ξ)| · |x − y| (10.46)

for some ξ ∈ (x, y). Since there is a positive constant M such that |fn′ (x)| ≤ M for all x ∈ [a, b] and n ∈ N, the equation (10.46) gives |fn (x) − fn (y)| ≤ M |x − y|.

Chapter 10. Sequences and Series of Functions

24

By Definition 10.10 (Equicontinuity), we conclude that {fn } is equicontinuous and our desired result follows immediately from Problem 10.26. This completes the proof of the problem.  Problem 10.28 ⋆ Let −∞ < a < b < ∞. Suppose that {fn } is a sequence of twice differentiable functions on [a, b]. Furthermore, we have fn (a) = fn′ (a) = 0 and |fn′′ (x)| ≤ 1 for all x ∈ [a, b] and n ∈ N. Prove that {fn } has a uniformly convergent subsequence on [a, b]. Proof. Applying the Mean Value Theorem for Derivatives to the sequence {fn′ }, we see that there exists a ξ ∈ (x, y) such that |fn′ (x) − fn′ (y)| = |fn′′ (ξ)| · |x − y| ≤ |x − y|

(10.47)

for all x, y ∈ [a, b] and all n ∈ N. Put y = a in the inequality (10.47), we get |fn′ (x)| ≤ |x − a| ≤ b − a for all x ∈ [a, b] and all n ∈ N. Similarly, we apply the Mean Value Theorem for Derivatives to the sequence {fn }, so there is a θ ∈ (x, y) such that |fn (x) − fn (y)| = |fn′ (θ)| · |x − y| ≤ (b − a) · |x − y| for all x, y ∈ [a, b] and all n ∈ N. Given ǫ > 0, if we take δ = estimate (10.48) that ǫ |fn (x) − fn (y)| ≤ < ǫ 2

ǫ 2(b−a) ,

(10.48)

then it follows from the

holds for all x, y ∈ [a, b] satisfying |x−y| < δ and all n ∈ N. By Definition 10.10 (Equicontinuity), the family {fn } is equicontinuous. By the estimate (10.48) again, if y = a, then |fn (x)| = |fn (x) − fn (a)| ≤ (b − a) · |x − a| ≤ (b − a)2 for all x ∈ [a, b] and all n ∈ N. Thus {fn } is uniformly bounded. Hence we follow from the Arzel`a-Ascoli Theorem that {fn } has a uniformly convergent subsequence on [a, b], completing  the proof of the problem. Problem 10.29 ⋆ Suppose that {fn } is a family of continuously differentiable functions on [0, 1] and it satisfies −1 |fn′ (x)| ≤ x p for all x ∈ (0, 1], where p > 1. Furthermore, we have Z

1

fn (x) dx = 0. 0

Prove that {fn } has a uniformly convergent subsequence.

(10.49)

25

10.4. Equicontinuous Families of Functions

Proof. For x, y ∈ [0, 1] with 0 ≤ x < y ≤ 1 and n ∈ N, we have Z y Z y Z y 1 ′ ′ t− p dt = |fn (t)| dt ≤ fn (t) dt ≤ |fn (x) − fn (y)| = x

x

x

1 1 p (y 1− p − x1− p ). p−1

(10.50)

1

Define the function g : [0, 1] → R by g(x) = x1− p which is clearly uniformly continuous on [0, 1]. In other words, given ǫ > 0, there is a δ > 0 such that |x

1− p1

−y

1− p1

| = |g(x) − g(y)|
0 or fn (x) < 0 for all x ∈ [0, 1]. In other words, for each positive integer n, this observation and the continuity of fn guarantee that there exists a xn ∈ [0, 1] such that fn (xn ) = 0. Using the estimate (10.50) with y = xn , we have |fn (x)| ≤

1− 1 2p p 1− p1 − xn p ≤ x p−1 p−1

for all x ∈ [0, 1], i.e., {fn } is uniformly bounded. Now the application of the Arzel`a-Ascoli Theorem implies that it contains a uniformly convergent subsequence on [0, 1]. We have completed  the proof of the problem. Problem 10.30 ⋆ Suppose that {fn } is a collection of continuous functions on [0, 1]. In addition, suppose that {fn } pointwise converges to 0 on [0, 1] and there exists a positive constant M such that Z

0

for all n ∈ N. Is it true that

lim

1

fn (x) dx ≤ M

Z

n→∞ 0

(10.52)

1

fn (x) dx = 0?

Proof. The answer is negative. In fact, for every n ∈ N, we consider fn : [0, 1] → R by  1 2 ); if x ∈ [0, 2n   2n x,    1 1 −2n2 x + 2n, if x ∈ [ 2n , n ); fn (x) =      0, if x ∈ [ n1 , 1].

It is easy to check that each fn is continuous on [0, 1]. Now we have fn (0) = fn (1) = 0

(10.53)

Chapter 10. Sequences and Series of Functions for every n ∈ N. If x ∈ (0, 1), then one can find an N ∈ N such that x ≥

26 1 N

which implies that

fn (x) = 0 for all n ≥ N . Therefore, it means that {fn } pointwise converges to 0 on [0, 1]. Since the graph 1 , n) and ( n1 , 0), we have of the fn is an isosceles triangle with vertices (0, 0), ( 2n Z

1

fn (x) dx =

0

1 2

which implies the truth of the bound (10.52), but the failure of the limit (10.53). Hence we have completed the proof of the problem.  Problem 10.31 (Abel’s Test for Uniform Convergence) ⋆ ⋆ Suppose that {gn } is a sequence of real-valued functions defined on E ⊆ R such that gn+1 (x) ≤ gnP (x) for every x ∈ E and every n ∈ N. If the family {gn } is uniformly bounded on E and if fn converges uniformly on E, prove that the series ∞ X

fn (x)gn (x)

n=1

also converges uniformly on E.

Proof. Since {gn } is uniformly bounded on E, there is a positive constant M such that

for all x ∈ E and all n ∈ N. Since N ∈ N such that n > m ≥ N implies

P

|gn (x)| ≤ M

(10.54)

fn converges uniformly on E, given ǫ > 0, there is an

n X ǫ fk (x) < 3M

(10.55)

k=m

for all x ∈ E. Let Fm,n (x) = n X

k=m

n X

fk (x). Then we have

k=m

fk gk = fm gm + fm+1 gm+1 + · · · + fn gn = Fm,m gm + (Fm,m+1 − Fm,m )gm+1 + · · · + (Fm,n − Fm,n−1 )gn

= Fm,m (gm − gm+1 ) + Fm,m+1 (gm+1 − gm+2 ) + · · · + Fm,n−1 (gn−1 − gn ) + Fm,n gn .

Since gn+1 (x) ≤ gn (x) for every x ∈ E and every n ∈ N, we have gk (x) − gk+1 (x) ≥ 0

(10.56)

27

10.4. Equicontinuous Families of Functions

for all x ∈ E and k = m, m + 1, . . . , n − 1. Combining the estimate (10.55) and the formula (10.56), if n > m ≥ N , then n X fk (x)gk (x) ≤ |Fm,m (x)| · [gm (x) − gm+1 (x)] + |Fm,m+1 (x)| · [gm+1 (x) − gm+2 (x)] k=m

+ · · · + |Fm,n−1 (x)| · [gn−1 (x) − gn (x)] + |Fm,n (x)| · |gn (x)| ǫ ǫ ǫ · [gm (x) − gm+1 (x)] + · · · + · [gn−1 (x) − gn (x)] + · |gn (x)| < 3M 3M 3M ǫ ǫ = · [gm (x) − gn (x)] + · |gn (x)|. (10.57) 3M 3M

Obviously, we have 0 ≤ gm (x) − gn (x) ≤ 2M . Therefore, we follow from the estimates (10.54) and (10.57) that n X fk (x)gk (x) < ǫ k=m

for all x ∈ E. By Theorem 10.3 (Cauchy Criterion for Uniform Convergence), we conclude that P fn (x)gn (x) converges uniformly on E. This completes the analysis of the proof.  Problem 10.32

⋆ ⋆ Let K be a compact metric space and C (K) be equicontinuous on K. Prove that C (K) is also equicontinuous on K.

Proof. Since K is compact, every element f ∈ C (K) must be bounded by the Extreme Value Theorem [25, p. 100]. Given ǫ > 0. Fix θ ∈ (0, ǫ). Since C (K) is equicontinuous on K, there exists a δ > 0 such that ǫ (10.58) |f (x) − f (y)| < 3 for all f ∈ C (K) and all x, y ∈ K with |x − y| < δ. Let F ∈ C (K). Then there is a sequence {fn } ⊆ C (K) such that sup |fn (x) − F (x)| = kfn − F k → 0 (10.59) x∈K

as n → ∞. In other words, there is an N ∈ N such that n ≥ N implies |fn (x) − F (x)|
ǫδ 2 . By Definition 10.2 (Uniform Convergence), Bn → f uniformly on  [0, 1] and we complete the analysis of the proof.

Chapter 10. Sequences and Series of Functions

32

CHAPTER

11

Improper Integrals

11.1 Fundamental Concepts In [25, Chap. 9], we study an integral whose integrand is a function f defined and bounded on a finite interval [a, b]. However, we can extend the concepts of integration to include functions tending to infinity at some certain points or unbounded intervals. This is the content of the socalled improper integrals. The main references we have used here are [1, §10.23], [2, §10.13], [16, Chap. 1] and [27, §6.5].

11.1.1 Improper Integrals of the First Kind Definition 11.1. Let a ∈ R. Suppose that f : [a, +∞) → R is a function integrable on every closed and bounded interval [a, b], where a ≤ b < +∞, i.e., f ∈ R on [a, b]. We set Z

Z

+∞

f (x) dx = lim

b

b→+∞ a

a

f (x) dx,

(11.1)

where the left-hand side is called the improper integral of the first kind. If the limit (11.1) exists and is finite, then we call the improper integral convergent. Otherwise, the improper integral is divergent. Similarly, we may define the following improper integral if the limit exists and is finite: Z

b

f (x) dx = lim

a→−∞ a

−∞

Finally, if both

Z

a

f (x) dx

Z

and

−∞

Z

b

f (x) dx.

+∞

f (x) dx a

are convergent for some a ∈ R, then we have Z Z a Z +∞ f (x) dx + f (x) dx = −∞

−∞

33

+∞

f (x) dx a

(11.2)

Chapter 11. Improper Integrals

34

so that the improper integral on the left-hand side of the expression (11.2) is convergent. If one of the integrals on the right-hand side of the expression (11.2) is divergent, then the corresponding improper integral is divergent. Remark 11.1 We must notice that it can be shown that the choice of a in the expression (11.2) is not important.

11.1.2 Improper Integrals of the Second Kind Definition 11.2. Let a, b ∈ R. Suppose that f : [a, b) → R is a function integrable on every closed and bounded interval [a, c], where a ≤ c < b, i.e., f ∈ R on [a, c]. We set Z c Z b f (x) dx, (11.3) f (x) dx = lim c→b− a

a

where the left-hand side is called the improper integral of the second kind. If the limit (11.3) exists and is finite, then we call the improper integral convergent. Otherwise, the improper integral is divergent. The key of the definition is that f may be unbounded in a neighborhood of b. Similarly, if we have f : (a, b] → R, then we define Z b Z b f (x) dx. f (x) dx = lim c→a+ c

a

Definition 11.3. Let a, b ∈ R and a < c < b. Suppose that f : [a, c) ∪ (c, b] → R is a function integrable on every closed and bounded interval of [a, c) ∪ (c, b] and f (x) → ±∞ as x → c. Then we have Z Z Z b

c

b

(11.4)

c

a

a

f (x) dx

f (x) dx +

f (x) dx =

and the improper integral on the left-hand side of (11.4) exists if and only if the two improper integrals on its right-hand side exist.

Examples of improper integrals of the first and the second kinds can be found in Problems 11.1 and 11.2 respectively.

11.1.3 Cauchy Principal Value We can define the so-called Cauchy Principal Value, or simply principal value, of an improper integral in some cases. Definition 11.4 (Cauchy Principal Value). Let a, b ∈ R and a < c < b. We suppose that f : [a, c) ∪ (c, b] → R is a function integrable on every closed and bounded interval of [a, c) ∪ (c, b] and f (x) → ±∞ as x → c. Then the principal value of the improper integral of f on [a, b] is given by Z b Z b   Z c−ǫ f (x) dx . (11.5) f (x) dx + f (x) dx = lim P.V. a

ǫ→0+

a

c+ǫ

35

11.1. Fundamental Concepts

Now if the improper integral (11.4) exists, then it is easy to see that it equals to its principal value (11.5). However, the converse is not true, see Problem 11.4 for an example.

11.1.4 Properties of Improper Integrals Theorem 11.5. Suppose that a ∈ R and f, g : [a, b) → R are integrable on every closed and bounded interval of [a, b), where b is either finite or +∞. (a) For any A, B ∈ R, the function Af + Bg is also integrable on every closed and bounded interval of [a, b) and furthermore, Z

Z

b

(Af + Bg)(x) dx = A a

b

f (x) dx + B

Z

b

g(x) dx.

a

a

(b) Let ϕ : [α, β) → [a, b) be a strictly increasing differentiable function such that ϕ(x) → b as x → β. Then the function F : [α, β) → R defined by F (t) = f (ϕ(t))ϕ′ (t) satisfies Z

b

f (x) dx =

a

Z

β

F (t) dt =

Z

β

f (ϕ(t))ϕ′ (t) dt.

α

α

Theorem 11.6. Suppose that a ∈ R and f : (a, b) → R is integrable on every closed and bounded interval of (a, b), where b is either finite or +∞. For a < c < b, we have Z

b

f (x) dx =

a

Hence the improper integral

Z

c

f (x) dx + a

Z

b

f (x) dx.

(11.6)

c

Z

b

f (x) dx

(11.7)

a

converges if and only if both improper integrals Z

c

f (x) dx and

Z

b

f (x) dx

c

a

converge. In particular, if b = +∞ in the formula (11.6), then the first and the second improper integrals on its right-hand side are of the first kind and the second kind respectively. In this case, the integral (11.7) is sometimes called the improper integral of the third kind. An example of this is the classical Gamma function Γ defined by Γ(x) =

Z

+∞

tx−1 e−t dt,

0

where x > 0. See Problem 11.15 for the proof of its convergence.

Chapter 11. Improper Integrals

36

11.1.5 Criteria for Convergence of Improper Integrals Theorem 11.7 (Comparison Test). Suppose that f, g : [a, b) → [0, +∞), 0 ≤ f (x) ≤ g(x) on [a, b), where b is either finite or +∞, and Z c f (x) dx a

exists for every a ≤ c < b. Then we have Z b Z b 0≤ f (x) dx ≤ g(x) dx. a

a

Furthermore, the convergence of the improper integral of g implies that of f and the divergence of the improper integral of f implies that of g. Theorem 11.8 (Limit Comparison Test). Suppose that f, g : [a, b) → [0, +∞) and both Z c Z c g(x) dx f (x) dx and a

a

exist for every a ≤ c < b. Furthermore, if lim

x→b

f (x) = A 6= 0, g(x)

(11.8)

where b is either finite or +∞, then the improper integrals Z b Z b g(x) dx f (x) dx and a

a

both converge or both diverge. If the value of A in the limit (11.8) is 0, then the convergence of Z b Z b f (x) dx only. g(x) dx implies the convergence of a

a

Theorem 11.9 (Absolute Convergence Test). Suppose that f : [a, b) → R is a function such that Z c f (x) dx a

exists for every a ≤ c < b and the improper integral Z b |f (x)| dx a

converges, where b is either finite or +∞. Then we have Z b Z b |f (x)| dx f (x) dx ≤ a

a

and the improper integral

converges.

Z

b

f (x) dx a

37

11.2. Evaluations of Improper Integrals

Theorem 11.10 (Integral Test for Convergence of Series). Let N be a positive integer. Suppose that f : [N, +∞) → [0, +∞) decreases on [N, +∞) and Z b f (x) dx N

converges for every N ≤ b < +∞. Then we have Z +∞ ∞ +∞ X X f (x) dx ≤ f (k) ≤ f (k). N

k=N +1

Particularly,

Z

k=N

+∞

f (x) dx

N

converges if and only if

+∞ X

f (k)

k=N

converges. Remark 11.2

Of course, there are analogies of convergence theorems for functions f : (a, b] → [0, +∞), where a is either finite or −∞, but we won’t repeat the statements here.

11.2 Evaluations of Improper Integrals Problem 11.1 ⋆ Evaluate the following improper integrals: Z +∞ dx (a) . 1 + x2 1 Z +∞ dx √ . (b) x 1 Proof. (a) By Definition 11.1, we obtain Z +∞ Z b  dx π π dx b arctan b − = . = lim = lim [arctan x] = lim 1 b→+∞ 1 + x2 b→+∞ 1 1 + x2 b→+∞ 4 4 1 (b) By Definition 11.1, we have Z b Z +∞ √ √ dx dx √ = lim √ = lim [2 x]b1 = lim (2 b − 2) = ∞. b→+∞ x b→+∞ 1 x b→+∞ 1 This completes the proof of the problem.



Chapter 11. Improper Integrals

38

Problem 11.2 ⋆ Evaluate the following improper integrals: Z 1 dx √ . (a) x 0 Z 3 1 √ (b) dx. 9 − x2 1 Proof. (a) By Definition 11.2, it is easy to check that Z

1

0

dx √ = lim x ǫ→0+

Z

1

ǫ

√ √ dx √ = lim [2 x]1ǫ = lim (2 − 2 ǫ) = 2. ǫ→0+ x ǫ→0+

(b) By Definition 11.2 and then the substitution x = 3 sin θ, we know that Z

1

3

1 √ dx = lim c→3− 9 − x2

Z

c 1

1 √ dx = lim c→3− 9 − x2

Z

arcsin

c 3

dθ = lim arcsin c→3−

0

Hence we complete the proof of the problem.

π c = . 3 2 

Problem 11.3 ⋆ Evaluate the improper integral

where α is real.

Z

+∞ 1

1 dx, xα

Proof. Clearly, if b ≥ 1, then we have Z

b 1

 h 1−α i b x    , 1 1−α 1 dx =  xα   [ln x]b ,  1−α 1 b −1    , 1−α =    ln b,

if α 6= 1; otherwise, if α 6= 1; otherwise.

On the one hand, if α 6= 1, then we get from Definition 11.1 that Z

+∞ 1

b1−α

  

−1 1 = dx = lim  b→+∞ 1 − α xα 

1 , if α > 1; α−1 +∞,

otherwise.

(11.9)

39

11.2. Evaluations of Improper Integrals

On the other hand, if α = 1, then we ahve Z +∞ 1 dx = lim ln b = +∞. b→+∞ x 1

(11.10)

Combining the two results (11.9) and (11.10), we see that Z

+∞ 1

  

1 , if α > 1; 1 α−1 dx =  xα  +∞, otherwise.

We have completed the proof of the problem.



Problem 11.4 ⋆ Evaluate the improper integral of

Z

1 −1

dx x

(11.11)

and its principal value.

Proof. We write

Z

1

−1

dx = x

Z

0 −1

dx + x

Z

1

0

dx . x

(11.12)

Since the two improper integrals on the right-hand side of (11.12) are divergent, the improper integral (11.11) is divergent too. However, we have P.V.

Z

1

−1

 dx = lim ǫ→0+ x

Z

−ǫ −1

dx + x

Z

1 ǫ

This completes the proof of the problem.

1  −ǫ  dx  = lim ln |x| + ln |x| = 0. ǫ→0+ x ǫ −1

Problem 11.5 ⋆ Let −∞ < a < b < +∞. Evaluate the following improper integral Z

b a

dx (b − x)α

if α < 1.

Proof. For a ≤ c < b, we have Z c h (b − x)1−α ic (b − c)1−α − (b − a)1−α dx = = . α α−1 α−1 a a (b − x) Since α < 1, we have lim (b − c)1−α = 0

c→b−



Chapter 11. Improper Integrals

40

and then we get from Definition 11.2 that Z

b a

(b − c)1−α − (b − a)1−α dx (b − a)1−α = lim = . (b − x)α c→b− α−1 1−α

We complete the proof of the problem.



Problem 11.6 ⋆ Prove that

Z

0

+∞

Z  b 1 +∞ p 2 f ax + dx = f ( x + 4ab) dx, x a 0

where a and b are positive. (Assume the improper integrals in the question are well-defined.)

Proof. Suppose that y = ax − Then we have ax +

b . x

p b = y 2 + 4ab. x

(11.13)

(11.14)

Adding the two expressions (11.13) and (11.14) to get x= so that

p 1 (y + y 2 + 4ab) = ϕ(y) 2a

p 1 y + y 2 + 4ab dx = ϕ′ (y) = · p . dy 2a y 2 + 4ab

When x → 0+, y → −∞; when x → +∞, y → +∞. By Theorem 11.5(b), we have Z

0

+∞

p Z +∞ p  y + 1 b y 2 + 4ab dx = f ( y 2 + 4ab) · p f ax + dy x 2a −∞ y 2 + 4ab p Z 0 p y + y 2 + 4ab 1 2 = dy f ( y + 4ab) · p 2a −∞ y 2 + 4ab p Z +∞ p y + 1 y 2 + 4ab + f ( y 2 + 4ab) · p dy 2a 0 y 2 + 4ab p Z +∞ p y 2 + 4ab − y 1 2 = dy f ( y + 4ab) · p 2a 0 y 2 + 4ab p Z +∞ p y + y 2 + 4ab 1 2 + f ( y + 4ab) · p dy 2a 0 y 2 + 4ab Z 1 +∞ p 2 f ( y + 4ab) dy. = a 0

This completes the proof of the problem.



41

11.2. Evaluations of Improper Integrals Problem 11.7 ⋆ Suppose that f : (0, 1) → R is an increasing function and Z

1 0

Z

1

f (x) dx exists. Prove that

0

1 X k  f (x) dx = lim . f n→+∞ n n n

k=1

Proof. Let n ∈ N. Since f is increasing, we see that Z

1 1− n

f (x) dx =

0

n−1 XZ k=1

k n k−1 n

f (x) dx ≤

n−1 XZ k=1

k 

1 X k dx = f f n n n

k n k−1 n

n−1 k=1

and Z

1 1 n

f (x) dx =

n−1 XZ k=1

k+1 n k n

f (x) dx ≥

n−1 XZ k=1

k+1 n k n

k

1 X k dx = . f f n n n n−1 k=1

Hence we conclude that Z Since

Z

1 1− n

0

1 X k  ≤ f (x) dx ≤ f n n n−1 k=1

Z

1 1 n

f (x) dx.

(11.15)

1

f (x) dx exists, Definition 11.2 ensures that 0

Z

0

1

f (x) dx = lim

Z

1 1− n

n→+∞ 0

f (x) dx = lim

n→+∞

Z

1 1 n

f (x) dx.

(11.16)

Using this fact (11.16) and applying [25, Theorem 5.6, p. 50] to the inequalities (11.15), we have the desired result which completes the proof of the problem.  Problem 11.8 ⋆

⋆ Suppose that 0 < θ < 1. We define fθ (x) =

hθi x

−θ

for all x ∈ (0, 1). Evaluate the improper integral Z

Proof. We notice that fθ (x) = −

θ

x

h1i x

1

fθ (x) dx. 0

−

h θ i x

+θ

1

x

−

h 1 i x

Chapter 11. Improper Integrals

42

which implies that Z 1  Z 1 Z 1 1 h 1 i θ h θ i θ dx + dx − − fθ (x) dx = − x x x x 0 0 0 Z θ Z 1 Z 1  θ h θ i θ h θ i 1 h 1 i =− θ dx − dx + dx. − − − x x x x x x 0 θ 0

(11.17)

If θ < x ≤ 1, then [ xθ ] = 0 so that Z

θ

1

Z 1 1 θ h θ i θ dx = − − dx = −θ ln x = θ ln θ. x x θ θ x

(11.18)

Furthermore, the first integral on the right-hand side of (11.17) can be written as Z

θ

θ

−

x

0

h θ i x

Z

dx = lim

θ

θ

x

ǫ→0+ ǫ

−

h θ i x

dx.

(11.19)

Applying the substitution x = θy to the integral on the right-hand side of (11.19), we obtain Z

θ ǫ

θ

x

−

h θ i x

dx = θ

Z

1

ǫθ −1

1 y

−

h 1 i y

dy.

Therefore, we have Z 1  Z 1 Z θ 1 h 1 i 1 h 1 i θ h θ i dx = lim θ dy = θ dy. − − − ǫ→0+ x x y y y ǫθ −1 y 0 0

(11.20)

By putting the expression (11.18) and (11.20) back into the expression (11.17), we establish immediately that Z 1 fθ (x) dx = θ ln θ, 0

completing the analysis of the problem.



11.3 Convergence of Improper Integrals Problem 11.9 ⋆ Prove that

Z

+∞

2

e−x dx

0

is convergent. 2

Proof. Let f (x) = e−x on [0, +∞) and g(x) =

  f (x), if x ∈ [0, 1]; 

e−x ,

otherwise.

43

11.3. Convergence of Improper Integrals 2

Since e−x ≤ e−x for all x ≥ 1, we have 0 ≤ f (x) ≤ g(x) on [0, +∞). Since f is bounded on [0, 1], the proper integral Z

1

2

e−x dx

0

is finite. Furthermore, we have Z

+∞

g(x) dx = 0

Z

1

Z

2

e−x dx +

0

+∞

e−x dx =

Z

0

1

1

2

= e−x dx − [e−x ]+∞ 1

Z

1

2

e−x dx + e−1

0

which means that the improper integral Z

+∞

g(x) dx 0

converges. By Theorem 11.7 (Comparison Test), the improper integral Z

+∞

2

e−x dx

0

is convergent. We end the analysis of the proof of the problem.



Problem 11.10 ⋆ Let p ∈ R. Prove that

Z

is convergent.

+∞

xp e−x dx

1

Proof. Consider f (x) = xp e−x and g(x) = x−2 . Obviously, by repeated use of L’Hˆospital’s Rule [25, Theorem 8.10, p. 130], we know that xp+2 f (x) = lim = 0. x→+∞ ex x→+∞ g(x) lim

Since we have

Z

1

+∞

x−2 dx = −[x−1 ]+∞ = 1, 1

Theorem 11.8 (Limit Comparison Test) implies the required result. Hence we have completed the proof of the problem.  Problem 11.11 ⋆ Prove Theorem 11.9 (Absolute Convergence Test).

Chapter 11. Improper Integrals Z Z b |f (x)| dx converges and Proof. Suppose that

that

a

44 x a

f (t) dt exists for all x ∈ [a, b). We know

0 ≤ f (x) + |f (x)| ≤ 2|f (x)|,

so Theorem 11.7 (Comparison Test) shows that Z b [f (x) + |f (x)|] dx a

converges. Hence we immediately deduce from Theorem 11.5(a) that the improper integral Z b f (x) dx a

converges. This completes the proof of the problem.



Problem 11.12 ⋆ Prove that the improper integral Z

+∞ 1

sin x dx x2

converges.

Proof. We notice that

sin x 1 2 ≤ 2 x x for all x ∈ [1, +∞). By Problem 11.3, the improper integral Z +∞ 1 dx x2 1 is convergent. By Theorem 11.7 (Comparison Test), Z +∞ sin x 2 dx x 1

is also convergent. Finally, it follows from Theorem 11.9 (Absolute Convergence Test) that the improper integral Z +∞ sin x dx x2 1  converges. We have completed the proof of the problem. Problem 11.13 ⋆ Prove that the improper integral Z

0

converges if and only if α > −1.

1

(− ln x)α dx

45

11.3. Convergence of Improper Integrals

Proof. Let t = − ln x. Then we have dt = − x1 dx. When x → 0+, t → +∞; when x = 1, t = 0. Thus Theorem 11.5(b) implies that Z

1 0

Z 0 tα e−t dt (− ln x)α dx = − +∞ Z +∞ tα e−t dt = 0 Z Z 1 α −t t e dt + =

+∞

tα e−t dt.

(11.21)

1

0

Now we are going to investigate the convergence of the two improper integrals of the expression (11.21). Obviously, we have tα e−t tα+2 = lim =0 t→+∞ t−2 t→+∞ et lim

for every α ∈ R. Therefore, the last integral of the expression (11.21) converges by Theorem 11.8 (Limit Comparison Test). Next, since tα e−t = lim e−t = 1, t→0+ t→0+ tα lim

Theorem 11.8 (Limit Comparison Test) again shows that the first integral of the expression (11.21) converges if and only if the improper integral Z

1

tα dt

(11.22)

0

converges. If α = −1, then the integral (11.22) becomes Z

0

1

dt = lim ǫ→0+ t

Z

1 ǫ

1 dt = lim ln t = − lim ln ǫ ǫ→0+ ǫ→0+ t ǫ

which is definitely divergent. For α 6= −1, we have Z

1

1 − ǫ1+α ǫ→0+ 1 + α

tα dt = lim

0

which is convergent if and only if α > −1. Hence the improper integral Z

1

(− ln x)α dx

0

converges if and only if α > −1, completing the proof of the problem.



Chapter 11. Improper Integrals

46

Problem 11.14 ⋆

⋆ Suppose that f ∈ R([0, 1]), f is periodic with period 1 and Z

1

f (x) dx = 0.

(11.23)

0

If α > 0, prove that the improper integral Z +∞

x−α f (x) dx

1

exists. Hint: It is true that

where b ≥ 1 and g(x) =

Z

Z

b

x

−α

f (x) dx =

Z

b

x−α dg,

1

1

x

f (t) dt. 1

Proof. Let 1 ≤ x ≤ b < +∞ and set g(x) =

Z

x

f (t) dt. 1

By the First Fundamental Theorem of Calculus [25, p. 161], g is continuous on [1, b]. By the hypothesis (11.23) and the periodicity of f , we get Z x Z 1 Z x Z x+1 f (t) dt = g(x) f (t) dt + f (t + 1) dt = f (t) dt = g(x + 1) = 1

1

0

0

so that g is also periodic with period 1. Consequently, g is bounded by a positive constant M on [1, +∞). Next, it follows from the hint that, for every b ≥ 1, we have Z b Z b −α x−α dg x f (x) dx = 1 1 Z b −α b x−α−1 g(x) dx = [x g(x)]1 + α 1 Z b −α x−α−1 g(x) dx. (11.24) = b g(b) − g(1) + α 1

Since g is bounded by M on [1, +∞) and α > 0, we have lim b−α g(b) = 0.

b→+∞

(11.25)

In addition, we have |x−α−1 g(x)| ≤ M x−α−1 for all x ∈ [1, +∞), so Problem 11.3, Theorems 11.7 (Comparison Test) and 11.9 (Absolute Convergence Test) imply that Z +∞ x−α−1 g(x) dx (11.26) 1

47

11.3. Convergence of Improper Integrals

is convergent for every α > 0. Applying the results (11.24) and (11.25) to the expression (11.26) to conclude that Z +∞ x−α f (x) dx 1

exists, completing the proof of the problem.



Problem 11.15 ⋆

⋆ Prove that Γ(x) =

Z

+∞

tx−1 e−t dt

(11.27)

0

converges for all x > 0.

Proof. We write Γ(x) =

Z

1

t

x−1 −t

e

dt +

Z

+∞

tx−1 e−t dt.

(11.28)

1

0

Now Problem 11.10 ensures that the second integral of the expression (11.28) converges for every x ∈ R. For the first integral, we notice that if a > 0 and t = 1s , then we obtain Z 1 Z 1 a 1 x−1 −t s−x−1 e− s ds. t e dt = 1

a

For s > 1, since

1

0 ≤ s−x−1 e− s ≤ s−x−1

and we know from Problem 11.3 that

Z

+∞

s−x−1 ds

1

is convergent for all x > 0, Theorem 11.7 (Comparison Test) now implies that Z +∞ 1 s−x−1 e− s ds 1

converges for all x > 0. Hence the first integral of the expression (11.28) and then the improper  integral (11.27) both converge for all x > 0. This completes the proof of the problem. Problem 11.16 (Cauchy Criterion for Improper Integrals) ⋆ ⋆ Suppose that a ∈ R and f : [a, b) → R is integrable on every closed and bounded interval of [a, b), where b is finite or +∞. Then the improper integral Z

b

f (x) dx

(11.29)

a

converges if and only if for every ǫ > 0, there exists a M ≥ a such that for all M ≤ A, B < b, we have Z B f (x) dx < ǫ. (11.30) A

Chapter 11. Improper Integrals

48

Proof. Suppose that the improper integral (11.29) converges to L. Given ǫ > 0. By Definition 11.1 or 11.2, there exists a M ≥ a such that M ≤ c < b implies ǫ Z c (11.31) f (x) dx − L < . 2 a If M ≤ A, B < b, then the estimate (11.31) gives Z

B

A

Z A Z B f (x) dx − L + L − f (x) dx f (x) dx = a a Z A Z B f (x) dx f (x) dx − L + L − ≤ a

a

< ǫ.

Conversely, we define F : [a, b) → R by F (r) =

Z

r

f (x) dx. a

Then our hypothesis implies that F exists on [a, b) and with this notation, the condition (11.30) is equivalent to |F (B) − F (A)| < ǫ. (11.32) It is easy to see that there exists a N ∈ N such that N1 < b − M . Therefore, if we write 1 An = b − n1 , Bm = b − m , an = F (An ) and am = F (Bm ), then we have M ≤ An , Bm < b for all n, m ≥ N and so the inequality (11.32) means that |am − an | < ǫ for all n, m ≥ N . In other words, {an } is Cauchy and it converges to the real number L. Hence this fact implies that Z

a

b

f (x) dx = lim

Z

n→+∞ a

1 b− n

 1 = lim an = L. f (x) dx = lim F b − n→+∞ n→+∞ n

This completes the proof of the problem.



Problem 11.17 (Abel’s Test for Improper Integrals) ⋆ Suppose that a ∈ R and f : [a, +∞) → R is a function which is integrable on every Z +∞ f (x) dx converges. closed and bounded interval of [a, +∞) and the improper integral ⋆

a

Let g : [a, +∞) → R be a monotonic bounded function. Then the improper integral Z +∞ f (x)g(x) dx a

converges.

49

11.3. Convergence of Improper Integrals

Proof. Let a ≤ A ≤ B < +∞. Without loss of generality, we may assume that g is increasing on [A, B]. Thus the Second Mean Value Theorem for Integrals [7] implies the existence of a p ∈ [A, B] such that Z B Z p Z B f (x) dx. (11.33) f (x) dx + g(B) f (x)g(x) dx = g(A) p

A

A

By the hypothesis, there exists a positive constant M1 such that |g(x)| ≤ M1 on [a, +∞). Given ǫ > 0. By Problem 11.16 (Cauchy Criterion for Improper Integrals), there is a positive constant M2 ≥ a such that Z B Z p ǫ ǫ f (x) dx and (11.34) f (x) dx ≤ ≤ 2M 2M 1 1 p A for all A ≥ M2 ≥ a. Combining the formula (11.33) and the inequalities (11.34), we see that Z B Z p Z B f (x) dx f (x) dx + |g(B)| · f (x)g(x) dx ≤ |g(A)| · p

A

A

ǫ ǫ ≤ M1 · + M1 · 2M1 2M1 =ǫ

for all A, B ≥ M2 ≥ a. By Problem 11.16 (Cauchy Criterion for Improper Integrals) again, we conclude that the improper integral Z +∞ f (x)g(x) dx a

converges, completing the proof of the problem.



Problem 11.18 (Dirichlet’s Test for Improper Integrals) ⋆ ⋆ Suppose that a ∈ R and f : [a, +∞) → R is a function which is integrable on every closed and bounded interval of [a, +∞) and there exists a positive constant M such that Z b f (x) dx ≤ M

(11.35)

a

for all b > a. Let g : [a, +∞) → R be a monotonic function such that g(x) → 0 as x → +∞. Then the improper integral Z +∞ f (x)g(x) dx a

converges.

Proof. Given that ǫ > 0. Since g(x) → 0 as x → +∞, there exists a M1 ≥ a such that |g(x)|
0. Since a

proper Integrals) guarantees that there exists a M ≥ a such that if Z

Furthermore, we have Z

x x 2

Z f (t) dt =

x x 2

x x 2

x 2

≥ M , then

ǫ f (t) dt < . 2

 x  xf (x) = . f (t) dt ≥ f (x) · x − 2 2

(11.40)

(11.41)

Combining the inequalities (11.40) and (11.41), we conclude that |xf (x)| < ǫ which is equivalent to saying that xf (x) → 0 as x → +∞. We complete the proof of the  problem.

53

11.4. Miscellaneous Problems on Improper Integrals Problem 11.23 (Frullani’s Integral) ⋆ Suppose that f is continuous on [0, +∞) and the limit

⋆

lim f (x)

x→+∞

exists. Let a, b > 0. Compute the improper integral Z +∞ f (ax) − f (bx) dx x 0

Proof. For 0 < r < R < +∞, using the method of substitution, we have Z

R r

Z R f (ax) f (bx) dx − dx x x r r Z bR Z aR f (x) f (x) dx − dx = x x br ar Z bR Z br f (x) f (x) dx − dx = x x aR ar Z bR Z br f (x) dα, f (x) dα − =

f (ax) − f (bx) dx = x

Z

R

(11.42)

aR

ar

where α(x) = ln x. It is clear that α is monotonically increasing on [ar, br] and [aR, bR]. Since f is continuous on [ar, br] and [aR, bR], it is bounded on these closed intervals and we follow from [25, Theorem 9.3(a)] that f ∈ R(α) on [ar, br] and [aR, bR]. Using the form of the First Mean Value Theorem for Integrals [2, Theorem 7.30, p. 160], we see that there exist p ∈ [ar, br] and q ∈ [aR, bR] such that Z

br ar

b br = f (p) ln f (x) dα = f (p) ln ar a

and

Z

bR

f (x) dα = f (q) ln aR

bR b = f (q) ln . (11.43) aR a

If we put the two formulas (11.43) back into the expression (11.42), then we get Z

r

R

f (ax) − f (bx) b dx = [f (p) − f (q)] ln . x a

(11.44)

As r → 0+ and R → +∞, we obtain p → 0+ and q → +∞. Recall that f (0+) = f (0) and f (+∞) is a real number, so we may take r → 0+ and R → +∞ in the expression (11.44), we achieve Z +∞ b f (ax) − f (bx) dx = [f (0) − f (+∞)] ln , x a 0 completing the proof of the problem.



Chapter 11. Improper Integrals

54

Problem 11.24 ⋆ Suppose that f : [0, +∞) → [0, +∞) is continuous and Z

+∞

f (x) dx

(11.45)

0

converges. Prove that

1 lim n→+∞ n

Z

n

xf (x) dx = 0. 0

Proof. Suppose that M= 1+

Z

1

.

+∞

f (x) dx

0

Given ǫ > 0. By the hypothesis (11.45), there is an N ∈ N such that n ≥ N implies that Z +∞ ǫ f (x) dx < 0≤ . (11.46) M n Now for all sufficiently large enough n such that n > nǫ ≥ N , we know from the inequality (11.46) that Z Z Z 1 n 1 nǫ 1 n 0≤ xf (x) dx = xf (x) dx + xf (x) dx n 0 n 0 n Z n nǫ Z nǫ ǫ f (x) dx f (x) dx + < M 0 nǫ Z nǫ ǫ ǫ < f (x) dx + M 0 M Z +∞   ǫ · f (x) dx + 1 < M 0 = ǫ. This means that

1 lim n→+∞ n completing the proof of the problem.

Z

n

xf (x) dx = 0,

0



Problem 11.25 ⋆

⋆ For every x > 1, prove that P.V.

exists.

Z

Proof. Given ǫ > 0 and x ∈ [0, 1). Now we have

x 0

dt ln t

1 = 0, t→0+ ln t lim

55

11.4. Miscellaneous Problems on Improper Integrals

so if we define f (t) =

    

1 , if t > 0; ln t 0,

if t = 0,

then f is well-defined and continuous on [0, x] so that Z

dt = ln t

Z

 dt = lim ln t ǫ→0+

Z

x 0

x

f (t) dt 0

exists. When x > 1, we notice that P.V.

Z

x 0

1−ǫ 0

dt + ln t

Z

x

1+ǫ

dt  . ln t

(11.47)

By Taylor’s Theorem [25, p. 131], we can show that h 2(t − 1) i (t − 1)2 (t − 1)2 (t − 1)3 + , = (t − 1) + − 1 · 2 3ξ 3 3ξ 3 2

ln t = (t − 1) − where ξ ∈ (1, t). Let α(t) =

2(t−1) 3ξ 3 .

Then it is clear that lim α(t) = 0.

t→1

Besides, we have the identity 1 1 α(t) − 1 = − ln t t − 1 2 + [α(t) − 1](t − 1)

(11.48)

so that α(t) − 1 = 0. t→1 2 + [α(t) − 1](t − 1) lim

Therefore, the second term on the right-hand side of the equation (11.48) is continuous and thus integrable around t = 1. This fact and the expression (11.48) indicate that the principal value 1 (11.47) depends on the evaluation of the principal value of t−1 on [0, x] which is P.V.

Z

x 0

Z x  Z 1−ǫ dt dt dt  + = lim t − 1 ǫ→0+ 0 t−1 1+ǫ t − 1 x  1−ǫ  + ln |t − 1| = lim ln |t − 1| ǫ→0+

0

1+ǫ

= ln(x − 1).

Hence we conclude that P.V.

Z

0

exists, completing the proof of the problem.

x

dt ln t 

Chapter 11. Improper Integrals

56

Problem 11.26 ⋆ ⋆ Suppose that the improper integral

Z

+∞

f (x) dx converges. Prove that the improper

0

integral

Z

converges for every ǫ > 0.

+∞

e−ǫx f (x) dx

0

Proof. Let 1 < A < C and fix ǫ > 0. Then e−ǫx ≥ 0 on [A, C]. By the Second Mean Value Theorem for Integrals [7], we see that Z

C

−ǫx

e

−Aǫ

f (x) dx = e

A

for some B ∈ [A, C]. Thus we have Z

C

A

Since that

Z

+∞

Z e−ǫx f (x) dx ≤

Z

B

f (x) dx A

B A

f (x) dx .

(11.49)

f (x) dx converges, Problem 11.16 (Cauchy Criterion for Improper Integrals) ensures 0

Z

B

A

f (x) dx < ǫ

(11.50)

for large enough A. Combining the inequalities (11.49) and (11.50), we obtain Z

C

A

e−ǫx f (x) dx < ǫ

for large enough A and our conclusion follows immediately from Problem 11.16 (Cauchy Criterion for Improper Integrals) again. This ends the analysis of the problem. 

CHAPTER

12

Lebesgue Measure

12.1 Fundamental Concepts In [25, Chap. 9], we recall the basic theory and main results of the Riemann integral of a bounded function f over a bounded and closed interval [a, b] with approximations associated with f and partitions of [a, b]. However, the techniques developed so far for Riemann integration is unsatisfactory for general sets. Consequently, only a small class of functions can be “integrable” in the sense of Riemann. This leads to the theories and the notions of the Lebesgue measure, Lebesgue measurable functions and Lebesgue integration. The discussion of these components will be presented in the successive three chapters and for simplicity, we only discuss the Lebesgue theory on the real line. The main references for this chapter are [5, Chap. 1], [9, Chap. 2 - 5], [17, Chap. 2], [21, Chap. 1] and [22, Chap. 7].

12.1.1 Lebesgue Outer Measure Definition 12.1 (Length of Intervals). Let I be a nonempty interval of R and a and b are the end-points of I.a We define its length, denoted by ℓ(I), to be infinite if I is unbounded. Otherwise, we define ℓ(I) = b − a. Definition 12.2 (Lebesgue Outer Measure). Let E ⊆ R and {Ik } be a countable collection of nonempty open and bounded intervals covering E, i.e., E⊆

∞ [

Ik .

k=1

Then the Lebesgue outer measure (or simply outer measure) of E, denoted by m∗ (E), is given by ∞ ∞ nX o [ m∗ (E) = inf ℓ(Ik ) E ⊆ Ik , k=1

k=1

where the infimum takes over all countable collections of nonempty open and bounded intervals covering E. a

Note that |a| or |b| can be infinite.

57

Chapter 12. Lebesgue Measure

58

Theorem 12.3 (Properties of Outer Measure). The outer measure m∗ satisfies the following properties: (a) m∗ (∅) = 0. (b) m∗ (I) = ℓ(I) for every interval I. (c) (Monotonicity) If E ⊆ F , then m∗ (E) ≤ m∗ (F ). (d) (Countably Subadditivity) If {Ek } is a countable collection of sets of R, then we have m∗

∞ [

n=1

∞  X m∗ (Ek ). Ek ≤ k=1

(e) (Translation invariance) For every E ⊆ R and p ∈ R, we have m∗ (p + E) = m∗ (E), where p + E = {p + x | x ∈ E}. Remark 12.1 The outer measure m∗ , however, is not countable additive. In other words, there exists a countable collection {Ek } of disjoint subsets of R such that m

∗

∞ [

k=1



Ek 6=

∞ X

m∗ (Ek ).

k=1

12.1.2 Lebesgue Measurable Sets Let S be any subset of R. If E ⊆ S, then we denote E c = S \ E. Definition 12.4 (Lebesgue Measurability). A set E ⊆ R is Lebesgue measurable, or simply measurable, if for every set A ⊆ R, we haveb m∗ (A) = m∗ (A ∩ E) + m∗ (A ∩ E c ). If E is measurable, then we define the Lebesgue measure of E to be m(E) = m∗ (E). If all the sets considered in Theorem 12.3(c) and (d) (Properties of Outer Measure) are measurable, then the corresponding results also hold with m∗ replaced by m. That is, if E ⊆ F , then m(E) ≤ m(F ) and if all Ek are measurable, then m

∞ [

n=1

∞  X m(Ek ). Ek ≤ n=1

Further properties of measurable sets are listed in the following theorem. b

This definition is due to Carath`eodory.

(12.1)

59

12.1. Fundamental Concepts

Theorem 12.5 (Properties of Measurable Sets). Let E ⊆ R. Then we have the following results: (a) If m∗ (E) = 0, then it is measurable. (b) Every interval I is measurable. (c) If E is measurable, then E c is also measurable. (d) If E is measurable and p ∈ R, then p + E is also measurable and m(p + E) = m(E). (e) The finite union or intersection of measurable sets is also measurable. Remark 12.2 We notice that any set E ⊆ R with positive outer measure contains a nonmeasurable subset. See, for examples, [9, Chap. 4, pp. 81 - 83] and [17, Theorem 17, p. 48].

Definition 12.6 (Almost Everywhere). Let E ⊆ R be a measurable set. We way that a property P holds almost everywhere on E if there exists a subset A ⊆ E such that m(A) = 0 and P holds for all x ∈ E \ A.

12.1.3 σ-algebras and Borel Sets In the following discussion, we suppose that S ⊆ R.

Definition 12.7 (σ-algebra). Let M be a collection of subsets of S. Then M is called an σ-algebra in S if M satisfies the following conditions: (a) S ∈ M. (b) If E ∈ M, then E c ∈ M. (c) If E =

∞ [

k=1

Ek and if Ek ∈ M for all k = 1, 2, . . ., then E ∈ M.

It follows easily from Definition 12.7 (σ-algebra) that if E, F, Ek ∈ M for all k = 1, 2, . . ., then we have ∞ \ (12.2) Ekc ∈ M and E \ F ∈ M. k=1

Theorem 12.8. The union of a countable collection of measurable sets is also measurable, i.e., the set ∞ [ Ek E= k=1

is measurable if all Ek are measurable. Particularly, the collection M of all measurable sets of S is an σ-algebra in S.

Chapter 12. Lebesgue Measure

60

Remark 12.3 Hence we deduce immediately from the first relation (12.2) and Theorem 12.8 that the intersection of a countable collection of measurable sets is also measurable.

Theorem 12.9. Every open or closed set in S is measurable. Theorem 12.10. Let T be a topology of R. (a) The intersection of all σ-algebras in R containing T is a smallest σ-algebra B in R such that T ⊆ B.c (b) Members of B are called Borel sets of R and they are Borel measurable.d (c) The σ-algebra B is generated by the collection of all open intervals or half-open intervals or closed intervals or open rays or closed rays of R.e It is well-known that every Borel measurable set is a Lebesgue measurable set, but there exist Lebesgue measurable sets which are not Borel measurable, see [5, Exercise 9, p. 48] and [19, Remarks 2.21, p. 53].

12.1.4 More Properties of Lebesgue Measure Theorem 12.11 (Countably Additivity). If {Ek } is a countable collection of disjoint measur∞ [ able sets, then Ek is measurable and k=1

m

∞ [



Ek =

k=1

∞ X

m(Ek ).

k=1

Theorem 12.12 (Continuity of Lebesgue Measure). Suppose that {Ek } is a sequence of measurable sets. (a) If Ek ⊆ Ek+1 for all k ∈ N, then we have m

∞ [

k=1

 Ek = lim m(Ek ). k→∞

(b) If Ek+1 ⊆ Ek for all k ∈ N and m(E1 ) < ∞, then we have m

∞ \

k=1

c

 Ek = lim m(Ek ). k→∞

The Borel σ-algebra B is sometimes called the σ-algebra generated by T . See also Problem 12.35. e In fact, it is [5, Propositin 1.2, p. 22].

d

61

12.2. Lebesgue Outer Measure

12.2 Lebesgue Outer Measure Problem 12.1 ⋆ Prove Theorem 12.5(a). Proof. Let m∗ (E) = 0 and A be any set of R. Since A ∩ E ⊆ E and A ∩ E c ⊆ A, we know from Theorem 12.3(c) (Properties of Outer Measure) that m∗ (A ∩ E) ≤ m∗ (E) = 0 and m∗ (A ∩ E c ) ≤ m∗ (A). Therefore, we get m∗ (A) ≥ m∗ (A ∩ E) + m∗ (A ∩ E c ).

(12.3)

m∗ (A) ≤ m∗ (A ∩ E) + m∗ (A ∩ E c )

(12.4)

Since A = (A ∩ E) ∪ (A ∩ E c ), we always have

by Theorem 12.3(d) (Properties of Outer Measure). Now the inequalities (12.3) and (12.3) give m∗ (A) = m∗ (A ∩ E) + m∗ (A ∩ E c ). By Definition 12.4 (Lebesgue Measurability), E is measurable and this completes the proof of  the problem. Problem 12.2 ⋆ Let E be open in R. If F and G are subsets of R such that F ⊆ E and G ∩ E = ∅, prove that m∗ (F ∪ G) = m∗ (F ) + m∗ (G). Proof. By Theorem 12.9, E is measurable. Since F ⊆ E, we have F ∩ E c = ∅. Since G ∩ E = ∅, we have G ⊆ E c . Hence we see from Definition 12.4 (Lebesgue Measurability) that m∗ (F ∪ G) = m∗ ((F ∪ G) ∩ E) + m∗ ((F ∪ G) ∩ E c ) = m∗ (F ) + m∗ (G), as desired. We have completed the proof of the problem.



Problem 12.3 ⋆ Construct an unbounded subset of R with Lebesgue outer measure 0.

Proof. We claim that the set Q satisfies our requirements. In fact, Q is clearly unbounded. Furthermore, if r ∈ Q, then m∗ ({r}) = 0 by Definition 12.2 (Lebesgue Outer Measure). Since we have [ Q= {r}, r∈Q

Chapter 12. Lebesgue Measure

62

Theorem 12.3(d) (Properties of Outer Measure) implies that X m∗ (Q) ≤ m∗ ({r}) = 0, r∈Q

completing the proof of the problem.



Problem 12.4 ⋆ Suppose that E is the set of irrationals in the interval (0, 1). Prove that m∗ (E) = 1.

Proof. By Theorem 12.3(b) (Properties of Outer Measure), we see that m∗ ((0, 1)) = ℓ((0, 1)) = 1. Let F = (0, 1) ∩ Q. By Problem 12.3, we have m∗ (F ) = 0.

(12.5)

Since (0, 1) = E ∪ F , it deduces from Theorem 12.3(c) (Properties of Outer Measure) and the result (12.5) that 1 = m∗ ((0, 1)) ≤ m∗ (E) + m∗ (F ) = m∗ (E) ≤ 1

so that m∗ (E) = 1. Hence we have the required result and it completes the proof of the  problem. Problem 12.5 ⋆ Suppose that E ⊆ R and m∗ (E) < ∞. Prove that for every ǫ > 0, there exists an open set V ⊆ R containing E such that m∗ (V ) − m∗ (E) < ǫ.

(12.6)

Proof. Given ǫ > 0. Then it follows from Definition 12.2 (Lebesgue Outer Measure) that there is a countable collection of intervals {Ik } covering E such that ∞ X

ℓ(Ik ) < m∗ (E) + ǫ.

(12.7)

k=1

Let V =

∞ [

k=1

Ik . We know that V is open in R and E ⊆ V . Using Theorems 12.9, 12.3(b) and

(d) (Properties of Outer Measure) and the inequality (12.7), we have ∗

m(V ) = m (V ) = m

∗

∞ [

k=1



Ik ≤

∞ X k=1

∗

m (Ik ) =

∞ X

ℓ(Ik ) < m∗ (E) + ǫ

k=1

which implies the inequality (12.6). We complete the proof of the problem. 

63

12.2. Lebesgue Outer Measure Problem 12.6

⋆ A set V ⊆ R is called a Gδ set if it is the intersection of a countable collection of open sets of R. Suppose that E ⊆ R and m∗ (E) < ∞. Prove that there exists a Gδ set V containing E such that m∗ (V ) = m∗ (E).

Proof. By Problem 12.5, for each k ∈ N, there exists an open set Vk in R containing E such that 1 m∗ (Vk ) < m∗ (E) + . k Let V =

∞ \

k=1

Vk . Then it is clearly a Gδ set by the definition. Furthermore, since E ⊆ V , for

each k ∈ N, we must have

E ⊆ V ⊆ Vk ,

so Theorems 12.3(c) (Properties of Outer Measure) and 12.9 imply that 1 m∗ (E) ≤ m∗ (V ) ≤ m∗ (Vk ) < m∗ (E) + . k

(12.8)

Taking k → ∞ in the inequalities (12.8), we obtain immediately that m∗ (V ) = m∗ (E). This completes the proof of the problem.



Problem 12.7 ⋆ ⋆ Let n ∈ N. Suppose that {E1 , E2 , . . . , En } is a finite sequence of disjoint measurable subsets of R. For an arbitrary set A ⊆ R, prove that n n  X  [ m∗ (A ∩ Ei ). Ei = m A∩ ∗

i=1

(12.9)

i=1

Proof. If n = 1, then the formula (12.9) is obviously true. Assume that the formula (12.9) is true for n = k for some positive integer k, i.e., k k  X  [ m∗ (A ∩ Ei ). Ei = m A∩ ∗

i=1

i=1

Suppose that {E1 , E2 , . . . , Ek , Ek+1 } is a finite sequence of disjoint measurable sets. Now we follow from the assumption that k k+1 k   [ X X ∗ ∗ ∗ m A∩ Ei + m (A ∩ Ek+1 ) = m∗ (A ∩ Ei ). (12.10) m (A ∩ Ei ) + m (A ∩ Ek+1 ) = ∗

i=1

i=1

i=1

Chapter 12. Lebesgue Measure Since

k [

64

Ei and Ek+1 are disjoint, we have

i=1 k [

i=1

 Ei ∩ Ek+1 = ∅

k [

and

i=1

k  [ c Ei = Ei ∩ Ek+1 i=1

which imply that  k+1 [ i=1

 Ei ∩ Ek+1 = Ek+1

 k+1 [

and

i=1

k  [ c Ei ∩ Ek+1 Ei . =

(12.11)

i=1

Substituting the identities (12.11) into the left-hand side of the equation (12.10) and then using the fact that Ek+1 is measurable, we obtain from Definition 12.4 (Lebesgue Measurability) that k   k+1    [ [  c Ei + m∗ (A ∩ Ek+1 ) = m∗ A ∩ Ei ∩ Ek+1 m∗ A ∩ i=1

i=1

  k+1  [  ∗ +m A∩ Ei ∩ Ek+1 i=1

= m∗



A∩ |

k+1 [

Ei

i=1 {z }



k+1     [ c Ei ∩ Ek+1 + m∗ A ∩ ∩ Ek+1

The new A.

k+1  [  Ei . = m∗ A ∩

|

i=1 {z }

The new A.

(12.12)

i=1

Compare the equations (12.10) and (12.12) to get m

∗



A∩

k+1 [ i=1



Ei =

k+1 X i=1

m∗ (A ∩ Ei ).

By induction, our formula (12.9) is true for every n ∈ N and we end the analysis of the problem.  Problem 12.8 ⋆ Suppose that E ⊆ R has positive outer measure. Prove that there exists a bounded subset F ⊆ E such that m∗ (F ) > 0. Proof. For each n ∈ Z, we denote In = [n, n + 1]. Then we have [ E= (E ∩ In ). n∈Z

By Theorem 12.3(d) (Properties of Outer Measure), we see that X 0 < m∗ (E) ≤ m∗ (E ∩ In ) n∈Z

65

12.3. Lebesgue Measurable Sets

which implies that at least one of E ∩ In has positive outer measure. Let F = E ∩ IN be such a set for some N ∈ Z. Since F ⊆ IN , it must be bounded. Since F ⊆ E, this F is the desired set. This completes the proof of the problem. 

12.3 Lebesgue Measurable Sets Problem 12.9 ⋆ Prove that if E and F are measurable, then m(E ∪ F ) + m(E ∩ F ) = m(E) + m(F ). Proof. Notice that E ∪ F = (E \ F ) ∪ (E ∩ F ) ∪ (F \ E).

Obviously, E \F , E ∩F and F \E are pairwise disjoint. By Theorem 12.11 (Countably Additive), it is true that m(E ∪ F ) + m(E ∩ F ) = [m(E \ F ) + m(E ∩ F )] + [m(F \ E) + m(E ∩ F )].

(12.13)

Since E = (E \ F ) ∪ (E ∩ F ) and F = (F \ E) ∪ (E ∩ F ), Theorem 12.11 (Countably Additivity) again shows that the equation (12.13) reduces to m(E ∪ F ) + m(E ∩ F ) = m(E) + m(F ). This completes the proof of the problem.



Problem 12.10 ⋆ Let E ⊆ [0, 1] and m(E) > 0. Prove that there exist x, y ∈ E such that |x − y| is irrational.

Proof. Assume that |x − y| ∈ Q for all x, y ∈ E. Then it means that E ⊆ x + Q for any x ∈ E. By Problem 12.3 and Theorem 12.3(e) (Properties of Outer Measure), we have m∗ (x + Q) = m∗ (Q) = 0, where x ∈ E. Thus Theorem 12.5(a) (Properties of Measurable Sets) implies that E is measurable and then 0 < m(E) = m∗ (E) ≤ m∗ (x + Q) = m(x + Q) = 0, a contradiction. Hence there exist x, y ∈ E such that |x − y| is irrational. We end the proof of the problem.  Problem 12.11 ⋆ Let E ⊆ [0, 1] and m(E) > 0. Prove that there exist x, y ∈ E with x 6= y such that x − y is rational.

Chapter 12. Lebesgue Measure

66

Proof. Denote the rationals in [−1, 1] by r1 , r2 , . . .. Define En = rn + E for every n ∈ N. By Theorem 12.5(d) (Properties of Measurable Sets), we see that m(En ) = m(rn + E) = m(E) > 0. Thus we must have

∞ X

m(En ) = ∞.

∞ [

En ⊆ [−1, 2].

n=1

(12.14)

Clearly, we have En ⊆ [−1, 2] for every n ∈ N so that

n=1

(12.15)

Assume that Ei ∩ Ej = ∅ if i 6= j. Then Theorem 12.11 (Countably Additivity) and the set relation (12.15) imply that ∞ X

m(En ) = m

∞ [

n=1

n=1

 En ≤ m([−1, 2]) = 3

which contradicts the result (12.14). In other words, there exist n, m ∈ N such that En ∩ Em 6= ∅. Let z ∈ En ∩ Em . By the definition, there are x, y ∈ E such that z = x + rn = y + rm which gives x − y = rm − rn ∈ Q. We have completed the proof of the problem.



Problem 12.12 ⋆ Suppose that {rn } is an enumeration of rationals Q. Prove that ∞  [ 1 1 rn − 2 , rn + 2 6= ∅. R\ n n

(12.16)

n=1

Proof. Since each (rn − n12 , rn + n12 ) is measurable, Theorem 12.8 and then Theorem 12.5(c) (Properties of Measurable Sets) imply that the set ∞  [ 1 1 rn − 2 , rn + 2 R\ n n n=1

is measurable. By the inequality (12.1), we obtain m

∞ ∞ ∞  [ 1 1  X  1  X 2 1 m rn − 2 , rn + 2 N − 1. Prove that and n=1

m

N \

n=1

 En > 0.

Proof. We have the identity N \

n=1

En′

= [0, 1] \ En . Since every where inequality (12.1) give m

N [

n=1

En′



=m

En = [0, 1] \

En′

N [

N [

En′ ,

n=1

(n = 1, . . . , N ) is measurable, the definition and the N  X m([0, 1] \ En ). ([0, 1] \ En ) ≤

n=1

n=1

(12.19)

Chapter 12. Lebesgue Measure

68

Furthermore, we apply the identity (12.17) to the right-hand side of the inequality (12.19) to get N N N  X [ X ′ m(En ) < N − (N − 1) = 1. (12.20) m([0, 1]) − m En ≤ n=1

n=1

n=1

Therefore, using the identity (12.17) and the result (12.20), we establish m

N \

En

n=1



N N   [  [ ′ En′ > 1 − 1 = 0. En = m([0, 1]) − m = m [0, 1] \ n=1

n=1

We complete the proof of the problem.



Problem 12.15 (The Borel-Cantelli Lemma) ⋆ Suppose that {Ek } is a countable collection of measurable sets and ∞ X k=1

m(Ek ) < ∞.

Prove that almost all x ∈ R lie in at most finitely many of the sets Ek .

Proof. For each n ∈ N, let An =

∞ [

k=n

Ek . Then we have A1 ⊇ A2 ⊇ · · · and Theorem 12.3(d)

(Properties of Outer Measure) implies that ∞ ∞ [  X m(A1 ) = m Ek ≤ m(Ek ) < ∞. k=1

(12.21)

k=1

Apply Theorem 12.12 (Continuity of Lebesgue Measure), we acquire that ∞ ∞  [ ∞ ∞  \  \ X m(Ek ) = 0. m Ek =m An = lim m(An ) ≤ lim n=1

n→∞

n=1

k=n

Hence almost all x ∈ R fail to lie in the set ∞  [ ∞ \ n=1

k=n

n→∞

k=n

 Ek .

In other words, it means that for almost all x ∈ R lie in at most finitely many Ek , completing  the proof of the problem. Problem 12.16 ⋆ Let {Ek } be a sequence of measurable sets such that m

∞ [

k=1

 Ek < ∞

and

inf {m(Ek )} = α > 0.

k∈N

Suppose that E is the set of points that lie in an infinity of sets Ek . Prove that E is measurable and m(E) ≥ α.

69

12.3. Lebesgue Measurable Sets

Proof. With the same notations as in the proof of Problem 12.15, we see that E=

∞ \

An

n=1

and so m(E) = m

 An = lim m(An ).

(12.22)

 Ek ≥ m(En ) ≥ α > 0

(12.23)

∞ \

n→∞

n=1

Obviously, we have m(An ) = m

∞ [

k=n

for every n ∈ N. Now if we combine the results (12.22) and (12.23), then we get immediately that m(E) ≥ α. We complete the proof of the problem.



Problem 12.17 ⋆ Prove Theorem 12.10(a) and (b).

Proof. Let Ω be the family of all σ-algebras M in R containing T . It is clear that the collection of all subsets of R satisfies Definition 12.7 (σ-algebra), so Ω 6= ∅. Let B=

\

(12.24)

M.

M∈Ω

Since T ⊆ M for all M ∈ Ω, T ⊆ B. If En ∈ B for every n = 1, 2, . . ., then the definition (12.24) ∞ [ En ∈ M for implies that En ∈ M for every M ∈ Ω. Since every M is an σ-algebra, we have n=1

all M ∈ Ω. By the definition (12.24) again, we achieve ∞ [

n=1

En ∈ B

so that B satisfies Definition 12.7(c) (σ-algebra). Now we can show the other two parts in a similar way. Hence we conclude that B is an σ-algebra and it follows from the definition (12.24) that B is actually a smallest σ-algebra containing T . For the second assertion, by Theorem 12.8, we know that the collection M′ of all Lebesgue measurable sets of R is an σ-algebra in R and by Theorem 12.9, M′ contains T . In other words, it means that M′ ∈ Ω and thus B ⊆ M′ by the definition (12.24). Consequently, a Borel set must be Lebesgue  measurable. Hence we have completed the proof of the problem.

Chapter 12. Lebesgue Measure

70

Problem 12.18 ⋆ Given ǫ > 0. Construct a dense open set V in R such that m(V ) < ǫ.

Proof. Let {rn } be a sequence of rational numbers of R. For every n, we consider the open interval In given by In = (rn − ǫ · 2−(n+1) , rn + ǫ · 2−(n+1) ). ∞ [ In which is evidently open in R. Since Q ⊆ V and Q is dense in R, V is also Define V = n=1

dense in R. Finally, we note from Theorem 12.3(d) (Properties of Outer Measure) that m(V ) = m

∞ [

n=1

completing the proof of the problem.



In ≤

∞ X

n=1

∞ X 1 m(In ) = ǫ · = ǫ, 2n n=1



Problem 12.19 ⋆ Let C be the Cantor set. Prove that m(C) = 0.

Proof. In [25, Problem 4.28, p. 45], we know that C=

∞ \

En ,

(12.25)

n=1

where En =

2n−1 [−1 h k=0

3k + 0 3k + 1 i h 3k + 2 3k + 3 i , , ∪ . 3n 3n 3n 3n

Since each En is the union of closed intervals, En is measurable by Theorem 12.8. Thus it follows from Theorem 12.3(d) (Properties of Outer Measure) that m(En ) ≤ = =
0. By Problem 12.5 and Theorem 12.9, there exists an open set V containing E ∪ F such that m(V ) = m∗ (V ) < m∗ (E ∪ F ) + ǫ. (12.29)

Since E ⊆ V ∩ E ′ , F ⊆ V ∩ F ′ and both V ∩ E ′ and V ∩ F ′ are measurable, we have m∗ (E) + m∗ (F ) ≤ m∗ (V ∩ E ′ ) + m∗ (V ∩ F ′ ) = m(V ∩ E ′ ) + m(V ∩ F ′ ).

(12.30)

By Problem 12.9 and the inequality (12.29), we further reduce the inequality (12.30) to m∗ (E) + m∗ (F ) ≤ m((V ∩ E ′ ) ∪ (V ∩ F ′ )) + m((V ∩ E ′ ) ∩ (V ∩ F ′ )) ≤ m(V ∩ (E ′ ∪ F ′ )) + m(E ′ ∩ F ′ ) ≤ m(V ) + 0

< m∗ (E ∪ F ) + ǫ. Since ǫ is arbitrary, we obtain m∗ (E) + m∗ (F ) ≤ m∗ (E ∪ F ) which implies the equation (12.27). Conversely, we suppose that the equation (12.27) holds. By Problem 12.6, there are Gδ sets E ′ and F ′ containing E and F respectively such that m(E ′ ) = m∗ (E ′ ) = m∗ (E)

and m(F ′ ) = m∗ (F ′ ) = m∗ (F ).

Assume that m(E ′ ∩ F ′ ) > 0. Then Problem 12.9 implies that m∗ (E ∪ F ) = m∗ (E) + m∗ (F ) = m(E ′ ) + m(F ′ )

= m(E ′ ∪ F ′ ) + m(E ′ ∩ F ′ )

Chapter 12. Lebesgue Measure

72 > m(E ′ ∪ F ′ )

≥ m∗ (E ′ ∪ F ′ )

which is a contradiction. Consequently, we obtain m(E ′ ∩ F ′ ) = 0. This completes the proof of the problem.  Problem 12.22 ⋆ Prove that Q and R \ Q are Borel sets in R.

Proof. Let q ∈ Q. Since T ⊆ B and B is an σ-algebra, Definition 12.7 (σ-algebra) shows that B also contains all closed subsets of R. Since {q} is closed in R, we have {q} ∈ B and the countability of Q implies that Q ∈ B. By Definition 12.7 (σ-algebra) again, we have R\Q∈B and hence we complete the proof of the problem.



Problem 12.23 ⋆

⋆ Define M = {E ⊆ R | E or E c is at most countable}.

Prove that M is an σ-algebra in R.

Proof. We check Definition 12.7 (σ-algebra). Since Rc = ∅, we have R ∈ M. Let E ∈ M. If E c is at most countable, then E c ∈ M. Otherwise, E is at most countable. In this case, since (E c )c = E is at most countable, we still have E c ∈ M. Suppose that Ek ∈ M for all k = 1, 2, . . .. If all Ek are at most countable, then the set E=

∞ [

Ek

k=1

must be at most countable which means that E ∈ M. Next, suppose that there exists an uncountable Ek . Without loss of generality, we may assume that it is E1 . Since E1 ∈ M, E1c is at most countable. Since ∞ \ Ec = Ekc ⊆ E1c , k=1

E c must be at most countable. In other words, E ∈ M and we follow from Definition 12.7  (σ-algebra) that M is an σ-algebra. This ends the proof of the problem.

73

12.3. Lebesgue Measurable Sets Problem 12.24

⋆ ⋆ Suppose that E ⊂ R is compact and Kn = {x ∈ R | d(x, E) < n1 } for all n ∈ N, where d(x, E) = inf{d(x, y) | y ∈ E}. Prove that m(E) = lim m(Kn ). n→∞

Proof. It is clear that E⊆ Let x ∈

∞ \

n=1

∞ \

Kn .

n=1

Kn \ E ⊆ E c . Since E is compact, it is closed in R and then E c is open in R. By

the definition, there is a ǫ > 0 such that Nǫ (x) ⊆ E c and this means that d(x, E) ≥ ǫ.

(12.31)

∞ \

Kn , so x ∈ KN . However, the definition

Pick an N ∈ N such that N > 1ǫ . Recall that x ∈ of KN shows that d(x, E)
0, there exists an open set W ⊆ [0, 1] such that Q ∩ [0, 1] ⊂ W

and m(W ) < ǫ.

Then construct a closed subset K of [0, 1] such that m(K) > 0 and

K ◦ = ∅.

Proof. By Problem 12.18, there exists an open set V containing Q in R such that m(V ) < ǫ. If we consider W = V ∩ [0, 1], then it is open in [0, 1] and it contains Q ∩ [0, 1]. It is clear that m(W ) ≤ m(V ) < ǫ.

(12.38)

This prove the first assertion. For the second assertion, we suppose that K = [0, 1] \ W . Then K is certainly closed in [0, 1]. By the estimate (12.38) and using Problem 12.13, we see that m(K) = m([0, 1] \ W ) = m([0, 1]) − m(W ) ≥ 1 − ǫ > 0. Assume that K ◦ 6= ∅. Then we pick p ∈ K ◦ . Since K ◦ is open in [0, 1] by [25, Problem 4.11, pp. 35, 36], there exists a δ > 0 such that Nδ (p) ⊆ K ◦ ⊆ K ⊆ [0, 1]. However, the neighborhood Nδ (p) must contain a rational r of [0, 1] which contradicts the fact  that K contains no rationals in [0, 1]. Hence we complete the proof of the problem.

77

12.3. Lebesgue Measurable Sets Remark 12.4

By the second assertion of Problem 12.28, we see that the concept of Lebesgue measure zero is different from empty interior.

Problem 12.29 ⋆ Suppose that {an } is a sequence of real numbers and {αn } is a sequence of positive ∞ X √ αn < ∞. Prove that there corresponds a measurable set E such numbers such that ⋆

n=1

that

∞ X

m(E c ) = 0 and

n=1

for all x ∈ E.

αn 0, there exists an open set V ⊆ R containing E such that m∗ (V \ E) < ǫ. (12.41)

Proof. Suppose that E is measurable. Given ǫ > 0. We consider the case that m(E) < ∞. Since m∗ (E) = m(E) < ∞, it follows from Problem 12.5 that there exists an open set V ⊆ R such that m(V ) − m(E) < ǫ. (12.42) By Problem 12.13, we have m(V \ E) = m(V ) − m(E).

(12.43)

Combining the inequality (12.42) and the result (12.43), we establish the expected result (12.41). Next, we consider the case that m∗ (E) = ∞. Now, for each k ∈ N, we define Ek = E ∩ [k, k + 1).

79

12.4. Necessary and Sufficient Conditions for Measurable Sets

By Theorem 12.3(c) (Properties of Outer Measure), we observe that m(Ek ) ≤ m∗ ([k, k + 1)) = 1 so that the result in the previous paragraph can be applied to each Ek . In other words, for each positive integer k, there is an open set Vk containing Ek such that m(Vk \ Ek ) < It is clear that the set V =

∞ [

ǫ . 2k

Vk is open in R containing E. Furthermore, we have

k=1

V \E ⊆ and this implies that m(V \ E) ≤

∞ X k=1

∞ [

(Vk \ Ek )

k=1

∞ X ǫ m(Vk \ Ek ) < =ǫ 2k k=1

which is the case (12.41) again. Conversely, we suppose that the hypothesis (12.41) holds. Thus for each k ∈ N, we can choose an open set Vk containing E such that m∗ (Vk \ E) < Now if we define V =

∞ [

ǫ . 2k

(12.44)

Vk , then it is an open set in R and we have

k=1

V \E =

∞ [

(Vk \ E)

k=1

so that we may apply Theorem 12.3(d) (Properties of Outer Measure) to the inequality (12.44) to get ∞ X ∗ m (V \ E) ≤ m∗ (Vk \ E) < ǫ. k=1

m∗ (V

Since ǫ is arbitrary, we have \E) = 0 and Theorem 12.5(a) (Properties of Measurable Sets) ensures that V \ E is measurable. Finally, we use the fact E = V \ (V \ E) and the set relation  (12.2) to conclude that E is measurable and hence we complete the proof of the problem. Remark 12.5 Similar to the proof of Problem 12.31, we can show that E is measurable if and only if for every ǫ > 0, there exists a closed set K in R contained in E such that m∗ (E \ K) < ǫ.

(12.45)

Chapter 12. Lebesgue Measure

80

Problem 12.32 ⋆ ⋆ Let E ⊆ R. Prove that E is measurable if and only if for every ǫ > 0, there exist an open set V and a closed set K such that K⊆E⊆V

and m(V \ K) < ǫ.

(12.46)

Proof. Suppose that E is measurable. Then we know from the inequalities (12.41) and (12.45) that there exist an open set V and a closed set K such that K ⊆ E ⊆ V,

m∗ (V \ E)
a} is measurable. (b) For every a ∈ R, the set f −1 ([a, ∞]) = {x ∈ R | f (x) ≥ a} is measurable. (c) For every a ∈ R, the set f −1 ([−∞, a)) = {x ∈ R | f (x) < a} is measurable. (d) For every a ∈ R, the set f −1 ([−∞, a]) = {x ∈ R | f (x) ≤ a} is measurable. Remark 13.1 If the sets considered in Definition 13.1 (Lebesgue Measurable Functions) and Theorem 13.2 (Criteria for Measurable Functions) are Borel, then the function f is called Borel measurable or simply a Borel function.

83

Chapter 13. Lebesgue Measurable Functions

84

Theorem 13.3. The function f : R → R is Lebesgue (resp. Borel) measurable if and only if f −1 (V ) is Lebesgue (resp. Borel) measurable for every open set V in R. Particularly, if f is continuous on R, then f is Borel measurable. Similar to the situation of Lebesgue measurable sets and Borel measurable sets, a Borel measurable function must be Lebesgue measurable, but not the converse. Theorem 13.4 (Properties of Measurable Functions). Suppose that f, g : R → R are measurable, {fn } is a sequence of measurable functions and k ∈ N. (a) If Φ : E → R is continuous on E, where f (R) ⊆ E, then the composition Φ ◦ f : R → R is also measurable.a (b) f k , f + g and f g are measurable. (c) The functions sup fn (x), n∈N

inf fn (x),

lim sup fn (x) and

n∈N

n→∞

lim inf fn (x) n→∞

are all measurable. (d) If f (x) = h(x) a.e. on R, then h is measurable.

13.1.2 Simple Functions and the Littlewood’s Three Principles Definition 13.5 (Simple Functions). Let E ⊆ R. The characteristic function of E is given by   1, if x ∈ E; χE (x) =  0, otherwise. If E1 , E2 , . . . , En are pairwise disjoint measurable sets, then the function s(x) =

n X

ak χEk (x),

(13.1)

k=1

where a1 , a2 , . . . , an ∈ R, is called a simple function. Theorem 13.6 (The Simple Function Approximation Theorem). Suppose that f : R → R is measurable and f (x) ≥ 0 on R. Then there exists a sequence {sn } of simple functions such that 0 ≤ s1 (x) ≤ s2 (x) ≤ · · · ≤ f (x) and

f (x) = lim sn (x) n→∞

(13.2)

for all x ∈ R. Remark 13.2 The condition that f is nonnegative can be omitted in Theorem 13.6 (The Simple Function Approximation Theorem). In this case, the set of inequalities (13.2) are replaced by 0 ≤ |s1 (x)| ≤ |s2 (x)| ≤ · · · ≤ |f (x)|. a

Here the result also holds if Φ is a Borel function.

85

13.2. Lebesgue Measurable Functions

If the measurable sets E1 , E2 , . . . , En considered in the expression (13.1) are closed intervals, then the function s will be called a step function. In this case, we have the following approximation theorem in terms of step functions: Theorem 13.7. Suppose that f : R → R is measurable on R. Then there exists a sequence {ϕn } of step functions such that f (x) = lim ϕn (x) n→∞

a.e. on R. Classically, Littlewood says that (i) every measurable set is nearly a finite union of intervals; (ii) every measurable function is nearly continuous and (iii) every pointwise convergent sequence of measurable functions is nearly uniformly convergent. The mathematical formulation of the first principle has been given in Problem 12.34. Now the precise statements of the remaining principles are given in the following two results: Theorem 13.8 (Egorov’s Theorem). Suppose that {fn } is a sequence of measurable functions defined on a measurable set E with finite measure. Given ǫ > 0. If fn → f pointwise a.e. on E, then there exists a closed set Kǫ ⊆ E such that m(E \ Kǫ ) < ǫ

and

fn → f uniformly on Kǫ .

Theorem 13.9 (Lusin’s Theorem). Suppose that f is a finite-valued measurable function on the measurable set E with m(E) < ∞. Given ǫ > 0. Then there exists a closed set Kǫ ⊆ E such that m(E \ Kǫ ) < ǫ and f is continuous on Kǫ .

13.2 Lebesgue Measurable Functions Problem 13.1 ⋆ Suppose that f : R → R is Lebesgue measurable. Prove that |f | is also Lebesgue measurable. Proof. Consider Φ : R → [0, ∞) defined by Φ(x) = |x| which is clearly continuous. Since |f | = Φ ◦ f , Theorem 13.4(a) (Properties of Measurable Functions) ensures that |f | is also Lebesgue measuable and it completes the proof of the problem.  Problem 13.2 ⋆ Suppose that f, g : R → R are measurable. Prove that max(f, g) and min(f, g) are measurable.

Chapter 13. Lebesgue Measurable Functions

86

Proof. Note that 1 max(f, g) = (f + g + |f − g|) 2

and

min(f, g) =

1 (f + g − |f − g|). 2

By Problem 12.1 and Theorem 13.4(b) (Properties of Measurable Functions), we see immediately that both max(f, g) and min(f, g) are measurable. This completes the proof of the problem.  Remark 13.3 As a particular case of Problem 13.2, the functions f + = max(f, 0) and f − = min(f, 0) are measurable.

Problem 13.3 ⋆ Does there exist a nonmeasurable nonnegative function f such that

√

f is measurable?

√ Proof. The answer is negative. Assume that f was measurable. We know that Φ(x) = x2 is continuous in R. Now we have p p f = ( f )2 = Φ ◦ f ,

so Theorem 13.4(a) (Properties of Measurable Functions) ensures that f is measurable, a con tradiction. Hence this completes the proof of the problem. Problem 13.4 ⋆ Suppose that f : [0, ∞) → R is differentiable. Prove that f ′ is measurable.

Proof. Let x ∈ [0, ∞). By the definition, we have f (x) = lim fn (x), n→∞

where fn (x) =

1 )−f (x) f (x+ n 1 n

. Since f is differentiable in [0, ∞), it is continuous on [0, ∞). By

Theorem 13.3, f is (Borel) measurable. For each n ∈ N, we notice that

 1 = (f ◦ g)(x), f x+ n

where g(x) = x + n1 which is clearly continuous on [0, ∞) and so measurable. Thus we follow from Theorem 13.4(a) (Properties of Measurable Functions)b that f (x + n1 ) is measurable. By Theorem 13.4(b) and then (c) (Properties of Measurable Functions), each fn and then f , as the limit of {fn }, are also measurable. This ends the proof of the problem.  b

With Φ and f replaced by f and g in our question respectively.

87

13.2. Lebesgue Measurable Functions Problem 13.5

⋆ Suppose that E ⊆ R is measurable and f : E → R is continuous on E. Prove that f is measurable.

Proof. Let a ∈ R and A = {x ∈ E | f (x) > a}. If A = ∅, then there is nothing to prove. Otherwise, pick x ∈ A. Then the Sign-preserving Property [25, Problem 7.15, p. 112] implies that there exists a δx > 0 such that f (y) > a holds for all y ∈ (x − δx , x + δx ) ∩ E. Therefore, we have i h[ [ (x − δx , x + δx ) ∩ E. A= [(x − δx , x + δx ) ∩ E] = x∈A

x∈A

It is trivial that the set

[

x∈A

(x − δx , x + δx )

is open in R, so it is measurable by Theorem 12.9. This fact and the measurability of E shows that A is measurable. By Definition 13.1 (Lebesgue Measurable Functions), f is measurable and  this completes the proof of the problem. Problem 13.6 ⋆ If f is measurable on E ⊆ R, prove that {x ∈ E |f (x) = a} is measurable for every a ∈ R.

Proof. We have {x ∈ E |f (x) = a} = {x ∈ E |f (x) ≥ a} ∩ {x ∈ E |f (x) ≤ a}.

(13.3)

By Theorem 13.4 (Properties of Measurable Functions), the sets on the right-hand side of the expression (13.3) are measurable. Finally, we apply Theorem 12.5(e) (Properties of Measurable Sets) to conclude that {x ∈ E |f (x) = a} is also measurable, so it ends the analysis of the problem.



Problem 13.7 ⋆ If f is measurable on E ⊆ R, prove that {x ∈ E |f (x) = ∞} is measurable.

Proof. We observe that {x ∈ E | f (x) = ∞} =

∞ \

{x ∈ E | f (x) > n}.

(13.4)

n=1

Since f is measurable on E, each set {x ∈ E | f (x) > n} is measurable. Recall from Remark 12.3 that the set on the right-hand side of the expression (13.4) is measurable and our desired  result follows. This completes the proof of the problem.

Chapter 13. Lebesgue Measurable Functions

88

Problem 13.8 ⋆ Suppose that f and g are measurable on E ⊆ R. Prove that the set {x ∈ E | f (x) > g(x)} is measurable.

Proof. For every x ∈ E, since f (x) > g(x), the density of rationals [25, Theorem 2.2, p. 10] implies that there is a r ∈ Q such that f (x) > r > g(x). Therefore, we have {x ∈ E | f (x) > r > g(x)} = {x ∈ E | f (x) > r} ∩ {x ∈ E | g(x) < r}.

(13.5)

By the hypotheses, both sets {x ∈ E | f (x) > r} and {x ∈ E | g(x) < r} are measurable. Thus it follows from this and the expression (13.5) that the set {x ∈ E | f (x) > r > g(x)} is also measurable. Now we know that {x ∈ E | f (x) > g(x)} =

[

{x ∈ E | f (x) > r > g(x)}

r∈Q

and then Theorem 12.8 guarantees that {x ∈ E | f (x) > g(x)} is measurable and we complete  the analysis of the problem. Problem 13.9 ⋆ Prove that (a) a characteristic function χE is measurable if and only if E is measurable. (b) a simple function s =

n X

ak χEk is measurable.

k=1

Proof. In the following discussion, suppose that E, E1 , . . . , En are measurable. (a) We deduce from Definition 13.5 (Simple Functions) that  ∅, if a ≤ 0;      R, if a > 1; {x ∈ R | χE (x) < a} =      R \ E, if 0 < a ≤ 1.

Consequently, the set {x ∈ R | χE (x) < a} is measurable for every a ∈ R and Definition 13.1 (Lebesgue Measurable Functions) implies that χE is measurable. Conversely, suppose that χE is measurable. Since E = R \ {x ∈ R | χE (x) < 21 } and {x ∈ R | χE (x) < 12 } is measurable, we conclude that E is also measurable.

(b) By part (a), each χEk is measurable. Then repeated applications of Theorem 13.4(b) (Properties of Measurable Functions) show that the simple function s is also measurable.

89

13.2. Lebesgue Measurable Functions

We complete the analysis of the proof.



Problem 13.10 ⋆ Construct a nonmeasurable function.

Proof. By Remark 12.2, R contains a nonmeasurable set S. Consider the characteristic function   1, if x ∈ S; χS (x) =  0, otherwise.

Then χS cannot be measurable by Problem 13.9. This completes the proof of the problem.  Problem 13.11 ⋆ Suppose that f : R → (0, ∞) is measurable. Prove that

1 f

is measurable.

Proof. Let g = f1 : R → (0, ∞). If a ≤ 0, then {x ∈ R | g(x) > a} = R which is definitely measurable. For a > 0, we note that g(x) > a if and only if 0 < f (x) < a1 so that n 1o {x ∈ R | g(x) > a} = x ∈ R 0 < f (x) < a n 1o = {x ∈ R | f (x) > 0} ∩ x ∈ R f (x) < . a

(13.6)

Since f is measurable, Theorem 13.2 (Criteria for Measurable Functions) ensures that the two sets on the right-hand side of the expression (13.6) are measurable. Thus the set {x ∈ R | g(x) > a} is measurable. By Definition 13.1 (Lebesgue Measurable Functions), the g is a measurable function, completing the proof of the problem.  Problem 13.12 ⋆ ⋆ Suppose that f : R → R is measurable and f (x + 1) = f (x) a.e. on R. Prove that there exists a function g : R → R such that f = g a.e. and g(x + 1) = g(x) on R. Proof. Suppose that E = {x ∈ R | f (x + 1) 6= f (x)}. By the hypothesis, we have m(E) = 0. Next, we define [ F = (E + n) (13.7) n∈Z

and

g(x) =

 / F;  f (x), if x ∈ 

0,

otherwise.

(13.8)

Chapter 13. Lebesgue Measurable Functions

90

By Theorem 12.5(d) (Properties of Measurable Sets), each E+n is measurable and m(E+n) = 0. Applying this fact to the inequality (12.1), we derive m(F ) ≤

∞ X

m(E + n) = 0.

n=1

Therefore, we follow from the definition (13.8) that f =g a.e. on R, proving the first assertion. Next, since g(x + 1) = g(x) on F c , it suffices to prove the second assertion on F . To see this, if x ∈ F , then we note from the definition (13.7) that x = E + m for some m ∈ Z which implies that x + 1 = E + m + 1 ∈ F. Hence it is obvious from the definition (13.8) that for every x ∈ F , we still have g(x + 1) = 0 = g(x) which proves the second assertion and this completes the proof of the problem.



Problem 13.13 ⋆

⋆ Let f : R → R be continuous a.e. on R. Prove that f is Lebesgue measurable.

Proof. Let a ∈ R and denote Ea = {x ∈ R | f (x) > a}. Let x ∈ Ea and A be the set of all discontinuities of f . There are two cases: • Case (1): f is discontinuous at x. In this case, A = 6 ∅. By the hypothesis, we know that m(A) = 0 so that A is measurable by Theorem 12.5(a) (Properties of Measurable Sets). • Case (2): f is continuous at x. In this case, the Sign-preserving Property [25, Problem 7.15, p. 112] ensures that there is a δx > 0 such that f (y) > a for all y ∈ (x − δx , x + δx ). In other words, we have (x − δx , x + δx ) ⊆ Ea . Therefore, we are able to write Ea = A ∪

[

x∈Ea \A

(x − δx , x + δx ).

Obviously, since every (x − δx , x + δx ) is open in R, the union [ (x − δx , x + δx )

(13.9)

x∈Ea \A

is also open in R. By Theorem 12.9, the set (13.9) is measurable. Hence we deduce from Case (1) that Ea is also measurable. Recall that a is arbitrary, so Theorem 13.2 (Criteria for Measurable Functions) concludes that f is Lebesgue measurable which completes the proof of  the problem.

91

13.2. Lebesgue Measurable Functions Problem 13.14 ⋆

⋆ Let f : R → R be monotone. Verify that f is Borel measurable.

Proof. Without loss of generality, we may assume that f is increasing. Let a ∈ R. Suppose that p ∈ {x ∈ R | f (x) < a}. Then f (p) < a and for all y < p, since f is increasing, we have f (y) < f (p) < a. Thus we have y ∈ {x ∈ R | f (x) < a}. Now we put α = sup{x ∈ R | f (x) < a}. On the one hand, if f (α) < a, then we have {x ∈ R | f (x) < a} = (−∞, α]. On the other hand, if f (α) ≥ a, then we have {x ∈ R | f (x) < a} = (−∞, α). In any case, we conclude from Theorem 12.10(c) that the set {x ∈ R | f (x) < a} is Borel measurable. By Remark 13.1, the function f is a Borel function, completing the proof of the problem.  Problem 13.15 ⋆ ⋆ Let S be dense in R and f : R → R. Prove that f is measurable if and only if the set {x ∈ R | f (x) > s} is measurable for every s ∈ S.

Proof. Suppose first that f is measurable. Then Theorem 13.2 (Criteria for Measurable Functions) says that {x ∈ R | f (x) > a} is Lebesgue measurable for every a ∈ R. In particular, the set {x ∈ R | f (x) > s} is measurable for every s ∈ S.

For the converse direction, it suffices to prove that {x ∈ R | f (x) > s} is measurable for every s ∈ S c . In fact, given p ∈ S c . Since S is dense in R, there exists a decreasing sequence {pn } ⊆ S such that pn → p as n → ∞. Next, we claim that {x ∈ R | f (x) > p} =

∞ [

{x ∈ R | f (x) > pn }.

(13.10)

n=1

To see this, for each n ∈ N, if y ∈ {x ∈ R | f (x) > pn }, then since pn > p, we have f (y) > p so that y ∈ {x ∈ R | f (x) > p}, i.e., ∞ [

{x ∈ R | f (x) > pn } ⊆ {x ∈ R | f (x) > p}.

(13.11)

n=1

On the other hand, if y ∈ {x ∈ R | f (x) > p}, then we have ǫ = f (y) − p > 0. Since pn converges to p decreasingly, there exists an N ∈ N such that n ≥ N implies pn − p < ǫ and this means that f (y) > pn ,

Chapter 13. Lebesgue Measurable Functions

92

i.e., y ∈ {x ∈ R | f (x) > pn } for all n ≥ N . Consequently, we have {x ∈ R | f (x) > p} ⊆

∞ [

{x ∈ R | f (x) > pn }.

(13.12)

n=1

Now our claim (13.10) follows immediately by combining the set relations (13.11) and (13.12). By the assumption, each {x ∈ R | f (x) > pn } is measurable, so the expression (13.10) shows that {x ∈ R | f (x) > p} is also measurable, where p ∈ S c . By Definition 13.1 (Lebesgue Measurable Functions), f is  measurable and it completes the analysis of the problem. Problem 13.16 ⋆ ⋆ Suppose that f : R → R is bijective and continuous and B denotes the collection of all Borel sets of R. Prove that f (B) ⊆ B.

Proof. Suppose that D = {E ⊆ R | f (E) ∈ B}. First of all, we want to show that D is an σ-algebra. To this end, recall a basic fact from [12, Exercise 2(g) and (h), p. 21] that if f is bijective, then f (E ∩ F ) = f (E) ∩ f (F ) and f (E \ F ) = f (E) \ f (F ).

(13.13)

Thus if E ∈ D, then the second set equation (13.13) implies that f (E c ) = f (R) \ f (E) = R \ f (E). Since f (E), R ∈ B, we have f (E c ) ∈ B and then E c ∈ D. Let Ek ∈ D so that f

∞ [



Ek =

k=1

∞ [

f (Ek ).

(13.14)

k=1

Since f (Ek ) ∈ B for every k = 1, 2, . . ., we have ∞ [

k=1

f (Ek ) ∈ B

and then the set relation (13.14) shows that ∞ [

k=1

Ek ∈ D.

Therefore, D is an σ-algebra by Definition 12.7 (σ-algebra) which proves our claim. Next, for every closed interval [a, b], the continuity of f ensures that f ([a, b]) is compact and connected by [25, Theorems 7.9 and 7.12, p. 100]. Note that f ([a, b]) ⊆ R, so [25, Theorem 4.16,

93

13.2. Lebesgue Measurable Functions

p. 31] implies that f ([a, b]) must be a closed interval. Since f is continuous and one-to-one, we know from [25, Problem 7.33, p. 122] that f is strictly monotonic and hence either f ([a, b]) = [f (a), f (b)]

or

f ([a, b]) = [f (b), f (a)].

Now both [f (a), f (b)] and [f (b), f (a)] are closed in R, so they belong to B. By the definition, we conclude that [a, b] ∈ D for every −∞ < a ≤ b < ∞. In other words, D contains every closed intervals of R and Theorem 12.10(c) says that B ⊆ D. Since D is an σ-algebra and B is a smallest σ-algebra containing all closed intervals of R, we establish that B=D which means f (B) ⊆ B, as desired. Hence we complete the analysis of the problem.



Problem 13.17 ⋆ Let E be measurable. Prove that f : E → R is measurable if and only if f −1 (V ) is measurable for every open set V in R.

Proof. We suppose that f −1 (V ) is measurable for every open set V in R. In particular, f −1 ((a, ∞)) = {x ∈ E | f (x) > a} is measurable for every a ∈ R. By Theorem 13.2 (Criteria for Measurable Functions), f is measurable. Conversely, suppose that f is measurable. By [18, Exercise 29, p. 45], V is an union of countable collection of disjoint open intervals {(an , bn )}, i.e., V =

∞ [

(an , bn ).

n=1

Recall the two

factsc

f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B) and f −1 (A ∩ B) = f −1 (A) ∩ f −1 (B), we have f −1 (V ) = = =

∞ [

n=1 ∞ [

n=1 ∞ [

n=1

f −1 ((an , bn )) f −1 ((−∞, bn ) ∩ (an , ∞)) f −1 ((−∞, bn )) ∩ f −1 ((an , ∞)).

By Theorem 13.2 (Criteria for Measurable Functions), both f −1 ((−∞, bn )) and f −1 ((an , ∞)) are measurable so that every f −1 ((an , bn )) is measurable. Thus Theorem 12.8 implies that f −1 (V )  is measurable, as desired. Hence we have completed the proof of the problem. c

See [12, Exercise 2(b) and (c), p. 20].

Chapter 13. Lebesgue Measurable Functions

94

Problem 13.18 ⋆ ⋆ Suppose that E ⊆ R is measurable and m(E) < ∞. Let f be measurable on E and |f (x)| < ∞ a.e. on E. Prove that for every ǫ > 0, there exists a measurable subset F ⊆ E such that m(E \ F ) < ǫ and f |F is bounded on F . Proof. For each n ∈ N, consider En = {x ∈ E | |f (x)| > n}

(13.15)

and A = {x ∈ E | f (x) = ±∞}. Since |f (x)| < ∞ a.e. on E, we have m(A) = 0. Since f is measurable on E and En = {x ∈ E | f (x) > n} ∪ {x ∈ E | f (x) < −n}, every En is also measurable. Furthermore, it is easy to check from the definition (13.15) that E1 ⊇ E2 ⊇ · · ·

and

A=

∞ \

En .

n=1

Since m(E1 ) ≤ m(E) < ∞, we apply Theorem 12.12 (Continuity of Lebesgue Measure) to conclude that lim m(En ) = m(A) = 0. (13.16) n→∞

Thus given ǫ > 0, there exists an N ∈ N such that m(EN ) < ǫ. Put F = E \ EN . Then we have EN = E \ F

and F = {x ∈ E | |f (x)| ≤ N }

which mean that f |F is definitely bounded by N on F . Therefore, this completes the proof of  the problem. Problem 13.19 ⋆ ⋆ Suppose that {fn } is a sequence of measurable functions on [0, 1] with |fn (x)| < ∞ a.e. on [0, 1]. Prove that there exists a sequence {an } of positive real numbers such that fn (x) →0 an

(13.17)

as n → ∞ a.e. on [0, 1]. Proof. Given ǫ > 0. Using the same notations and argument as in Problem 13.18, we still have the limit (13.16). In other words, for each n ∈ N, there corresponds a kn ∈ N such that m(Ek (n)) = m({x ∈ [0, 1] | |fn (x)| > kn }) < ǫ, where Ek (n) = {x ∈ [0, 1] | |fn (x)| > k}.

(13.18)

95

13.2. Lebesgue Measurable Functions

Now we pick ǫ = 2−n and an = nkn > 0 in the estimate (13.18) to get |f (x)| n 1 o n m(Ek (n)) = m x ∈ [0, 1] < 2−n > an n which implies that

∞ X

m(Ek (n))
0 and f (x) − f (y) ∈ V for every pair x, y ∈ E. Proof. Since V is an open set in R containing 0, there exists a ǫ > 0 such that (−ǫ, ǫ) ⊆ V . Let I = (− 2ǫ , 2ǫ ) and Ep = f −1 (p + I) (13.19) for every p ∈ R. Note that every Ep is measurable by Theorem 13.3. [ We claim that (p + I) = R. Otherwise, there was a x ∈ R such that p∈Q

x∈ /

[

(p + I).

p∈Q

However, there exists a rational q such that q ∈ (x, x + 4ǫ ) and this implies that  ǫ ǫ . x ∈ q − ,q + 2 2 Therefore, no such x exists and we have the claim. By this, we obtain [  [ [ (p + I) = f −1 (R) = R. Ep = f −1 (p + I) = f −1 p∈Q

(13.20)

p∈Q

p∈Q

Assume that m(Ep ) = 0 for all p ∈ Q. Then it follows from the set relation (13.20) and the inequality (12.1) that X ∞ = m(R) ≤ m(Ep ) = 0, p∈Q

Chapter 13. Lebesgue Measurable Functions

96

a contradiction. Thus there exists a p0 ∈ Q such that m(Ep0 ) > 0 which implies that Ep0 6= ∅. Now for every pair x, y ∈ Ep0 , the definition (13.19) shows that f (x), f (y) ∈ (p0 − 2ǫ , p0 + 2ǫ ) which certainly gives f (x) − f (y) ∈ (−ǫ, ǫ). It completes the proof of the problem.



13.3 Applications of Littlewood’s Three Principles Problem 13.21 ⋆

⋆

⋆ Prove Theorem 13.8 (Egorov’s Theorem).

Proof. For each pair of s, t ∈ N, we consider n Et (s) = x ∈ E |fi (x) − f (x)|
N . Prove that if fn → f pointwise a.e. on E, then fn → f in measure. Proof. Given that ǫ > 0. Let A = {x ∈ E | fn (x) → f (x)}. Then we have m(E \ A) = 0. Since m(E) < ∞, we have m(A) < ∞ and we are able to apply Theorem 13.8 (Egorov’s Theorem). Therefore, there is a closed set Kǫ ⊆ A such that m(A \ Kǫ ) < ǫ

and

fn → f uniformly on Kǫ .

By the definition of uniform convergence, there exists an N ∈ N such that n > N implies that |fn (x) − f (x)| < ǫ on Kǫ . In other words, if n > N , then |fn (x) − f (x)| > ǫ only possibly on A \ Kǫ or on E \ A. Hence, for all n > N , we obtain m({x ∈ E | |fn (x) − f (x)| > ǫ}) ≤ m(A \ Kǫ ) + m(E \ A) < ǫ. By the definition, fn → f in measure and it completes the proof of the problem.



99

13.3. Applications of Littlewood’s Three Principles Problem 13.25

⋆ ⋆ Prove Theorem 13.9 (Lusin’s Theorem) in the special case that f is a simple function defined on E.

Proof. Suppose that f takes the form f (x) =

n X

ai χEi (x),

(13.29)

i=1

where a1 , a2 , . . . , an are distinct and E1 , E2 , . . . , En are pairwise disjoint subsets of E whose union is E. By Remark 12.5, there exist closed sets K1 , K2 , . . . , Kn such that Ki ⊆ Ei where i = 1, 2, . . . , n. Set Kǫ =

n [

and

m(Ei \ Ki )
0 such that Ki = j=1 j6=i

Otherwise, for each s ∈ N, we have

ci = ∅. (p − δ, p + δ) ∩ K  1 1 c xs ∈ p − , p + ∩ Ki s s

ci is closed in R, it is also true that p ∈ K ci . This so that xs → p as s → ∞. Since the set K contradicts the fact that K1 , K2 , . . . , Kn are pairwise disjoint. Therefore, we have (p − δ, p + δ) ∩ Kǫ = (p − δ, p + δ) ∩ Ki

Chapter 13. Lebesgue Measurable Functions

100

so that g(x) = ai

(13.31)

for all x ∈ (p−δ, p+δ)∩Kǫ . Since (p−δ, p+δ)∩Kǫ is equivalent to |x−p| < δ and x ∈ Kǫ . Hence the result (13.31) actually means that g is continuous at p. As Ki and then p are arbitrary, we have shown that g is continuous on Kǫ and thus our second assertion is proven. Hence we have completed the proof of the problem.  Problem 13.26 ⋆

⋆ Prove Theorem 13.9 (Lusin’s Theorem) with the aid of Problem 13.25.

Proof. Given ǫ > 0. By Theorem 13.6 (The Simple Function Approximation Theorem), we see that there corresponds a sequence {sn } of simple functions defined on E converging to f pointwise on E. For each n ∈ N, Problem 13.25 ensures that there exists a closed set Kn ⊆ E such that ǫ (13.32) m(E \ Kn ) < n+1 and sn is continuous on Kn . 2 Since sn → f pointwise on E, it follows from Theorem 13.8 (Egorov’s Theorem) that there is a closed set K0 ⊆ E such that m(E \ K0 )
0. We claim that there is a compact set K ⊆ E such that m(E \ K) < ǫ. To see this, we know from Remark 12.5 that there is a closed set F of R contained in E such that ǫ m(E \ F ) < . (13.34) 2 Since m(E) < ∞, we must have m(F ) < ∞. Define Fn = F ∩ [−n, n] for every n = 1, 2, . . .. It is evident that each Fn is measurable and compact. Furthermore, we have F1 ⊆ F2 ⊆ · · ·

and F =

∞ [

Fn .

n=1

As a consequence of Theorem 12.12 (Continuity of Lebesgue Measure), we find that lim m(Fn ) = m(F ).

n→∞

Thus there exists an N ∈ N such that m(F \ [−N, N ]) = m(F \ FN )
0 such that 0 ≤ f (x) ≤ M on [0, 1]. Define g : [0, 1] → R by g =M −f which is a bounded (by 2M ) and nonnegative measurable function defined on [0, 1]. Let t be a simple function such that 0 ≤ f ≤ t. If we consider t′ = min(t, M ), then we have 0 ≤ t′ ≤ M

and f ≤ t′ ≤ t.

Since t′ = 21 (t + M − |t − M |), Problem 14.3 implies that t′ is simple. Note that Z Z t dm t′ dm ≤ [0,1]

[0,1]

and so inf

f ≤t′ ≤M

Z

′

t dm = inf [0,1]

f ≤t

Z

t dm. [0,1]

Chapter 14. Lebesgue Integration

112

Hence, in order to prove the representation (14.3), it suffices to consider simple functions t′ such that f ≤ t′ ≤ M . Next, we know from Definition 14.2 (Integration of Nonnegative Functions) that Z Z Z Z Z s dm, g dm = sup f dm = M dm − f dm = M− [0,1]

[0,1]

[0,1]

[0,1]

(14.7)

[0,1]

where the supremum is taken over all simple measurable functions s with 0 ≤ s ≤ g. Now t = M − s is simple by Problem 14.3 and f ≤ t ≤ M . Conversely, if t is a simple function such that f ≤ t ≤ M , then s = M − t is also a simple function with 0 ≤ s ≤ g. Thus it follows from the formula (14.4) and the expression (14.7) that Z Z (M − g) dm f dm = [0,1] [0,1] Z = M − sup s dm 0≤s≤g [0,1] Z   Z s dm M dm + inf − = inf 0≤s≤g 0≤s≤g [0,1] [0,1] Z (M − s) dm = inf 0≤s≤g [0,1] Z t dm = inf f ≤t≤M

[0,1]

which is our desired result. Hence, this completes the proof of the problem.



Problem 14.5 (Chebyshev’s Inequality) ⋆ Suppose that f is a nonnegative and integrable function on a measurable set E ⊆ R. If α > 0, then prove that Z 1 f dm. (14.8) m({x ∈ E | f (x) ≥ α}) ≤ α E Proof. Let F = {x ∈ E | f (x) ≥ α}. Then F is measurable. If m(F ) = ∞, then m(E) = ∞ and Theorem 14.3(b) implies that Z E

f dm ≥ α · m(E) = ∞

which contradicts Theorem 14.5(d) (Properties of Integrable Functions). In other words, we have m(F ) < ∞. Now it is easy to see from Theorem 14.3(e) that Z Z Z Z f dm ≥ α · m(F ) f dm ≥ f dm + f dm = E

F

E\F

F

which is exactly the inequality (14.8). This ends the analysis of the problem.



Problem 14.6 ⋆ Suppose that f : E → [0, ∞) is measurable, where E is a measurable set. If f ∈ L1 (E), prove that f is finite a.e. on E.

113

14.2. Properties of Integrable Functions

Proof. Let α > 0. Denote Fα = {x ∈ E | f (x) ≥ α} and F∞ = {x ∈ E | f (x) = ∞}. Obviously, we have F∞ ⊂ Fn and Fn+1 ⊆ Fn for all n ∈ N. In addition, we observe from Problem 14.5 and the integrability of f that Z m(F1 ) ≤

E

f dm < ∞.

Hence Theorem 12.12 (Continuity of Lebesgue Measure) and Problem 14.5 imply that Z 1 m(F∞ ) ≤ lim m(Fn ) ≤ lim f dm = 0, n→∞ n→∞ n E

i.e., f is finite a.e. on E which ends the proof of the problem.



Problem 14.7 ⋆ Suppose that E ⊆ R is measurable and f ∈ L1 (E) satisfies Z f dm < Cm(F ) F

for every measurable subset F of E with finite measure, where C is a positive constant. Prove that |f (x)| < C a.e. on E.

Proof. If m(E) = 0, then there is nothing to prove. Thus we assume that m(E) > 0. Since f ∈ L1 (E), we have |f | ∈ L1 (E) by Theorem 14.5(d) (Properties of Integrable Functions). Now Problem 14.5 guarantees that Z 1 m({x ∈ E | |f (x)| ≥ C}) ≤ |f | dm < ∞. (14.9) C E Consider the sets F1 = {x ∈ E | f (x) ≥ C} and F2 = {x ∈ E | f (x) ≤ −C}. By the estimate (14.9), both m(F1 ) and m(F2 ) are finite. On the set F1 , the hypothesis implies that Z Z f dm = Cm(F1 ) ≤

F1

F1

f dm < Cm(F1 )

so that m(F1 ) = 0. On the set F2 , the hypothesis gives Z f dm < Cm(F2 ) Cm(F2 ) ≤ F2

which shows again that m(F2 ) = 0. Equivalently, we conclude that |f (x)| < C a.e. on E and  this completes the proof of the problem. Problem 14.8 ⋆ Suppose that E and f are measurable and E.

Z

E

|f | dm = 0. Prove that f (x) = 0 a.e. on

Chapter 14. Lebesgue Integration

114

Proof. For every n = 1, 2, . . ., we let En = {x ∈ E | |f (x)| ≥ n1 }. On the one hand, we have {x ∈ E | f (x) 6= 0} =

∞ [

En ,

∞ X

m(En ).

n=1

so we deduce from the inequality (12.1) that m({x ∈ E | f (x) 6= 0}) ≤

(14.10)

n=1

On the other hand, we know from Problem 14.5 that Z |f | dm = 0. m(En ) ≤ n

(14.11)

E

Hence we conclude from the inequalities (14.10) and (14.11) that m({x ∈ E | f (x) 6= 0}) = 0 which is equivalent to the condition f (x) = 0 a.e. on E, completing the proof of the problem.  Problem 14.9 ⋆

⋆ Let f : [0, 1] → R be a measurable function. Prove that f ∈ L1 ([0, 1]) if and only if ∞ X

n=1

2n · m({x ∈ [0, 1] | |f (x)| ≥ 2n }) < ∞.

(14.12)

Proof. Define E0 = {x ∈ [0, 1] | |f (x)| < 2} and g(x) = 1 on E0 . Next, for each n ∈ N, we define En = {x ∈ [0, 1] | 2n ≤ |f (x)| < 2n+1 } and g(x) = 2n on En .

(14.13)

It is easy to check that m(E0 ) ≤ 1 and m(En ) ≤ m({x ∈ [0, 1] | |f (x)| ≥ 2n })

(14.14)

for all n ∈ N. Furthermore, it is also true that 0 ≤ g(x) − 1 < |f (x)| ≤ 2g(x) for all x ∈ [0, 1], thus we get from Theorem 14.3(a) that Z Z Z g dm |f | dm < 2 (g − 1) dm ≤ [0,1]

[0,1]

[0,1]

and these mean that f ∈ L1 ([0, 1]) if and only if g ∈ L1 ([0, 1]).

We claim that g ∈ L1 ([0, 1]) if and only if the condition (14.12) holds. To see this, we note from the definition (14.13) that E1 , E2 , . . . are pairwise disjoint and {x ∈ [0, 1] | |f (x)| ≥ 2n } = En ∪ En+1 ∪ · · · , where n = 1, 2, . . .. Therefore, we follow from Theorem 12.11 (Countably Additivity) that m({x ∈ [0, 1] | |f (x)| ≥ 2n }) =

∞ X

k=n

m(Ek ),

(14.15)

115

14.2. Properties of Integrable Functions

where n = 1, 2, . . .. Now we establish from Theorem 14.3(e), the relations (14.14) and finally the hypothesis (14.12) that Z

g dm = [0,1]

=

∞ Z X

g dm

n=0 En ∞ X n n=0

≤1+

2 · m(En ) ∞ X

n=1

2n · m({x ∈ [0, 1] | |f (x)| ≥ 2n }).

(14.16)

On the other hand, we observe from the expression (14.15) that ∞ X

n=1

n

n

2 · m({x ∈ [0, 1] | |f (x)| ≥ 2 }) = =

∞ X

n=1 ∞ X

n

2 ·

≤

=2

k X

2n

n=1

2k+1 m(Ek )

k=1 ∞ X

≤2

m(Ek )

k=n

m(Ek )

k=1

∞ X

∞ X

k=0 Z

2k m(Ek ) g dm.

(14.17)

[0,1]

By the inequalities (14.16) and (14.17), we see that ∞

1X n 2 · m({x ∈ [0, 1] | |f (x)| ≥ 2n }) ≤ 2 n=1

Z

[0,1]

g dm ≤ 1 +

∞ X

n=1

2n · m({x ∈ [0, 1] | |f (x)| ≥ 2n })

which implies that g ∈ L1 ([0, 1]) if and only if the hypothesis (14.12) holds. This completes the  proof of the problem. Problem 14.10 ⋆ ⋆ Let E be measurable with m(E) < ∞ and f be nonnegative and integrable on E. Given ǫ > 0, define En = {x ∈ E | nǫ ≤ f (x) < (n + 1)ǫ} for every n = 0, 1, 2, . . . and A(ǫ) = ǫ

∞ X

n=0

Prove that

Z

E

n · m(En ).

f dm = lim A(ǫ). ǫ→0

Chapter 14. Lebesgue Integration

116

Proof. Define F = {x ∈ E | f (x) = ∞}. By the definition, we know that E1 , E2 , . . . are pairwise disjoint and ∞ [ En . E=F∪ n=0

On the one hand, we obtain from Theorem 14.3(e) and Problem 14.6 that A(ǫ) = ǫ

∞ X

n=0

n · m(En ) =

∞ Z X

n=0 En

(nǫ) dm ≤

∞ Z X

f dm =

Z

f dm =

E\F

n=0 En

Z

f dm.

(14.18)

E

On the other hand, we see from Theorem 14.3(e) and Problem 14.6 again that ∞ X

A(ǫ) + ǫm(E) = ǫ = = >

n=0 ∞ X

n · m(En ) + ǫ ·

∞ X

m(En )

n=0

(n + 1)ǫm(En )

n=0 ∞ Z X

(n + 1)ǫ dm

n=0 En ∞ Z X

f dm

n=0 En

=

Z

f dm

E\F

=

Z

f dm.

(14.19)

E

Hence we deduce from the inequalities (14.18) and (14.19) that Z Z f dm f dm − ǫm(E) < A(ǫ) ≤ E

E

which implies the desired result by letting ǫ → 0. This completes the proof of the problem.  Problem 14.11 ⋆ Let E be a measurable set and f be measurable on E. If there is a nonnegative function g ∈ L1 (E) such that |f (x)| ≤ g(x) on E, prove that f ∈ L1 (E).

Proof. By Theorem 14.3(a) and the hypothesis, we have Z Z g dm < ∞, |f | dm ≤ E

E

i.e., f ∈ L1 (E) which completes the proof of the problem.



117

14.2. Properties of Integrable Functions

Problem 14.12 ⋆ Let E ∈ B and f be integrable on E. Define Z f dm ϕ(E) = E

for every E ∈ B. Prove that ϕ(E) = 0 for all E ∈ B if and only if f = 0 a.e. on R.

Proof. If f = 0 a.e. on R, then it follows form Theorem 14.3(f) that ϕ(E) =

Z

f dm = 0

E

for every E ∈ B. Conversely, consider E = {x ∈ R | f (x) ≥ 0}. Then we have f + = f · χE so that Z Z + f dm = ϕ(E) = 0. f dm = E

R

In other words, we have f + = 0 a.e. on R by Theorem 14.3(g). Similarly, we can show that Z

f − dm = 0 R

so that f − = 0 a.e. on R. By the definition, f = f + − f − = 0 a.e. on R which completes the proof of the problem. 

Problem 14.13 ⋆ Suppose that E is measurable and f, g, h : E → R∗ are measurable. If g and h are integrable on E and g ≤ f ≤ h a.e. on E, prove that f is integrable on E.

Proof. Since h and g are integrable on E, h−g is integrable on E by Theorem 14.5(a) (Properties of Integrable Functions). Obviously, we have 0≤f −g ≤h−g a.e. on E, so Problem 14.11 implies that f − g is also integrable on E. By Theorem 14.5(a) (Properties of Integrable Functions) again, the function f = (f − g) + g is integrable on E. This completes the proof of the problem.



Chapter 14. Lebesgue Integration

118

Problem 14.14 ⋆ Suppose that E is measurable, E1 , E2 , . . . are measurable subsets of E and f : E → R∗ ∞ Z ∞ X [ En . If f is integrable on each En and |f | dm is a measurable function. Let F = n=1 En

n=1

converges, prove that f is integrable on F and

∞ Z X Z f dm ≤ F

n=1 En

|f | dm.

Proof. Without loss of generality, we may assume that E1 , E2 , . . . are pairwise disjoint.b Recall from Remark 14.1 that |f ± | ≤ |f | on each En , so we have Z Z ± |f | dm (14.20) f dm ≤ En

En

for every n = 1, 2, . . .. Since f ± ≥ 0 on every En , we obtain from Theorem 14.3(e), the estimate (14.20) and then the hypothesis that Z

f ± dm = F

∞ Z X

n=1 En

f ± dm ≤

∞ Z X

n=1 En

|f | dm < ∞.

(14.21)

Recall that |f | = f + + f − , so the estimates (14.21) give Z Z Z + f − dm < ∞. f dm + |f | dm = F

F

F

By Definition 14.4 (Lebesgue Integrable Functions), the function f is integrable on F which proves the firs assertion. For the second assertion, we notice from Theorem 14.5(d) (Properties of Integrable Functions) and then Theorem 14.3(e) that ∞ Z Z Z X |f | dm = f dm ≤ F

F

n=1 En

|f | dm.

This completes the analysis of the problem.



Problem 14.15 ⋆ ⋆ Suppose that E is measurable with m(E) < ∞ and f, g : E → R are measurable. If f, g ∈ L1 (E) and Z Z g dm f dm ≤ F

F

for every measurable set F ⊆ E. Verify that f (x) ≤ g(x) a.e. on E. b

See the proof of Problem 12.25.

119

14.2. Properties of Integrable Functions

Proof. Consider

n 1o , En = x ∈ E f (x) ≥ g(x) + n

where n = 1, 2, . . .. Then we follow from Problems 13.6 and 13.8 that each n 1o n 1o En = x ∈ E f (x) > g(x) + ∩ x ∈ E f (x) − g(x) = n n

(14.22)

is measurable. Given n ∈ N fixed, since g ∈ L1 (E), Theorem 14.5(c) and (d) (Properties of Integrable Functions) say that Z Z |g| dm < ∞. (14.23) g dm ≤ En

En

Furthermore, we see immediately from the hypothesis and the expression (14.22) that Z Z Z 1 g dm + m(En ) f dm ≥ g dm ≥ (14.24) n En En En

for every n ∈ N. Consequently, we conclude from the inequalities (14.23) and (14.24) that m(En ) = 0 for each n = 1, 2, . . .. Thus we get from the inequality (12.1) that m

∞ [

n=1

 En = 0.

(14.25)

Let E ′ = {x ∈ E | f (x) > g(x)}. If p ∈ E ′ , then we must have f (p) > g(p) + n1 for some large enough n ∈ N, i.e., p ∈ En . This means, with the aid of the result (14.25), that m(E ′ ) = 0 and this is equivalent to f (x) ≤ g(x) a.e. on E. We have completed the proof of the problem.  Problem 14.16 ⋆

⋆ Suppose that f : [0, 1] → [0, ∞) is measurable. Consider 0 < ǫ ≤ 1 and o nZ f dm E is a measurable subset of [0, 1] with m(E) ≥ ǫ . S= E

Prove that inf(S) > 0.

Proof. Define E0 = {x ∈ [0, 1] | f (x) ≥ 1} and for each n ∈ N, we define

Then it is trivial that

n En = x ∈ [0, 1]

1o 1 . ≤ f (x) < n+1 n

[0, 1] = E0 ∪

∞ [

n=1

En .

(14.26)

Chapter 14. Lebesgue Integration

120

Since E0 , E1 , E2 , . . . are pairwise disjoint measurable subsets of [0, 1], we apply Theorem 12.11 (Countably Additivity) to the representation (14.26) to get ∞ X

m(En ) = 1

n=0

which implies the existence of an N ∈ N such that ∞ X

n=N

ǫ m(En ) ≤ . 2

(14.27)

By Problem 12.27, there exists a measurable subset F ⊆ [0, 1] such that m(F ) = ǫ. Let n F1 = F ∩ x ∈ [0, 1] f (x) ≥

∞ [ 1 o En =F ∩ N +1 n=N

and

F2 = F \ F1 .

Then we see from Problem 12.13 and the estimate (14.27) that m(F2 ) = m(F \ F1 ) = m(F ) − m(F1 ) ≥ ǫ − Hence it yields from Theorem 14.3(a) that Z Z f dm ≥ f dm ≥ F

F2

∞ X

n=N

m(En ) ≥

ǫ > 0. 2

1 ǫ m(F2 ) ≥ > 0. N +1 2(N + 1)

In other words, we have inf(S) > 0 which completes the proof of the problem.



Problem 14.17 ⋆ Is it true that if f, g : R → R are integrable, then f ◦ g : R → R is integrable?

Proof. The answer is negative. Consider f = χ[0,1] and g = χ{1} . Clearly, we have f, g ∈ L1 (R). However, we note that f (g(x)) = χ[0,1] (χ{1} (x))   χ[0,1] (1), if x = 1; =  χ[0,1] (0), otherwise =1

which implies that f ◦ g ∈ / L1 (R) because Z |f (g(x))| dm = m(R) = ∞. R

This completes the proof of the problem.



121

14.2. Properties of Integrable Functions

Problem 14.18 ⋆ Suppose that E is measurable with m(E) > 0 and f is measurable on E with Z Z |f | dm. f dm = E

E

Prove that either f ≥ 0 a.e. on E or f ≤ 0 a.e. on E.

Proof. Suppose that

Z

E

By the hypothesis, we have

Z

E

f dm ≥ 0.

(14.28)

(f − |f |) dm = 0,

so Theorem 14.3(g) implies that f (x) − |f (x)| = 0 a.e. on E and this means that f (x) = |f (x)| ≥ 0 a.e. on E. Next, we suppose that

Z

E

f dm ≤ 0.

In this case, we consider the function −f which is also measurable on E and it satisfies condition (14.28). Thus our previous analysis can be applied to −f to conclude that −f (x) = | − f (x)| ≥ 0 a.e. on E and this is equivalent to f (x) ≤ 0 a.e. on E. This completes the proof of the problem. Problem 14.19 ⋆ Suppose that f ∈ L1 ((0, ∞)). Prove that there exists a sequence xn → ∞ such that lim xn f (xn ) = 0.

n→∞

Proof. We put α = lim inf x|f (x)|. x→∞

If α = 0, then since −x|f (x)| ≤ xf (x) ≤ x|f (x)| on (0, ∞), we have lim inf xf (x) = 0 x→∞



Chapter 14. Lebesgue Integration

122

and, by the definition, such a sequence exists. Next, we suppose that α > 0. Then there is an N ∈ N such that x|f (x)| > α2 for all x ≥ N . Consequently, we conclude from Theorem 14.3(c) that Z ∞ Z Z α α |f | dm ≥ |f | dm ≥ dx = ln x|∞ N = ∞, 2 N 2x (N,∞) (0,∞) a contradiction. Hence we have completed the proof of the problem.



Problem 14.20 ⋆ ⋆ Suppose that E is measurable and f is integrable on E. Given ǫ > 0. Prove that there is a δ > 0 such that Z f dm < ǫ (14.29) F

for every F ⊆ E with m(F ) < δ.

Proof. Given ǫ > 0. If f is bounded on E, then there exists a M > 0 such that |f (x)| ≤ M . Since f ∈ L1 (E), Theorem 14.5(d) (Properties of Integrable Functions) gives Z f dm ≤ M · m(F ). (14.30) F

If F is a measurable subset of E such that m(F ) < (14.30) that the inequality (14.29) holds in this case.

ǫ M,

then we conclude from the inequality

Suppose next that f is unbounded on E. For each n = 0, 1, 2, . . ., we consider the sets En = {x ∈ E | n ≤ |f (x)| < n + 1},

Fn =

n [

Ek

k=0

and

Gn = E \ Fn =

n [

Ek .

(14.31)

k=N +1

Since |f (x)| ≥ 0 on E and E0 , E1 , . . . are pairwise disjoint subsets of E whose union is E, we know from Theorem 14.3(e) and the hypothesis that Z ∞ Z X |f | dm < ∞. |f | dm = n=0 En

E

Therefore, there exists an N ∈ N such that Z Z ∞ X |f | dm = GN

n=N +1

ǫ |f | dm < . 2 En

(14.32)

ǫ 2(N +1) .

Let δ = Thus if F is a measurable subset of E with m(F ) < δ, then it yields from Theorem 14.5(d) (Properties of Integrable Functions), the facts E = FN ∪ GN and FN ∩ GN = ∅ that Z Z Z Z |f | dm. (14.33) |f | dm + |f | dm = f dm ≤ F

F

F ∩FN

F ∩GN

By applying the definition (14.31) and the inequality (14.32) to the expression (14.33), we obtain immediately that Z Z ǫ ǫ ǫ |f | dm < (N + 1)δ + = + = ǫ. f dm ≤ (N + 1)m(F ) + 2 2 2 GN F

This completes the proof of the problem.



123

14.3. Applications of Fatou’s Lemma

14.3 Applications of Fatou’s Lemma Problem 14.21 ⋆ Prove that we may have the strict inequality in Fatou’s Lemma. Proof. For each n = 1, 2, . . ., we consider the function fn : R → R defined by   1, if x ∈ [n, n + 1); fn (x) =  0, otherwise.

(14.34)

If x ∈ R, then it is easy to check that x ∈ / [n, n + 1) for all large enough n so that lim inf fn (x) = 0 n→∞

which gives

Z  R

 lim inf fn dm = 0. n→∞

However, the definition of fn implies that Z Z fn dm = R

n+1

dm = 1. n

Hence the sequence of functions (14.34) shows that the strict inequality in Fatou’s Lemma can  actually occur, completing the proof of the problem. Problem 14.22 ⋆ ⋆ Suppose that fn , gn , f, g ∈ L1 (R) for every n ∈ N. If fn → f , gn → g pointwise a.e. on R, |fn (x)| ≤ gn (x) a.e. on R and Z Z g dm, gn dm → R

R

prove that

Z

R

fn dm →

Z

f dm. R

Proof. By the triangle inequality, we know that |fn − f | ≤ |fn | + |f | ≤ gn + |f | so that gn + |f | − |fn − f | ≥ 0. By Fatou’s Lemma, we have Z Z Z |f | dm = (g + |f |) dm g dm + R R ZR lim (g + |f | − |fn − f |) dm = R n→∞

Chapter 14. Lebesgue Integration Z lim inf (g + |f | − |fn − f |) dm = R n→∞ Z ≤ lim inf (g + |f | − |fn − f |) dm n→∞ R Z Z  Z |fn − f | dm |f | dm − g dm + = lim inf n→∞ R RZ Z R Z = |f | dm − lim sup |fn − f |) dm g dm + n→∞

R

R

so that lim sup n→∞

Z

R

124

R

|fn − f |) dm = 0.

(14.35)

Since fn , f ∈ L1 (E), we deduce from Theorem 14.5(a) and (d) (Properties of Integrable Functions) that fn − f ∈ L1 (E) and Z Z Z Z |fn − f | dm. (14.36) f dm = (fn − f ) dm ≤ fn dm − R

R

R

R

Applying the result (14.35) to the inequality (14.36), we see that Z Z f dm f dm − →0 n R

R

as n → ∞ which implies that

lim

Z

n→∞ R

This ends the proof of the problem.

fn dm =

Z

f dm.

R



Problem 14.23 ⋆ Let −∞ < a < b < ∞. Let {fn } be a sequence of nonnegative measurable functions on (a, b) such that fn → f a.e. on (a, b). Define Z x Z x fn dm, f dm and Fn (x) = F (x) = a

a

where n = 1, 2, . . .. Prove that Z Z (f + F ) dm ≤ lim inf n→∞

[a,b]

(fn + Fn ) dm.

Proof. Since {fn } satisfies the hypothesis of Fatou’s Lemma, so we know that Z x Z x Z x fn dm = lim inf Fn (x). lim inf fn dm ≤ lim inf f dm = F (x) = a

a

n→∞

n→∞

(14.37)

[a,b]

a

n→∞

(14.38)

Since fn is nonnegative, each Fn is also nonnegative. Now we apply Fatou’s Lemma to the inequality (14.38) one more time, we see that Z Z (f + lim inf Fn ) dm (f + F ) dm ≤ [a,b]

[a,b]

n→∞

125

14.3. Applications of Fatou’s Lemma Z

lim inf (fn + Fn ) dm Z (fn + Fn ) dm ≤ lim inf =

[a,b] n→∞

n→∞

[a,b]

which is exactly the expected result (14.37). It completes the proof of the problem.



Problem 14.24 ⋆ Suppose that E is measurable and {fn } is a sequence of measurable functions defined on E. If g is integrable on E and |fn | ≤ g on E for all n = 1, 2, . . ., prove that Z Z Z Z lim sup fn dm. (14.39) fn dm ≤ lim sup fn dm ≤ lim inf fn dm ≤ lim inf n→∞

E

n→∞

E

E

E

Proof. Since |fn | ≤ g on E and for all n = 1, 2, . . ., we must have g − fn ≥ 0 and g + fn ≥ 0 on E and for all n = 1, 2, . . .. Applying Fatou’s Lemma to both g − fn and g + fn , we get Z Z lim inf (g + fn ) dm ≤ lim inf (g + fn ) dm E n→∞

and

Z

n→∞

lim inf (g − fn ) dm ≤ lim inf

E n→∞

Then they imply that Z

n→∞

lim inf fn dm +

E n→∞

and

Z

E

g dm −

Z

E

Z

E

g dm ≤ lim inf n→∞

lim sup fn dm ≤ n→∞

Z

E

E

Z

E

Z

(g − fn ) dm.

fn dm + E

g dm − lim sup n→∞

Z

Z

g dm

(14.40)

fn dm.

(14.41)

E

E

Combining the results (14.40) and (14.41), we conclude immediately that the set of inequalities (14.39) hold and we complete the proof of the problem.  Problem 14.25 ⋆ ⋆ Suppose that E is a measurable set and {fn } is a sequence of integrable functions defined on E that converges pointwise almost everywhere to an integrable function f defined on E. If we have Z Z |f | dm, (14.42) |fn | dm = lim n→∞ E

prove that

lim

Z

n→∞ F

for every measurable subset F of E.

E

|fn | dm =

Z

F

|f | dm

Chapter 14. Lebesgue Integration

126

Proof. Let F be a measurable subset of E. Notice that |fn − f |χF ≤ |fn − f | ≤ |fn | + |f | on E. Thus if we set gn = |fn | + |f | − |fn − f |χF , then they are nonnegative and measurable so that we can apply Fatou’s Lemma to get Z Z gn dm lim inf gn dm ≤ lim inf n→∞ E E n→∞ Z Z (14.43) lim inf (|fn | + |f | − |fn − f |χF ) dm ≤ lim inf (|fn | + |f | − |fn − f |χF ) dm. E n→∞

n→∞

E

Since fn → f pointwise a.e. on E, we deduce from the hypothesis (14.42) and the inequality (14.43) that 2

Z

E

|f | dm −

Z

E

Z

Z

|f | dm |fn | dm + E Z − lim sup |fn − f |χF dm n→∞ E Z Z Z Z |f | dm − lim sup |fn − f | dm |f | dm + |f | dm − 0 ≤ 2 n→∞ E F E E Z |fn − f | dm ≤ 0 lim sup lim sup |fn − f |χF dm ≤ lim inf n→∞

n→∞

n→∞

E

F

which implies 0 ≤ lim inf n→∞

Z

F

|fn − f | dm ≤ lim sup n→∞

Z

F

|fn − f | dm ≤ 0.

(14.44)

Consequently, the inequalities (14.44) mean that lim

Z

n→∞ F

|fn − f | dm = 0.

(14.45)

By Theorem 14.5(a) and (d) (Properties of Integrable Functions) and the triangle inequality, we obtain Z Z Z Z Z |fn − f | dm. ||fn | − |f || dm ≤ |f | dm = (|fn | − |f |) dm ≤ |fn | dm − F

F

F

F

F

Therefore, the limit (14.45) implies that

Z Z |f | dm = 0 |fn | dm − lim

n→∞

F

F

which gives the desired result. This completes the proof of the problem. Problem 14.26 ⋆

⋆ Use Fatou’s Lemma to prove the Lebesgue’s Monotone Convergence Theorem.



127

14.3. Applications of Fatou’s Lemma

Proof. Since fn → f pointwise a.e. on E and every fn is nonnegative, Fatou’s Lemma gives Z

f dm ≤ lim inf n→∞

E

Z

fn dm.

(14.46)

E

Furthermore, we also have fn ≤ f a.e. on E and for all n = 1, 2, . . ., so Theorem 14.3(a) implies that Z Z f dm fn dm ≤ E

E

for all n ∈ N and then lim sup n→∞

Z

E

Z

fn dm ≤

f dm.

(14.47)

E

Hence we combine the two inequalities (14.46) and (14.47) to obtain lim

Z

n→∞ E

fn dm =

Z

f dm,

E

completing the proof of the problem.



Problem 14.27 ⋆

⋆ Use Fatou’s Lemma to prove the Lebesgue’s Dominated Convergence Theorem.

Proof. By Problem 14.11, we observe that fn , f ∈ L1 (E) for every n ∈ N. On the one hand, since fn + g ≥ 0 on E, Fatou’s Lemma is applicable to gain Z

(f + g) dm = E

or equivalently,

Z

lim inf (f + g) dm ≤ lim inf

E n→∞

Z

E

n→∞

f dm ≤ lim inf n→∞

Z

Z

(fn + g) dm

E

fn dm.

(14.48)

E

On the other hand, since g − fn ≥ 0, Fatou’s Lemma again shows that Z

E

(g − f ) dm =

so that

Z

lim inf (g − fn ) dm ≤ lim inf

E n→∞

n→∞

Z

E

Z

E

(g − fn ) dm =

f dm ≥ lim sup n→∞

Z

f dm.

Z

E

g dm − lim sup n→∞

Z

fn dm E

(14.49)

E

Hence our desired result follows by combining the inequalities (14.48) and (14.49), completing the proof of the problem. 

Chapter 14. Lebesgue Integration

128

Problem 14.28 ⋆ Suppose that {fn } is a sequence of integrable functions defined on R and fn → f pointwise a.e. on R. Furthermore, for every ǫ > 0, there is a measurable set E ⊆ R such that Z |fn | dm ≤ ǫ (14.50) Ec

for all n ≥ N and as n → ∞. Prove that

Z lim

E

|fn − f | dm → 0

Z

n→∞ R

(14.51)

|fn − f | dm = 0.

Proof. According to Fatou’s Lemma and the hypothesis (14.50), we see that Z Z Z |fn | dm ≤ ǫ. lim inf |fn | dm ≤ lim inf |f | dm = E c n→∞

Ec

n→∞

(14.52)

Ec

Next, using the inequality (14.52), we have Z Z Z |fn − f | dm |fn − f | dm + |fn − f | dm ≤ c R Z ZE ZE |f | dm |fn | dm + |fn − f | dm + ≤ Ec Ec ZE |fn − f | dm + 2ǫ. ≤ E

Hence we follow from the hypothesis (14.51) that Z lim sup |fn − f | dm ≤ 2ǫ. n→∞

R

Since ǫ is arbitrary, we establish lim

Z

n→∞ R

|fn − f | dm = 0

which completes the proof of the problem.



Problem 14.29 ⋆ Suppose that E is a measurable set, {fn }, {gn }, {hn } ⊆ L1 (E) and gn ≤ fn ≤ hn a.e. on E. Furthermore, we have fn → f , gn → g, hn → h pointwise on E. If g, h ∈ L1 (E) satisfy Z Z Z Z h dm hn dm → g dm and gn dm → E

as n → ∞, prove that

lim

E

E

E

Z

n→∞ E

fn dm =

Z

E

f dm.

129

14.4. Applications of Convergence Theorems

Proof. By the hypotheses, it is clear that g(x) ≤ f (x) ≤ h(x) a.e. on E. Since g and h are integrable on E, Problem 14.13 ensures that f ∈ L1 (E). Now fn (x) − gn (x) ≥ 0 a.e. on E for all positive integers n, so Fatou’s Lemma implies that Z Z Z g dm = (f − g) dm f dm − E E ZE lim inf (fn − gn ) dm = E n→∞ Z ≤ lim inf (fn − gn ) dm n→∞ E Z Z g dm fn dm − = lim inf n→∞

so that

Z

E

f dm ≤ lim inf n→∞

E

E

Z

fn dm.

(14.53)

E

Similarly, Fatou’s Lemma again shows that Z Z Z f dm = (h − f ) dm h dm − E E ZE lim inf (hn − fn ) dm = E n→∞ Z ≤ lim inf (hn − fn ) dm n→∞ E Z Z h dm − lim sup fn dm = n→∞

E

and then

Z

E

f dm ≥ lim sup n→∞

Z

fn dm.

E

(14.54)

E

Hence we derive from the inequalities (14.53) and (14.54) that Z Z f dm. fn dm = lim n→∞ E

E

This completes the proof of the problem.



14.4 Applications of Convergence Theorems Problem 14.30 ⋆ Suppose that fn : E → R is integrable for each n = 1, 2, . . ., where E is measurable with m(E) < ∞. Prove that if {fn } converges uniformly to f on E, then Z Z f dm. (14.55) fn dm = lim n→∞ E

E

Chapter 14. Lebesgue Integration

130

Proof. Since {fn } converges uniformly to f on E, there exists an N ∈ N such that |fn (x) − fN (x)| < 1 for all n ≥ N and x ∈ E. Therefore, we have |fn (x)| ≤ |fN (x)| + 1 on E. Since m(E) < ∞ and fN ∈ L1 (E), we conclude that |fN | + 1 ∈ L1 (E). Hence it follows from the Lebesgue’s Dominated Convergence Theorem that the limit (14.55) holds, completing the proof of the problem.  Problem 14.31 ⋆ Prove that lim R

n→∞

Proof. Let fn (x) =

1+nx2 (1+x2 )n

Z

0

1

1 + nx2 dx = 0. (1 + x2 )n

for every n = 1, 2, . . .. By the binomial theorem, we know that

(1 + x2 )n = 1 + nx2 +

n(n − 1) 4 n(n − 1) 4 x + · · · + x2n ≥ 1 + nx2 + x 2 2

on [0, 1] so that |fn (x)| ≤

1 + nx2 1 + nx2 +

n(n−1) 4 x 2

≤1

on [0, 1]. Therefore, each fn is bounded on [0, 1] and Theorem 14.6 ensures that fn ∈ L1 ([0, 1]) and then Z 1 Z fn dx. fn dm = R 0

[0,1]

Now, since we have lim

n→∞

1 + nx2 1 + nx2 +

n(n−1) 4 x 2

=0

for every x ∈ (0, 1], we have fn → f = 0 pointwise a.e. on [0, 1]. Hence we conclude from the Lebesgue’s Dominated Convergence Theorem that Z 1 Z Z 1 + nx2 lim R lim fn dm = 0 fn dm = dx = lim 2 n n→∞ n→∞ [0,1] 0 (1 + x ) [0,1] n→∞ which ends the proof of the problem. Problem 14.32 ⋆ Let −∞ < a < b < ∞ and f ∈ L1 ([a, b]). Prove that lim

n→∞

Z

 |f (x)|2  dm = 0. n ln 1 + n2 [a,b]



131

14.4. Applications of Convergence Theorems

Proof. Suppose that

 |f (x)|2  fn (x) = n ln 1 + , n2 where n = 1, 2, . . .. Since f ∈ L1 ([a, b]), Problem 14.6 ensures that |f | is finite a.e on [a, b]. Now it is easy to see that fn (x) =

1  |f (x)|2 n2 1 1 ln 1 + ≤ ln exp(|f (x)|2 ) = |f (x)|2 → 0 2 n n n n

(14.56)

pointwise a.e. on [a, b]. Next, if y ≥ 0, then we have √

e2

y

√ 4y =1+2 y+ + ··· ≥ 1 + y 2

√ so that ln(1 + y) ≤ 2 y. Thus we obtain  |f (x)|2  fn (x) = n ln 1 + ≤ 2n n2

r

|f (x)|2 = 2n · n2

p

|f (x)|2 = 2|f (x)| n

for all n ∈ N and all x ∈ [a, b]. By the fact f ∈ L1 ([a, b]) and the limit (14.56), we can apply the Lebesgue’s Dominated Convergence Theorem to conclude that Z Z Z  |f (x)|2  lim fn (x) dm = 0, fn (x) dm = n ln 1 + dm = lim lim n→∞ [a,b] n→∞ [a,b] n2 [a,b] n→∞ as required. This completes the proof of the problem.



Problem 14.33 ⋆ Suppose that fn , f ∈ L1 (R) for all n ∈ N, fn → f pointwise a.e. on R and Z Z |f | dm. |fn | dm = lim n→∞ R

Prove that lim

Z

n→∞ E

for every measurable subset E of R.

(14.57)

R

fn dm =

Z

f dm E

Proof. Notice from the triangle inequality that ||fn | − |fn − f || ≤ |f |, so the Lebesgue’s Dominated Convergence Theorem implies that Z Z  Z |fn − f | dm = lim (|fn | − |fn − f |) dm |fn | dm − lim n→∞ n→∞ R R R Z lim (|fn | − |fn − f |) dm = n→∞ R Z |f | dm. = R

(14.58)

Chapter 14. Lebesgue Integration

132

By the hypothesis (14.57) and the limit (14.58), we get Z Z Z Z |fn − f | dm |fn | dm − lim |fn − f | dm = lim |f | dm − lim n→∞ R n→∞ R n→∞ R R Z Z   = lim |fn − f | dm |fn | dm − n→∞ R R Z |f | dm = R

which implies that lim

Z

n→∞ R

|fn − f | dm = 0.

(14.59)

Since fn − f ∈ L1 (R) for every positive integer n, Theorem 14.5(c) and (d) (Properties of Integrable Functions) ensures that, for every measurable subset E of R, Z Z Z Z |fn − f | dm. (14.60) |fn − f | dm ≤ f dm ≤ fn dm − E

E

E

R

Using the limit (14.59) directly to the inequality (14.60), we acquire the desired result which completes the proof of the problem.  Problem 14.34 ⋆

⋆

⋆ Prove the Bounded Convergence Theorem.

Proof. It is clear that the theorem holds when m(E) = 0. Without loss of the generality, we may assume that m(E) > 0 in the following discussion. Since each fn is measurable on E and fn → f pointwise on E, f is measurable on E by Theorem 13.4(c) (Properties of Measurable Functions). Since {fn } is uniformly bounded, there exists a M > 0 such that |fn (x)| ≤ M (14.61) for all n ∈ N and x ∈ E. Consequently, we also have |f (x)| ≤ M

(14.62)

for all x ∈ E. Now since {fn } and f are bounded on E, each fn − f is also boundedc on E and the paragraph following Theorem 14.5 (Properties of Integrable Functions) guarantees that Z Z |fn − f | dm. (14.63) (fn − f ) dm ≤ E

E

If F is any measurable subset of E, then we have Z Z Z f dm = (fn − f ) dm fn dm − E E Z ZE = (fn − f ) dm + F

c

In fact, the sequence {fn − f } is uniformly bounded.

E\F

(fn − f ) dm,

133

14.4. Applications of Convergence Theorems

so we deduce from the bounds (14.61), (14.62) and the estimate (14.63) that Z Z Z Z (f − f ) dm + (f − f ) dm ≤ f dm f dm − n n n E\F F E E Z Z Z |f | dm |fn | dm + |fn − f | dm + ≤ E\F E\F F Z |fn − f | dm + 2M · m(E \ F ). ≤

(14.64)

F

Next, given ǫ > 0. Since m(E) < ∞, Egorov’s Theorem ensures that there exists a closed set ǫ Kǫ ⊆ E such that m(E \ Kǫ ) < 4M and fn → f uniformly on Kǫ . In other words, there exists an N ∈ N such that n ≥ N implies |fn (x) − f (x)|
0. Prove that there exists a measurable set F with finite measure such that Z |f | dm < ǫ. sup |f (x)| < ∞ and Fc

x∈F

Proof. For each n ∈ N, let Fn = {x ∈ E | n1 ≤ |f (x)| ≤ n} ⊆ E. Now we deduce from Problem 14.5 that Z Z n 1 o |f | dm |f | dm ≤ n m(Fn ) ≤ m x ∈ E |f (x)| ≥ ≤n n E Fn

which reduces to

m(Fn ) ≤ n

Z

E

|f | dm < ∞

for every n = 1, 2, . . .. Thus for each fixed positive integer n, if we set F = Fn , then we have m(F ) < ∞ and

sup |f (x)| ≤ n < ∞.

x∈F

To find the estimate of the integral, we let fn = f χFnc : E → R for each n = 1, 2, . . .. Then we have |fn | ≤ |f | on E. By the definition, we see that n o Fnc = x ∈ E 0 ≤ |f (x)| < n1 or |f (x)| > n .

Recall from Problem 14.6 that f is finite a.e. on E, so we conclude that fn (x) → 0 pointwise a.e. on E. Consequently, the Lebesgue’s Dominated Convergence Theorem is applicable and we get Z Z lim

n→∞ E

fn dm =

lim fn dm = 0.

E n→∞

139

14.4. Applications of Convergence Theorems

Therefore, there exists an N ∈ N such that Z

fN dm < ǫ.

(14.72)

E

Finally, we know from the definition of fN that the integral (14.72) is equal to Z |f | dm. c FN

We have completed the proof of the problem.



Chapter 14. Lebesgue Integration

140

CHAPTER

15

Differential Calculus of Functions of Several Variables

15.1 Fundamental Concepts The definition and basic properties of derivatives of f : R → R have been introduced and discussed in [25, Chap. 8]. This chapter extends the differential calculus theory to functions from Rn to Rm and the main references for this chapter are [2, Chap. 12], [18, Chap. 9], [22, Chap. 6 ] and [27, Chap. 7 & 8]. Here we assume that you are familiar with elementary linear algebra which can be found, for examples, in [10] or [13].

15.1.1 The Directional Derivatives and the Partial Derivatives Let S be a subset of Rn , x be an interior point of S, v be a point of S and f : S → Rm be a function. Definition 15.1 (Directional Derivatives). The directional derivative of f at x in the direction v, denoted by fv′ (x) or ∇v f (x), is defined by the limit f (x + hv) − f (x) h→0 h

fv′ (x) = lim whenever the limit exists.a

Let {e1 , e2 , . . . , en } and {u1 , u2 , . . . , um } be the standard bases of Rn and Rm respectively, xj be the jth component of x with respect to the standard basis {e1 , e2 , . . . , en } and f = (f1 , f2 , . . . , fm ). Then we have f (x) =

m X

fi (x)ui

and

i=1

a

fi (x) = f (x) · ui .

Since x ∈ S ◦ , it is true that x + hv ∈ S for all small h > 0 so that fv′ (x) is well-defined.

141

Chapter 15. Differential Calculus of Functions of Several Variables

142

Besides, it is clear from Definition 15.1 (Directional Derivatives) that fv′ (x) exists if and only if all (fi )′v (x) exist, where i = 1, 2, . . . , m. In this case, we have fv′ (x) = ((f1 )′v (x), (f2 )′v (x), . . . , (fm )′v (x)). Definition 15.2 (Partial Derivatives). For 1 ≤ i ≤ m and 1 ≤ j ≤ n, we define fi (x + hej ) − fi (x) h→0 h

(Dj fi )(x) = lim

whenever the limit exists. Then each (Dj fi )(x) is called the partial derivative of fi with respect to xj .b Remark 15.1 It is very easy to check from the above two definitions that the existence of fv′ (x) for all v ∈ Rn implies the existence of (Dj fi )(x), where 1 ≤ i ≤ m and 1 ≤ j ≤ n. Particularly, if v = ej , then we have fe′ j (x) = ((Dj f1 )(x), (Dj f2 )(x), . . . , (Dj fm )(x)), where 1 ≤ j ≤ n. However, the converse is false, see Problem 15.2 below.

15.1.2 Differentiation of Functions of Several Variables Definition 15.3. The function f : S ⊆ Rn → Rm is said to be differentiable at x ∈ Rn if there exists a linear transformation Tx : Rn → Rm such that |f (x + h) − f (x) − Tx (h)| = 0. h→0 |h| lim

(15.1)

In this case, we write f ′ (x) = Tx . If f is differentiable at every point of S, then we say that f is differentiable in S.c The derivative f ′ (x) (or equivalently the transformation Tx ) is called the differential of f at x or the total derivative of f at x. Furthermore, the limit (15.1) is equivalent to the form f (x + h) − f (x) = f ′ (x)h + R(h),

(15.2)

where the remainder R(h) satisfies |R(h)| = 0. h→0 |h| lim

(15.3)

Here we have some basic properties of the total derivative f ′ (x). Theorem 15.4 (Uniqueness of the Total Derivative). Suppose that both the linear transformations Tx and Tbx satisfy Definition 15.3. Then we have Tx = Tbx . ∂fi . ∂xj c Reader should pay attention that the h is a vector, not a scalar.

b

Another notation for Dj fi is

143

15.1. Fundamental Concepts

Theorem 15.5. If f is differentiable at p ∈ S ⊆ Rn , then f is continuous at p. Remark 15.2 A consequence of Theorem 15.5 is that directional differentiability does not imply total differentiability, see Problem 15.3. However, they are connected in some way. In fact, Problem 15.4 ensures that the existence of f ′ (x) implies that of fv′ (x) for every v ∈ Rn . Theorem 15.6 (The Chain Rule). Suppose that S is open in Rn , f : S → Rm is differentiable at a and g : f (S) ⊆ Rm → Rk is differentiable at f (a). Then the mapping F = g ◦ f : S → Rk is differentiable at a and F′ (a) = g′ (f (a)) × f ′ (a). (15.4) We note that on the right-hand side of the equation (15.4), we have the product of two linear transformations.

15.1.3 The Total Derivatives and the Jacobian Matrices Theorem 15.7. Let {e1 , e2 , . . . , en } and {u1 , u2 , . . . , um } denote the standard bases of Rn and Rm respectively. Suppose that S is open in Rn and f : S ⊆ Rn → Rm is differentiable at x. Then all the partial derivatives (Dj fi )(x) exist, where 1 ≤ i ≤ m and 1 ≤ j ≤ n, and furthermore the total derivative f ′ (x) satisfies m X ′ (Dj fi )(x)ui , (15.5) f (x)ej = i=1

where 1 ≤ j ≤ n.

This theorem says that, once we know that f ′ (x) exists, we can express it in terms of the partial derivatives (Dj fi )(x). In addition, we recall from Definition 15.3 that f ′ (x) is in fact a linear transformation Tx : Rn → Rm , so elementary linear algebra tells us that there is a (unique) matrix associated with f ′ (x). This matrix is called the Jacobian matrix of f at x and it is denoted by Jf (x). We see from the forms (15.5) that Jf (x) is related to the partial derivatives (Dj fi )(x) as follows:   (D1 f1 )(x) (D2 f1 )(x) · · · (Dn f1 )(x)  (D1 f2 )(x) (D2 f2 )(x) · · · (Dn f2 )(x)    Jf (x) =  . (15.6)  .. .. .. ..   . . . . (D1 fm )(x) (D2 fm )(x) · · · (Dn fm )(x)

m×n

In terms of the Jacobian matrices, the matrix form of Theorem 15.6 (The Chain Rule) can be written as JF (x) = Jg (f (x)) × Jf (x), (15.7) where this time the product on the right-hand side of the equation (15.7) refers to matrix product. In addition, if m = 1, then f = f is real-valued so the formula (15.5) gives f ′ (x)v = v1 f ′ (x)e1 + v2 f ′ (x)e2 + . . . + vn f ′ (x)en

Chapter 15. Differential Calculus of Functions of Several Variables

144

= v1 (D1 f )(x) + v2 (D2 f )(x) + · · · + vn (Dn f )(x)

= ∇f (x) · v, where

∇f (x) = ((D1 f )(x), (D2 f )(x), . . . , (Dn f )(x))

(15.8)

is called the gradient of f at x.

15.1.4 The Mean Value Theorem for Differentiable Functions The Mean Value Theorem for Differentiable Functions. Suppose that S is open in Rn and f : S → Rm is differentiable in S. If λx + (1 − λ)y ∈ S, where 0 ≤ λ ≤ 1, then for every a ∈ Rm , there exists a point z lying on the line segment of x and y such that a · [f (y) − f (x)] = a · [f ′ (z)(y − x)]. Theorem 15.8. Suppose that the open set S in the previous theorem is convex and there is a constant M such that sup {|f ′ (x)y|} ≤ M |y|≤1

for every x ∈ S.d Then we have |f (b) − f (a)| ≤ M |b − a| for every a, b ∈ S. Theorem 15.9. Let S ⊆ Rn be open and convex. If f : S → Rm is differentiable in S and f ′ (x) = 0 for all x ∈ S, then f is a constant on S.

15.1.5 Continuously Differentiable Functions Definition 15.10 (Continuously Differentiable Functions). Denote L(Rn , Rm ) to be the set of all linear transformations of Rn into Rm . Let S ⊆ Rn be open and f : S → Rm be differentiable. We say that f is continuously differentiable in S if f ′ : S → L(Rn , Rm ) is a continuous mapping. The class of all continuously differentiable functions in S is denoted by C ′ (S). Now the following theorem tells us a necessary and sufficient condition for f ∈ C ′ (S). Theorem 15.11. Suppose that S is open in Rn . Then f : S → Rm belongs to C ′ (S) if and only if all Dj fi exist and are continuous on S for all 1 ≤ i ≤ m and 1 ≤ j ≤ n. d

Note that f ′ (x)y is a vector in Rm .

145

15.1. Fundamental Concepts

15.1.6 The Inverse Function Theorem and the Implicit Function Theorem The Inverse Function Theorem. Let E be an open subset of Rn and f : E → Rn be a continuously differentiable function on E. Let a ∈ E be a point such that the linear transformation f ′ (a) is invertible, i.e., det Jf (a) 6= 0. Then there exists an open set U ⊆ E containing a and an open set V ⊆ Rn containing b = f (a) such that f : U → V is a bijection. Furthermore, the inverse mapping f −1 : V → U is differentiable at x and 1 (f −1 )′ (y) = ′ f (x) for every x ∈ U and y = f (x). See Figure 15.1 for the idea of this theorem.

Figure 15.1: The Inverse Function Theorem.

The Implicit Function Theorem. Let E be an open subset of Rn+m and f : E → Rm be a continuously differentiable function on E. We write (x, y) = (x1 , x2 , . . . , xn , y1 , y2 , . . . , ym ). Let (a, b) ∈ E be a point such that f (a, b) = 0, where a ∈ Rn and b ∈ Rm . Suppose that   ∂f1 ∂f1 ∂f1 (a, b) (a, b) · · · (a, b)   ∂y1 ∂y2 ∂ym       ∂f ∂f2 ∂f2   2 (a, b) (a, b) · · · (a, b)     ∂y1 ∂y2 ∂ym  6= 0. det      . . . ..   .. .. .. .         ∂fm ∂fm ∂fm (a, b) (a, b) · · · (a, b) ∂y1 ∂y2 ∂ym m×m

Then there exists an open set U ⊆ Rn containing a such that there exists a unique continuously differentiable function g : U → Rm satisfying g(a) = b

and

f (x, g(x)) = 0

Chapter 15. Differential Calculus of Functions of Several Variables

146

for all x ∈ U . Furthermore, we have  ∂f 1 (x, g(x))  ∂y1    .. ′ g (x) = −  .     ∂f m (x, g(x)) ∂y1  ∂f1 (x, g(x))  ∂x1    .. × .     ∂f m (x, g(x)) ∂x1

−1 ∂f1 (x, g(x))  ∂ym    .. ..  . .     ∂fm ··· (x, g(x)) ∂ym m×m  ∂f1 ··· (x, g(x))  ∂xn    .. ..  . .     ∂fm ··· (x, g(x)) ∂xn m×n ···

for all x ∈ U . Remark 15.3 Classically, mathematicians prove first the Inverse Function Theorem and then derive from it the Implicit Function Theorem. In addition, the usual ingredients of proofs of the two theorems consist of compactness of balls in Rn and the Contraction Principle (see [18, Theorem 9.23, p. 220] or [22, Theorem 6.6.4, p. 150]). In 2013, Oliveira [14] employed Dini’s approach to prove first the Implicit Function Theorem by induction and then he derived from it the Inverse Function Theorem. The tools he used are the Intermediate Value Theorem and the Mean Value Theorem for Derivatives. For an extensive treatment of the Implicit Function Theorem, we refer the reader to the book [8].

15.1.7 Higher Order Derivatives The partial derivatives D1 f , D2 f , . . . , Dn f of a function f : Rn → Rm are also functions from to Rm . If the partial derivatives Dj f are differentiable, then they are called the secondorder partial derivatives of f which are denoted by

Rn

Dij f = Di (Dj f ) =

∂2f ∂  ∂f  , = ∂xi ∂xj ∂xi ∂xj

where 1 ≤ i, j ≤ n.

Theorem 15.12 (Clairaut’s Theorem). If both partial derivatives Di f and Dj f exist in an open set S ⊆ Rn and both Dij f and Dji f are continuous at p ∈ S, then we have (Dij f )(p) = (Dji f )(p).

147

15.2. Differentiation of Functions of Several Variables

15.2 Differentiation of Functions of Several Variables Problem 15.1 ⋆ Consider the function f : R2 → R defined by f (x, y) =

  1, if xy = 0; 

0, otherwise.

Show that (D1 f )(0, 0) = (D2 f )(0, 0), but f is discontinuous at (0, 0). Proof. We note that 1−1 f (h, 0) − f (0, 0) = lim =0 h→0 h h→0 h

(D1 f )(0, 0) = lim and

1−1 f (0, h) − f (0, 0) = lim = 0. h→0 h→0 h h Thus (D1 f )(0, 0) = (D2 f )(0, 0). However, we notice that (D2 f )(0, 0) = lim

lim f (x, x) = 0 6= 1 = f (0, 0)

x→0

which means that f is discontinuous at (0, 0). This completes the proof of the problem.



Remark 15.4 Problem 15.1 asserts that the existence of the partial derivatives D1 f, D2 f, . . . , Dn f at p does not necessarily imply the continuity of f at p. Fortunately, this really happens when all partial derivatives are bounded in an open set E ⊆ Rn , see [18, Exercise 7, p. 239]. Problem 15.2 ⋆ Consider f : R2 → R which is given by f (x, y) =

  −x − y, if xy = 0; 

1,

otherwise.

Prove that both (D1 f )(0) and (D2 f )(0) exist, but not fv (0) for any v = (a, b) with ab 6= 0. Proof. By Definition 15.2 (Partial Derivatives), we have −h f (h, 0) − f (0, 0) = lim = −1 h→0 h h→0 h

(D1 f )(0) = lim

Chapter 15. Differential Calculus of Functions of Several Variables

148

and f (0, h) − f (0, 0) −h = lim = −1. h→0 h→0 h h

(D2 f )(0) = lim

However, if v = (a, b), where ab 6= 0, then a 6= 0 and b 6= 0. Therefore, we obtain from Definition 15.1 (Directional Derivatives) that 1 f (ha, hb) − f (0, 0) = lim h→0 h h→0 h

fv′ (0) = lim

which does not exist. This completes the proof of the problem.



Problem 15.3 ⋆ Prove that the function f : R2 → R defined by f (x, y) =

      

xy 2 , if x 6= 0; x2 + y 6 0,

otherwise.

has well-defined fv′ (0) for every v but f is discontinuous at 0.

Proof. Suppose that v = (a, b). By Definition 15.1 (Directional Derivatives), we have f (0 + hv) − f (0) h→0 h f (ha, hb) = lim h→0 h ha · h2 b2 = lim h→0 h(h2 a2 + h6 b6 ) ab2 = lim 2 h→0 a + h4 b6   b2   , if a 6= 0; a =    0, otherwise.

fv′ (0) = lim

Thus fv′ (0) is well-defined for every v.

To show that f is discontinuous at 0, we first note that f (0, 0) = 0. Next, if y 6= 0, then we have 1 y4 = f (y 2 , y) = 4 y + y6 1 + y2 which implies that lim f (y 2 , y) = 1 6= f (0, 0).

y→0

Hence f is discontinuous at 0 and we complete the proof of the problem.



149

15.2. Differentiation of Functions of Several Variables

Problem 15.4 ⋆ Suppose that x is an interior point of S ⊆ Rn , f : S ⊆ Rn → Rm is a function and v ∈ Rn . If f is differentiable at x, prove that fv′ (x) = f ′ (x)v.

(15.9)

Proof. Fix v and consider h = tv for some sufficiently small real t. Since f is differentiable at x, it follows from the approximation (15.2) that f (x + tv) − f (x) − tf ′ (x)v = f (x + tv) − f (x) − f ′ (x)(tv) = f (x + h) − f (x) − f ′ (x)h

= R(h)

= R(tv). By the limit (15.3), we see that lim

t→0

|R(tv)| = 0. |t|

Thus we get f (x + tv) − f (x) − tf ′ (x)v R(tv) = lim = 0, t→0 t→0 t t

fv′ (x) − f ′ (x)v = lim

i.e., fv′ (x) = f ′ (x)v. This completes the proof of the problem.



Remark 15.5 We notice that the left-hand side of the formula (15.9) is a limit, while its right-hand side is a matrix product. In other words, this gives an easier way (matrix product) to compute the directional derivative fv′ (x) whenever f is differentiable at x.

Problem 15.5 ⋆ Suppose that fv′ (x) and gv′ (x) exist. Prove that (f + g)′v (x) = fv′ (x) + gv′ (x).

Proof. By Definition 15.1 (Directional Derivatives), we have (f + g)(x + hv) − (f + g)(x) h g(x + hv) − g(x) f (x + hv) − f (x) + lim = lim h→0 h→0 h h ′ ′ = fv (x) + gv (x),

(f + g)′v (x) = lim

h→0

completing the proof of the problem.



Chapter 15. Differential Calculus of Functions of Several Variables

150

Problem 15.6 ⋆ Prove Theorem 15.5.

Proof. Since f ′ (p) = Tp , we rewrite the formula (15.2) as f (p + h) − f (p) = Tp (h) + R(h).

(15.10)

Let h = h1 e1 + h2 e2 + · · · + hn en . Then h → 0 if and only if hi → 0 for all i = 1, 2, . . . , n. Since Tp is linear, we have Tp (h) = h1 Tp (e1 ) + h2 Tp (e2 ) + · · · + hn Tp (en ) so that Tp (h) → 0 as h → 0. By the limit (15.3), we see that R(h) → 0 as h → 0. Hence we conclude from the formula (15.10) that lim f (p + h) = f (p),

h→0

i.e., f is continuous at p. This completes the proof of the problem.



Problem 15.7 ⋆ Prove Theorem 15.7.

Proof. By Problem 15.4 and then Remark 15.1, we obtain f ′ (x)ej = fe′ j (x) = ((Dj f1 )(x), (Dj f2 )(x), . . . , (Dj fm )(x)) =

m X (Dj fi )(x)ui . i=1

We have completed the proof of the problem.



Problem 15.8 ⋆ ⋆ Let S ⊆ Rn and x be an interior point of S. Suppose that f : S → Rm is given by f = (f1 , f2 , . . . , fm ). Use Definition 15.3 to prove that f is differentiable at x if and only if every fi : S → R is differentiable at x. Proof. We notice that for any y = (y1 , y2 , . . . , ym ) ∈ Rm , we have |yi | ≤ |y| ≤

√

m max |yi |

(15.11)

1≤i≤m

for all i = 1, 2, . . . , m. Now the Jacobian matrices for the component functions fi , where i = 1, 2, . . . , m, at x are given by Jfi (x) =

(D1 fi )(x) (D2 fi )(x) · · · (Dn fi )(x)




.

151

15.2. Differentiation of Functions of Several Variables

If we compare these with the matrix (15.6), then we assert that 

  Jf (x) =  

Jf1 (x) Jf2 (x) .. . Jfm (x)

    

or



  f ′ (x) =  

f1′ (x) f2′ (x) .. . ′ (x) fm



  . 

Therefore, the ith component function of f (x + h) − f (x) − f ′ (x)h is exactly fi (x + h) − fi (x) − fi′ (x)h and then we follow from the inequalities (15.11) that |f (x + h) − f (x) − f ′ (x)h| |fi (x + h) − fi (x) − fi′ (x)h| ≤ |h| |h| √ |fi (x + h) − fi (x) − fi′ (x)h| . ≤ m× |h|

(15.12)

Hence, by taking h → 0 to the inequalities (15.12), we get the desired result. This completes  the proof of the problem. Problem 15.9 ⋆ Let T : Rn → Rm be linear. Prove that T is differentiable everywhere on Rn and T ′ (x) = T .

Proof. Since T is linear, we have T (x + h) − T (x) − T (h) = 0 which is in the form (15.2). Next Theorem 15.4 (Uniqueness of the Total Derivative) implies that T ′ (x) = T which ends the proof of the problem.



Problem 15.10 ⋆ Use Theorem 15.6 (The Chain Rule) and Problem 15.9 to prove Problem 15.8 again.

Proof. Suppose that f is differentiable at x. Let πi : Rm → R be the projection πi (y) = πi (y1 , y2 , . . . , ym ) = yi , where 1 ≤ i ≤ m. Since each πi is linear, every πi is differentiable everywhere on Rm by Problem 15.9. Obviously, we have fi = πi ◦ f , so Theorem 15.6 (The Chain Rule) implies that fi is also differentiable at x.

Chapter 15. Differential Calculus of Functions of Several Variables

152

Conversely, we suppose that every fi is differentiable at x ∈ S. Define φi : R → Rm by φi (y) = (0, . . . , 0, y, 0, . . . , 0), where the y is the ith coordinate. It is obvious that each φi is linear so that it is differentiable everywhere on R by Problem 15.9. By Theorem 15.6 (The Chain Rule), the composite function φi ◦ fi : S → Rm is differentiable at x. Since f = (f1 , f2 , . . . , fm ) =

m X i=1

φi ◦ fi ,

we have the desired result that f is differentiable at x. This completes the proof of the problem.  Problem 15.11 ⋆ Suppose that f , g : Rn → Rm are functions such that f is differentiable at p, f (p) = 0 and g is continuous at p. If h(x) = g(x) · f (x), prove that h′ (p)v = g(p) · [f ′ (p)v],

(15.13)

where v ∈ Rn . Proof. Since f is differentiable at p and f (p) = 0, we get from the formula (15.2) that f (p + h) = f ′ (p)h + R(h) which gives h(p + h) − h(p) = g(p + h) · f (p + h) − g(p) · f (p) = g(p + h) · [f ′ (p)h + R(h)]

= g(p + h) · [f ′ (p)h + R(h)] + g(p) · f ′ (p)h − g(p) · f ′ (p)h

= g(p) · f ′ (p)h + g(p + h) · R(h) + [g(p + h) − g(p)] · f ′ (p)h.

Since g is continuous at p, the limit (15.3) implies |R(h)| |g(p + h) · R(h)| = |g(p)| lim = 0. h→0 h→0 |h| |h| lim

Furthermore, we see that  h [g(p + h) − g(p)] · f ′ (p)h = lim [g(p + h) − g(p)] · f ′ (p) h→0 h→0 |h| |h| h = 0 · lim f ′ (p) h→0 |h| = 0. lim

Thus the equation (15.14) can be expressed as b h(p + h) − h(p) = g(p) · f ′ (p)h + R(h),

(15.14)

153

15.2. Differentiation of Functions of Several Variables

where

b |R(h)| = 0. h→0 |h| lim

By Definition 15.3, we have the expected result (15.13), completing the proof of the problem.  Problem 15.12 ⋆ Suppose that f : R3 → R2 is given by f (x, y, z) = (x2 − y + z, e−xy sin z + xz). Prove that f is differentiable at x ∈ R3 and find Jf (p). Proof. We notice that the component functions f1 and f2 of f are given by f1 (x, y, z) = x2 − y + z

and

f2 (x, y, z) = e−xy sin z + xz

so that D1 f1 = 2x,

D2 f1 = −1,

−xy

D1 f2 = −ye

sin z + z,

D3 f1 = 1, D2 f2 = −xe−xy sin z,

D3 f2 = e−xy cos z + x.

(15.15)

Since all Dj fi exist and are continuous on R3 , we deduce from Theorem 15.11 that f is differentiable in R3 . Furthermore, we gain immediately from the partial derivatives (15.15) that   2x −1 1 Jf (x) = . −ye−xy sin z + z −xe−xy sin z e−xy cos z + x This completes the analysis of the problem.



Problem 15.13 ⋆ Suppose that f, g, h : R2 → R are differentiable such that z = f (x, y), x = g(r, θ) and y = h(r, θ). Verify that ∂z ∂x ∂z ∂y ∂z = · + · ∂r ∂x ∂r ∂y ∂r

and

∂z ∂z ∂x ∂z ∂y = · + · . ∂θ ∂x ∂θ ∂y ∂θ

(15.16)

Proof. Let z = ψ(r, θ) = f (g(r, θ), h(r, θ)) and φ = (g, h). Since z = f (x, y), the matrix form (15.6) or the gradient form (15.8) shows that     ∂z ∂z ∂f ∂f = . Jf = ∂x ∂y ∂x ∂y Similarly, we have Jψ =



∂z ∂r

∂z ∂θ





∂x  ∂r  and Jφ =   ∂y ∂r

 ∂x ∂θ   . ∂y  ∂θ

Chapter 15. Differential Calculus of Functions of Several Variables

154

Since ψ = f ◦ φ, the matrix form (15.7) gives Jψ = Jf × Jφ or explicitly,   ∂x ∂x       ∂z ∂z ∂z ∂z  ∂r ∂θ  = ×  ∂x ∂y  ∂y ∂y  ∂r ∂θ ∂r ∂θ   ∂z ∂x ∂z ∂y ∂z ∂x ∂z ∂y · + · · + · = ∂x ∂r ∂y ∂r ∂x ∂θ ∂y ∂θ which is exactly the formulas (15.16), completing the analysis of the problem.



Problem 15.14 ⋆ Let p ∈ S ⊆ Rn . Prove that there is no function f : S → R such that fv′ (p) > 0 for every v ∈ Rn \ {0}. Proof. Assume that there was a function f : S → R such that fv′ (p) > 0 for every v ∈ Rn \ {0}. Let α ∈ R \ {0}. Then we notice from Definition 15.1 (Directional Derivatives) that ′ fαv (p) = lim

h→0

f (p + hαv) − f (p) f (p + hαv) − f (p) = α lim = αfv′ (p). αh→0 h αh

In particular, we have ′ f−v (p) = −fv′ (p) < 0

for every v ∈ Rn \ {0}, a contradiction. Hence we end the proof of the problem.



Problem 15.15 ⋆ Suppose that f, g, h, k : S ⊆ Rn → R are functions such that h = f g and k = ∇h and ∇k.

f g.

Find

Proof. By the definition (15.8), we have ∇h = (D1 h, D2 h, . . . , Dn h).

(15.17)

For 1 ≤ i ≤ n, since Di h = Di (f g) = (Di f )g + f (Di g), we obtain from the expression (15.17) that ∇h = ((D1 f )g + f (D1 g), (D2 f )g + f (D2 g), . . . , (Dn f )g + f (Dn g)) = (D1 f, D2 f, . . . , Dn f )g + f (D1 g, D2 g, . . . , Dn g)

= (∇f )g + f (∇g).

(15.18)

Write k = f g−1 , so the formula (15.18) gives ∇k = (∇f )g −1 + f [∇(g −1 )].

(15.19)

By the definition (15.8) again, we know that ∇(g−1 ) = (D1 (g−1 ), D2 (g−1 ), . . . , Dn (g−1 )) =

 −D g −D g −Dn g  ∇g 2 1 , , . . . , =− 2. 2 2 2 g g g g

(15.20)

155

15.3. The Mean Value Theorem for Differentiable Functions

If we substitute the expression (15.20) into the formula (15.19), then we assert that ∇k =

f ∇g 1 ∇f − 2 = 2 (g∇f − f ∇g). g g g

This completes the analysis of the problem.



Problem 15.16 ⋆ Suppose that f : R → R is differentiable in R and g : R3 → R is defined by g(x, y, z) = xk + y k + z k , where k ∈ N. Denote h = f ◦ g. Prove that |∇h(x, y, z)|2 = k2 [x2(k−1) + y 2(k−1) + z 2(k−1) ] × [f ′ (g(x, y, z))]2 .

(15.21)

Proof. We have h(x, y, z) = f (xk + y k + z k ), so Theorem 15.6 (The Chain Rule) implies D1 h = kxk−1 f ′ (xk + y k + z k ). Similarly, we have D2 h = ky k−1 f ′ (xk + y k + z k )

and D3 h = kz k−1 f ′ (xk + y k + z k ).

Thus it follows from the formula (15.8) that ∇h(x, y, z) = (kxk−1 f ′ (xk + y k + z k ), ky k−1 f ′ (xk + y k + z k ), kz k−1 f ′ (xk + y k + z k )) and then |∇h(x, y, z)|2 = k2 [x2(k−1) + y 2(k−1) + z 2(k−1) ] × [f ′ (xk + y k + z k )]2 = k2 [x2(k−1) + y 2(k−1) + z 2(k−1) ] × [f ′ (g(x, y, z)]2

which is the desired result (15.21). This ends the proof of the problem.



15.3 The Mean Value Theorem for Differentiable Functions Problem 15.17 ⋆

⋆ Prove the Mean Value Theorem for Differentiable Functions.

Proof. Fix a ∈ Rm . Define the function F : [0, 1] → R by F (t) = a · f (x + t(y − x)).

(15.22)

Chapter 15. Differential Calculus of Functions of Several Variables

156

The hypothesis shows that x + t(y − x) = (1 − t)x + ty ∈ S for all t ∈ [0, 1] so that the function F is well-defined. Since f is differentiable in S, Theorem 15.5 guarantees that f is continuous on S and then F must be continuous on [0, 1]. Furthermore, we apply Theorem 15.6 (The Chain Rule) to the definition (15.22) to get F ′ (t) = a · f ′ (x + t(y − x))(y − x). Now the Mean Value Theorem for Derivatives [25, p. 129] implies the existence of a ξ ∈ (0, 1) such that F (1) − F (0) = F ′ (ξ) = a · [f ′ (x + ξ(y − x))(y − x)]. (15.23)

We know that F (1) = a · f (y) and F (0) = a · f (x), so if we let z = x + ξ(y − x) ∈ S, then the formula (15.23) can be rewritten as a · [f (y) − f (x)] = a · [f ′ (z)(y − x)]. This completes the proof of the problem.



Problem 15.18 ⋆ Let S be open and convex in Rn and f : S → R. Let f be differentiable in S and a, a + h ∈ S. Prove that there exists a λ ∈ (0, 1) such that f (a + h) − f (a) = f ′ (a + λh)h. Proof. Let c 6= 0. Since S is convex, we have a + λh = λ(a + h) + (1 − λ)a ∈ S, where 0 ≤ λ ≤ 1. By the Mean Value Theorem for Differentiable Functions, we get c[f (a + h) − f (a)] = c[f ′ (z)(ah − a)] = c[f ′ (z)h]

(15.24)

for some z = a + λh. Hence our result follows from dividing the expression (15.24) by the nonzero constant c and we have completed the proof of the problem.  Problem 15.19 ⋆ Prove Theorem 15.9.

Proof. Since f ′ (x) = 0 on S, we have |f ′ (x)y| = 0 for every |y| ≤ 1. Thus we may take M = 0 in Theorem 15.8 to conclude that |f (b) − f (a)| = 0 for all a, b ∈ S. This means that f is constant on S, completing the proof of the problem.



Problem 15.20 ⋆ ⋆ Let a ∈ S ⊆ Rn and Br (a) = {x ∈ S | |x − a| < r}, where r > 0. Suppose that f : S → R satisfies fv′ (x) = 0 for every x ∈ Br (a) and every v ∈ Rn . Prove that f is constant on Br (a).

157

15.4. The Inverse Function Theorem and the Implicit Function Theorem

Proof. We first claim that if the directional derivative fv′ (a + tv) exists for each t ∈ [0, 1], then there exists a λ ∈ (0, 1) such that f (a + v) − f (a) = fv′ (a + λv).

(15.25)

To this end, we define g : [0, 1] → R by g(t) = f (a + tv).

(15.26)

By Definition 15.1 (Directional Derivatives), we see that g(t + h) − g(t) f (a + tv + hv) − f (a + tv) = lim = fv′ (a + tv), h→0 h→0 h h

g′ (t) = lim

(15.27)

where t ∈ [0, 1]. Therefore, the Mean Value Theorem for Derivatives shows that there exists a λ ∈ (0, 1) such that g(1) − g(0) = g′ (λ)(1 − 0) = g ′ (λ). (15.28) By the definition (15.26), we have g(1) = f (a + v) and g(0) = f (a). Hence our desired result (15.25) follows directly from the comparison of the expressions (15.27) and (15.28). Next, for every x ∈ Br (a), let v = x − a so that |v| < r. Since |a + λv − a| = λ|v| < λr < r, the hypothesis shows that fv′ (a + λv) = 0 and then f (x) = f (a + v) = f (a). In other words, f is constant on Br (a) which completes the proof of the problem.



15.4 The Inverse Function Theorem and the Implicit Function Theorem Problem 15.21 ⋆ Suppose that E = {(x, y) ∈ R2 | x > y}. Define f : E → R2 by f (x, y) = (f1 (x, y), f2 (x, y)) = (x + y, x2 + y 2 ). Prove that f is locally bijective. Proof. Since D1 f1 = 1, D2 f1 = 1, D1 f2 = 2x and D2 f2 = 2y are all continuous on E, Theorem 15.11 implies that f ∈ C ′ (E). Furthermore, we have   1 1 Jf (x, y) = . 2x 2y Since x > y if (x, y) ∈ E, it is true that det Jf (x, y) = 2(y − x) 6= 0. Hence it deduces from the Inverse Function Theorem that f is locally bijective. This completes the analysis of the  problem.

Chapter 15. Differential Calculus of Functions of Several Variables

158

Problem 15.22 ⋆ Suppose that f : Br (a) → Rn is differentiable at a and its inverse f −1 exists and is differentiable at f (a). Prove that det Jf (a) 6= 0. Proof. Assume that det Jf (a) = 0. By Problem 15.9 and Theorem 15.6 (The Chain Rule), we can establish In = (f −1 ◦ f )(a) = (f −1 ◦ f )′ (a) = (f −1 )′ (f (a)) × f ′ (a) so that 1 = det In = det Jf −1 (f (a)) × det Jf (a) = 0, a contradiction. Hence we must have det Jf (a) 6= 0 and we complete the proof of the problem.  Remark 15.6 In other words, Problem 15.22 says that the hypothesis det Jf (a) 6= 0 cannot be omitted in the Inverse Function Theorem.

Problem 15.23 ⋆ Show, by a counterexample, that the hypothesis f ∈ C ′ (E) cannot be dropped in the Inverse Function Theorem.

Proof. We consider the function f : R → R defined by  1   x + 2x2 sin , if x 6= 0; x f (x) =   0, otherwise.

Obviously, f is differentiable in (−1, 1) and

 f (h) − f (0) 1 = 1 6= 0. = lim 1 + 2h sin h→0 h→0 h h

f ′ (0) = lim However, for x 6= 0, we have

1 1 − 2 cos x x so that lim f ′ (x) does not exist. In other words, f ∈ / C ′ ((−1, 1)). f ′ (x) = 1 + 4x sin

x→0

Let V be any neighborhood of 0. Then there exists an N ∈ N such that 2 2 clear that V also contains (4N −1)π and (4N +1)π because

2 (4N −3)π

∈ V . It is

2 2 2 < < . (4N + 1)π (4N − 1)π (4N − 3)π By direct computation, we see that       2 2 2 0 and a unique continuously function x : (−δ, δ) → R such that x(0) = 0 and

F (x(t), t) = 0

which mean that x(t) is a solution of the equation (15.31). Thus we complete the proof of the problem.  Problem 15.26 ⋆ Consider the system of equations x2 + 3y 2 + z 2 − w = 9,

x3 + 4y 2 + z + w2 = 22. Prove that z and w can be written as differentiable functions of x and y around (1, 1, 1, −4). Proof. We write f1 (x, y, z, w) = x2 + 3y 2 + z 2 − w − 9 and

f2 (x, y, z, w) = x3 + 4y 2 + z + w2 − 22.

Next, we define F : R4 → R2 by F (x, y, z, w) = (f1 (x, y, z, w), f2 (x, y, z, w)) = (x2 + 3y 2 + z 2 − w − 9, x3 + 4y 2 + z + w2 − 22). By direct computation, we have F (1, 1, 1, −4) = (0, 0) and D1 f1 = 2x, 2

D1 f2 = 3x ,

D2 f1 = 6y,

D3 f1 = 2z,

D2 f2 = 8y,

D3 f2 = 1,

D4 f1 = −1,

D4 f2 = 2w.

Since all the partial derivatives exist and continuous on R4 , we deduce from Theorem 15.11 that F ∈ C ′ (R4 ). Besides, we know that     D3 f1 (1, 1, 1, −4) D4 f1 (1, 1, 1, −4) 2 −1 det = det = −15 6= 0. D3 f2 (1, 1, 1, −4) D4 f2 (1, 1, 1, −4) 1 −8 Hence it follows from the Implicit Function Theorem that z and w can be written as differentiable functions of x and y around (1, 1, 1, −4). This completes the proof of the problem.  Problem 15.27 ⋆ Show that there exist functions f, g : R4 → R which are functions of x, y, z, w, continuously differentiable in Bδ (2, 1, 1, −2) for some δ > 0 such that f (2, 1, 1, −2) = 4, g(2, 1, 1, −2) = 3 and the equations f 2 + g2 + w2 = 29 hold on Bδ (2, 1, 1, −2).

and

w2 f 2 g2 + + = 17 x2 y2 z2

161

15.5. Higher Order Derivatives

Proof. Let F1 (f, g, x, y, z, w) = f 2 + g2 + w2 − 29 and F2 (f, g, x, y, z, w) = Define F : R6 → R2 by

f2 x2

+

g2 y2

+

w2 z2

− 17.

F(f, g, x, y, z, w) = (F1 (f, g, x, y, z, w), F2 (f, g, x, y, z, w))   f 2 g 2 w2 = f 2 + g2 + w2 − 29, 2 + 2 + 2 − 17 . x y z It is clear that F(4, 3, 2, 1, 1, −2) = (0, 0). Furthermore, we have D1 F1 = 2f,

D2 F1 = 2g,

D3 F1 = 0,

D4 F1 = 0, 2f 2

2g 2f , D2 F2 = 2 , D3 F2 = − 3 , 2 x y x 2w2 2w D5 F2 = − 3 , D6 F2 = 2 . z z D1 F2 =

D5 F1 = 0,

D4 F2 = −

2g2

D6 F1 = 2w,

,

y3

If we take δ = 21 , then any point (x, y, z, w) in B 1 (2, 1, 1, −2) satisfy xyzw 6= 0. Otherwise, 2 assume for example that y = 0, then we obtain (x − 2)2 + (0 − 1)2 + (z − 1)2 + (w + 2)2
0. Since Dij f and Dji f are continuous at 0, there exists a δ > 0 such that |(Dij f )(x) − A|
0, there exists a countable collection of n-dimensional intervals I1 , I2 , . . . such that ∞ ∞ X [ µ(Ik ) < ǫ. Ik and S⊆ k=1

k=1

Theorem 16.10 (The Lebesgue’s Integrability Condition). Suppose that I is an interval in Rn and f : I → R is a bounded function. Then we have f ∈ R on I if and only if the set of discontinuities of f in I has n-measure zero. Particularly, we have f ∈ R on I for every continuous function f : I → R. Remark 16.2 (a) In particular, if µ(Sk ) = 0 for k = 1, 2, . . . and S ⊆ S1 ∪S2 ∪· · · , then we have µ(S) = 0. (b) It is easy to see from Definition 16.7 (Lower and Upper Riemann Integrals) that if f (x) = 0 on I, then L(P, f ) = U (P, f ) = 0 for every partition P of I. Hence Theorem 16.8 (The Riemann Integrability Condition) implies that f ∈ R on I and Z Z (16.6) f dx = 0 dx = 0. I

I

(c) We suggest the reader to compare our Theorems 16.8 (The Riemann Integrability Condition) and 16.10 (The Lebesgue’s Integrability Condition) with [25, Theorems 9.2 and 9.4, p. 159] respectively.

16.1.4 Jordan Measurable Sets in Rn Roughly speaking, the Jordan measure (or Jordan content) in Rn is an extension of the notion of size (length, area and volume) to more complicated shapes other than triangles or disks. Definition 16.11 (Jordan Measure). Suppose that E ⊆ I, where I is an interval in the form (16.1). We define the outer Jordan measure of E to be J ∗ (E) = inf

N X k=1

µ(Ik ),

171

16.1. Fundamental Concepts

where the infimum runs through over all finite coverings E ⊆

N [

the inner Jordan measure of E is defined to be J∗ (E) = sup

N X

Ik by intervals in Rn . Similarly,

k=1

µ(Ik ),

k=1

where the supremum runs through over all unions of finite intervals in Rn which are contained in E. In the case that J ∗ (E) = J∗ (E), the set E is said to be Jordan measurable and we denote this common value by J(E) which is called the Jordan measure or Jordan content of E. For examples, every finite subset of Rn is Jordan measurable and all open and closed balls in are Jordan measurable too. In Figure 16.2, the sum of the areas of the orange rectangles is an inside approximation of the area of the Jordan measurable set in R2 . Similarly, the sum of the areas of the green rectangles is an outside approximation of the area of the Jordan measurable set in R2 . Rn

Figure 16.2: The outer and the inner Jordan measures.

Remark 16.3 Readers should compare the similarities between Definition 16.11 (Jordan Measure) and Definition 12.2 (Lebesgue Outer Measure).

Definition 16.12 (Boundary Points). We call that a point x ∈ E ⊆ Rn is a boundary point of E if Nr (x) ∩ E 6= ∅ and Nr (x) ∩ (Rn \ E) 6= ∅

Chapter 16. Integral Calculus of Functions of Several Variables

172

for every r > 0. The set of all boundary points of E is called the boundary of E and it is denoted by ∂E. Theorem 16.13. Suppose that E is a bounded set of Rn . Then we have J ∗ (∂E) = J ∗ (E) − J∗ (E). In addition, E is Jordan measurable if and only if J(∂E) = 0. It is easy to see that if J(E) = 0, then µ(E) = 0. However, the converse is not true. See Problem 16.2 for proofs of these. The following theorem tells us some basic properties of Jordan measurable sets. Theorem 16.14 (Properties of Jordan Measurable Sets). Suppose that E and F are Jordan measurable. Then E ∪ F , E ∩ F and E \ F are also Jordan measurable. Furthermore, we have (a) J(E ∪ F ) = J(E) + J(F ) − J(E ∩ F ). (b) if E ∩ F = ∅, then J(E ∪ F ) = J(E) + J(F ). (c) if F ⊆ E, then J(E \ F ) = J(E) − J(F ).

16.1.5 Integration on Jordan Measurable Sets Definition 16.15 (Riemann Integrable on E). Suppose that E is a Jordan measurable set of Rn , f is bounded on E and I is an interval in the form (16.1) containing E. Define g : I → R by   f (x), if x ∈ E; g(x) =  0, if x ∈ I \ E. We say that f ∈ R on E if and only if g ∈ R on I and in this case, we have Z Z f dx = g dx. E

I

Theorem 16.16. Suppose that E is a Jordan measurable set of Rn . Then we have f ∈ R on E if and only if the set of discontinuous points of f in E has n-measure 0. In particular, f ∈ R on E for every continuous function f : E → R. Theorem 16.17 (Properties of Integration on E). Suppose that E is a Jordan measurable set in Rn , f, g : E → R and α ∈ R. (a) If f, g ∈ R on E, then so are f + g and αf . In fact, we have Z Z Z g dx f dx + (f + g) dx = and

E

E

E

Z

E

αf dx = α

Z

f dx. E

173

16.1. Fundamental Concepts

(b) If f, g ∈ R on E and f (x) ≤ g(x) for all x ∈ E, then we have Z Z g dx. f dx ≤ E

E

(c) If A and B are Jordan measurable sets satisfying A ∩ B = ∅ and E = A ∪ B and f ∈ R on E, then we have f ∈ R on A and on B and furthermore, Z Z Z f dx. f dx + f dx = B

A

E

The following famous theorem suggests a practical way to evaluate double integrals on a rectangle in terms of singles integrals. For multiple integrals on an interval in Rn , please read Problem 16.11. Fubini’s Theorem. Suppose that R = [a, b]×[c, d] and f : R → R is well-defined. Furthermore, suppose that for each x ∈ [a, b], g(y) = f (x, ·) ∈ R on [c, d] and for each y ∈ [c, d], h(x) = f (·, y) ∈ R on [a, b]. If f ∈ R on R, then we have Z d Z b Z bZ d ZZ Z   f (x, y) dx dy. f (x, y) dy dx = f (x, y) dx dy = f (x, y) d(x, y) = R

c

c

a

R

a

For evaluation of integrals on a general Jordan measurable set in terms of iterated integrals, we consider the concept of a projectable region first. We call a set E ⊆ Rn a projectable region if there exists a closed Jordan measurable set F ⊆ Rn−1 , k ∈ {1, 2, . . . , n} and continuous functions φ, ϕ : F ⊆ Rn−1 → R such that bk ∈ F and φ(b E = {x ∈ Rn | x xk ) ≤ xk ≤ ϕ(b xk )},

(16.7)

bk = (x1 , . . . , xk−1 , xk+1 , . . . , xn ). Then we have where x

Theorem 16.18. Suppose that E is a projectable region in Rn with k, F, φ and ϕ as given in the definition (16.7). Then E is a Jordan measurable set and if f ∈ R on E, then Z  Z ϕ(bxk ) Z  f (b xk ) dxk db xk . f dx = φ(b xk )

F

E

16.1.6 Two Important Theorems In [25, Chap. 9], we study the Mean Value Theorems for Integrals and the Change of Variables Theorem in the case of a single variable. Here we present their multi-dimensional versions as follows: The Mean Value Theorem for Multiple Integrals. Suppose that E is a Jordan measurable set of Rn . Let f, g ∈ R on E and g(x) ≥ 0 on E. Denote M = sup{f (x) | x ∈ E}

and

m = inf{f (x) | x ∈ E}.

Then there is a number λ ∈ [m, M ] such that Z Z g dx. f g dx = λ E

E

Particularly, we have

mJ(E) ≤

Z

E

f dx ≤ M J(E).

Chapter 16. Integral Calculus of Functions of Several Variables

174

The Change of Variables Theorem. Suppose that V is open in Rn , φ : V → Rn is continuously differentiable and injective on V . If det Jφ (x) 6= 0 for all x ∈ V , E is a Jordan measurable set with E ⊆ V , f ◦ φ ∈ R on E and f ∈ R on φ(E), then we have Z Z f (φ(x)) · | det Jφ (x)| dx. (16.8) f (y) dy = E

φ(E)

Recall that polar coordinates in R2 have the form x = r cos θ

and y = r sin θ,

where r is the distance between (x, y) and the origin and θ is the angle between the positive x-axis and the line connecting (x, y) and (0, 0). Set φ(r, θ) = (r cos θ, r sin θ) so that   cos θ −r sin θ det Jφ ((r, θ)) = det = r. sin θ r cos θ Obviously, φ is continuously differentiable on any open set V in R2 by Theorem 15.11. Furthermore, φ is injective and det Jφ (r, θ) 6= 0 on V if V does not intersect the set {(0, θ) | θ ∈ R}. Hence the formula (16.8) becomes Z Z f (r cos θ, r sin θ)r dr dθ. (16.9) f (x, y) dx dy = E

φ(E)

Remark 16.4 (a) There are two common, helpful and practical changes of variables in R3 . They are the cylindrical coordinates and the spherical coordinates which are given by x = r cos θ,

y = r sin θ,

z=z

and x = r sin ϕ cos θ,

y = r sin ϕ sin θ,

z = r cos θ

respectively. (b) In [24, p. 426], Wade points out that the formula (16.9) holds even though φ is not injective or det Jφ ((r, θ)) = 0 in the whole rθ-plane. Similar situations happen for the cylindrical coordinates and the spherical coordinates. (c) Wade calls this Change of Variables Theorem a global version which can be shown by a local version of the theorem, see [24, Lemma 12.45, p. 424].

16.2 Jordan Measurable Sets Problem 16.1 ⋆ Prove that J ∗ (E) = J ∗ (E) for every E ⊆ R.

175

16.2. Jordan Measurable Sets

Proof. Since E ⊆ E, we always have E⊆E⊆

N [

Ik

k=1

so that J ∗ (E) ≤ J ∗ (E). To prove the reverse direction, it suffices to show that if E ⊆ then E ⊆

N [

Ik . To this end, let I =

k=1

N [

N [

Ik ,

k=1

/ I. Particularly, p ∈ / E. Ik . Assume that p ∈ E but p ∈

k=1

Since E = E ∪ E ′ , we have p ∈ E ′ which implies that (p − ǫ, p + ǫ) ∩ E 6= ∅ for every ǫ > 0 and then (p − ǫ, p + ǫ) ∩ I 6= ∅ for every ǫ > 0. Thus p is a limit point of I. By Definition 16.1 (Intervals in Rn ), I is closed in Rn , so p ∈ I which is a contradiction. Therefore, it is true that E ⊆ I and then J ∗ (E) ≤ J ∗ (E). Consequently, we get the desired result that J ∗ (E) = J ∗ (E). This completes the proof of the problem.



Problem 16.2 ⋆ Prove that if J(E) = 0, then µ(E) = 0. Show also that the converse is not true.

Proof. Suppose that J(E) = 0. Given ǫ > 0. By Definition 16.11 (Jordan Measure), there is a finite collection of intervals {I1 , I2 , . . . , IN } whose union covers E such that N X

µ(Ik ) < ǫ.

(16.10)

k=1

If we consider the countable collection {I1 , I2 , . . . , IN , IN +1 , . . .}, where In = ∅ for all n ≥ N +1, then its union also covers E. Now the estimate (16.10) and the fact µ(∅) = 0 certainly give ∞ X

µ(Ik ) < ǫ.

k=1

By Theorem 16.3(a), we see that µ(E) = 0. Let F = Q ∩ [0, 1]. We know from Problem 12.3 and Theorem 12.5(a) (Properties of Measurable Sets) that m(F ) = 0. Since F is dense in [0, 1], we observe from Problem 16.1 that J ∗ (F ) = J ∗ (F ) = J ∗ ([0, 1]) = 1. However, the density of irrationals in [0, 1] implies that F has only empty interval so that J∗ (F ) = 0. Since J∗ (F ) 6= J ∗ (F ), Definition 16.11 (Jordan Measure) shows that F is not Jordan  measurable. This completes the proof of the problem.

Chapter 16. Integral Calculus of Functions of Several Variables

176

Remark 16.5 (a) By Problem 16.2, the same result (16.6) holds for any bounded f if J(E) = 0. (b) In addition, we recall that a rational q ∈ [0, 1] is Jordan measurable because {q} is a finite set, but their union is F = Q ∩ [0, 1] in Problem 16.2 is not Jordan measurable. In other words, this counterexample shows that countable union of Jordan measurable sets is not necessarily Jordan measurable.

Problem 16.3 ⋆ If E ⊆ F ⊆ Rn , prove that J ∗ (E) ≤ J ∗ (F ).

Proof. Suppose that {I1 , I2 , . . . , IN } is a finite collection of intervals in Rn covering F . Then we have N [ Ik . (16.11) E⊆F ⊆ k=1

By Definition 16.11 (Jordan Measure), we have ∗

J (E) ≤

N X

µ(Ik )

k=1

for every finite collection of intervals satisfying the set relations (16.11). By Definition 16.11 (Jordan Measure) again, we conclude that J ∗ (E) ≤ J ∗ (F ) which completes the proof of the  problem. Problem 16.4 ⋆ Suppose that E is a Jordan measurable subset of Rn and J(E) = 0. If F ⊆ E, prove that F is also Jordan measurable and J(F ) = 0.

Proof. Since J(E) = 0, we have J ∗ (E) = 0 by Definition 16.11 (Jordan Measure). By Problem 16.3, we know that J ∗ (F ) = 0. By the definition again, we always have J∗ (F ) ≤ J ∗ (F ). Thus we get J∗ (F ) = J ∗ (F ) = 0 and the definition again shows that F is Jordan measurable  and J(F ) = 0. We complete the proof of the problem. Problem 16.5 ⋆ Let E ⊆ R, prove that

J ∗ (E) = inf J ∗ (V ),

where the infimum takes over all open sets V containing E.

177

16.2. Jordan Measurable Sets

Proof. Let S = {J ∗ (V ) | V is an open set containing E}. If J ∗ (E) = ∞, then Problem 16.4 implies that J ∗ (V ) = ∞ and our result follows. Therefore, we may assume that J ∗ (E) < ∞. By Problem 16.4 again, J ∗ (E) is a lower bound of S. Now we want to show that for every ǫ > 0, there is an open sets V containing E such that J ∗ (V ) ≤ J ∗ (E) + ǫ.

(16.12)

To see this, given ǫ > 0, then the definition of the infimum implies that there exists a finite collection of intervals {I1 , I2 , . . . , IN } in Rn covering E such that N X k=1

ǫ µ(Ik ) ≤ J ∗ (E) + . 2

(16.13)

Recall from Definition 16.1 (Intervals in Rn ) that there are ak1 , bk1 , ak2 , bk2 , . . . , akn , bkn such that Ik = [ak1 , bk1 ] × [ak2 , bk2 ] × · · · × [akn , bkn ], where k = 1, 2, . . . , N . Let δ > 0 and we consider the open intervals Vk = (ak1 − δ, bk1 + δ) × (ak2 − δ, bk2 + δ) × · · · × (akn − δ, bkn + δ). Then we have Ik ⊆ Vk and µ(Vk ) =

N Y

(bkj − akj + 2δ)

j=1

by the definition (16.2). Now we can make δ as small as possible so that µ(Vk ) ≤ µ(Ik ) + If we define V =

N [

ǫ . 2k+1

(16.14)

Vk , then it is open in Rn and

k=1

E⊆

N [

k=1

Ik ⊆ V,

so we follow from Definition 16.11 (Jordan Measure) and the inequalities (16.14) that J ∗ (V ) ≤

N X k=1

µ(Vk ) ≤

N h X k=1

µ(Ik ) +

ǫ i

2k+1

=

N X k=1

ǫ µ(Ik ) + . 2

(16.15)

Combining the inequalities (16.13) and (16.15), we gain J ∗ (V ) ≤ J ∗ (E) + ǫ which is exactly the required result (16.12). Hence we obtain J ∗ (E) = inf J ∗ (V ) and this  completes the proof of the problem. Problem 16.6 ⋆ Prove that there exists a non-Jordan measurable subset of S = [0, 1] × [0, 1].

Chapter 16. Integral Calculus of Functions of Several Variables

178

Proof. We consider the subset E = {(x, y) | x, y ∈ Q ∩ [0, 1]}. We note that an interval in R2 is actually a rectangle. On the one hand, it is clear that there is no rectangle contained in E because a rectangle must contain a point with irrational coordinates. Thus it means that J∗ (E) = 0. On the other hand, let R be an interval contained in S. Then the density of Q shows that R ∩ E 6= ∅. This observation means that if E ⊆

N [

(16.16)

Ik , then we have

k=1 N [

Ik = S.

k=1

Otherwise, since all Ik are rectangles, one can find a rectangle R such that R⊆S\

N [

Ik

k=1

and then R ∩ E = ∅ which contradicts the observation (16.16). Therefore, we have J ∗ (E) = µ(S) = 1. Consequently, J ∗ (E) 6= J∗ (E) and it follows from Definition 16.11 (Jordan Measure) that E is  not Jordan measurable set. This ends the analysis of the problem.

16.3 Integration on Rn Problem 16.7 ⋆ Suppose that E is a compact Jordan measurable set in Rn . Prove that Z dx. J(E) = E

Proof. Let I be an interval containing E in the form (16.1). By the definition, we have   1, if x ∈ E; χE (x) =  0, if x ∈ I \ E.

Clearly, the set of discontinues points of χE in I are exactly ∂E. Since E is Jordan measurable, it follows from Theorem 16.13 that J(∂E) = 0. By Problem 16.2, we have µ(∂E) = 0 and

16.3. Integration on Rn

179

then Theorem 16.10 (The Lebegus’s Integrability Condition) ensures that χE ∈ R on I. By Definition 16.15 (Riemann Integrable on E), we conclude that Z dx E

exists and then Theorem 16.8 (The Riemann Integrability Condition) guarantees that Z Z Z dx. (16.17) dx = dx = E

E

E

Next, we let P be a partition of I into subintervals I1 , I2 , . . . , Ip and S = {k ∈ {1, 2, . . . , p} | Ik ∩ E 6= ∅}. Therefore, we obtain E⊆ and

[

Ik

k∈S

Mk (χE ) = sup{χE (x) | x ∈ Ik } =

  1, if k ∈ S; 

0, if k ∈ / S.

By Definition 16.7 (Lower and Upper Riemann Integrals) and then Definition 16.11 (Jordan Measure), we see that p X X Mk (χE )µ(Ik ) = µ(Ik ) U (P, χE ) = k=1

k∈S

which implies

Z

I

χE dx = inf{U (P, χE ) | P is a partition of I} nX [ o = inf µ(Ik ) E ⊆ Ik k∈S

k∈S

∗

= J (E).

Since E is Jordan measurable, the expression (16.18) reduces to Z χE dx = J(E).

(16.18)

(16.19)

I

By combining the expressions (16.17) and (16.19), we establish that Z Z Z dx, J(E) = χE dx = dx = I

E

E

completing the proof of the problem. Remark 16.6 In fact, the condition that E is compact can be dropped in Problem 16.7.



Chapter 16. Integral Calculus of Functions of Several Variables

180

Problem 16.8 ⋆ Prove Theorem 16.16.

Proof. Let I be an interval containing E. Define   f (x), if x ∈ E; g(x) =  0, if x ∈ I \ E.

Let Df (E) be the set of all discontinuities of f on E. By Theorem 16.10 (The Lebesgue’s Integrability Condition) or Theorem 16.16, g ∈ R on I if and only if µ(Dg (I)) = 0. We notice that the discontinuities of f are also discontinuities of g and g may have more discontinuities on ∂E, so we obtain Dg (I) = Df (E) ∪ Dg (∂E). Since E ◦ ∩ (∂E)◦ = ∅, Theorem 16.3(b) implies that µ(Dg (I)) = µ(Df (E)) + µ(Dg (∂E)).

(16.20)

Next, it is obvious that Dg (∂E) ⊆ ∂E. Since E is Jordan measurable, Theorem 16.13 shows that J(∂E) = 0. Applying Problems 16.2 and 16.3, we conclude that µ(Dg (∂E)) = 0. Hence it follows from the expression (16.20) that µ(Dg (I)) = 0 if and only if µ(Df (E)) = 0. Hence our desired result follows directly from Definition 16.15 (Riemann Integrable on E). We have  completed the proof of the problem. Problem 16.9 ⋆ Prove Theorem 16.17(c).

Proof. We remark that χE = χA∪B = χA + χB − χA∩B , so we have Z Z Z Z Z f χE dx = f χA dx + f χB dx − f χA∩B dx. f dx = E

I

E⊆I

I

(16.21)

I

By Remark 16.2(b), since A ∩ B = ∅, f χA∩B = 0 on I and then Z f χA∩B dx = 0. I

By this, the expression (16.21) reduces to Z Z Z f dx = f χA dx + f χB dx E

I

I

which is our required result, completing the proof of the problem.



16.3. Integration on Rn

181 Problem 16.10 ⋆ Let S = [0, 1] × [0, 1]. Prove that Z e−2 2 . y 3 exy d(x, y) = 2 S

Proof. It is evident that f is Riemann integrable with respect to each variable. Furthermore, 2 since f (x, y) = y 3 exy is continuous on S, Theorem 16.10 (The Lebesgue’s Integrability Condition) ensures that f satisfies the hypotheses of Fubini’s Theorem. Hence we obtain Z 1Z 1 Z 2 3 xy 2 y e d(x, y) = y 3 exy dx dy S 0 0 Z 1 Z 1  2 exy dx dy y3 = 0 0 Z 1 2 y(ey − 1) dy = 0

e−2 = . 2

This completes the proof of the problem.



Problem 16.11 ⋆ Suppose that fk ∈ R on Ik = [ak , bk ], where k = 1, 2, . . . , n. Verify that Z

I

f1 (x1 ) · · · fn (xn ) dx1 · · · dxn =

Z

b1 a1



f1 (x1 ) dx1 × · · · ×

Z

bn an

 fn (xn ) dxn ,

where I = I1 × · · · × In . Proof. For each k = 1, 2, . . . , n, let Dk be the set of discontinuities of fk on Ik . Let D be the set of discontinuities of the function f = f1 f2 · · · fn on I. By Theorem 16.10 (The Lebesgue’s Integrability Condition), we see that µ(Dk ) = 0, where k = 1, 2, . . . , n. It is evident that D⊆

n [

k=1

I1 × · · · × Ik−1 × Dk × Ik+1 × · · · × In .

Thus we know from Theorem 16.3(a) and then Definition 16.2 (The Measure of Intervals in Rn ) that µ(D) ≤ =

n X k=1

n X k=1

µ(I1 × · · · × Ik−1 × Dk × Ik+1 × · · · × In ) µ(I1 ) × · · · × µ(Ik−1 ) × µ(Dk ) × µ(Ik+1 ) × · · · × µ(In )

Chapter 16. Integral Calculus of Functions of Several Variables

182

= 0. By Theorem 16.10 (The Lebesgue’s Integrability Condition), it yields that f ∈ R on I. By repeated use of Fubini’s Theorem, we establish that Z b1  Z Z  f1 (x1 ) · · · fn (xn ) dx2 · · · dxn dx1 f1 (x1 ) · · · fn (xn ) dx1 · · · dxn = I2 ×···×Ik

a1

I

Z

=

Z

=

×

b1

a1 b1 a1

Z

= ··· Z =



f1 (x1 ) dx1 × 

f1 (x1 ) dx1 ×

I3 ×···×Ik

Z

I2 ×···×Ik b2

f2 (x2 ) · · · fn (xn ) dx2 · · · dxn

f2 (x2 ) dx2 a2



f3 (x3 ) · · · fn (xn ) dx3 · · · dxn

b1

f1 (x1 ) dx1 a1

Z

 Z



b2

a2

f2 (x2 ) dx2 · · ·





Z

bn an

 fn (xn ) dxn .

We have completed the proof of the problem.



Problem 16.12 ⋆ Suppose that Q = [0, 1] × · · · × [0, 1] and y = (1, 1, . . . , 1). Prove that Z  e − 1 n e−x·y dx = . e Q Proof. If x = (x1 , x2 , . . . , xn ), then we have −x · y = −(x1 + x2 + · · · + xn ) and thus e−x·y = e−x1 e−x2 · · · e−xn . Since e−xk ∈ R on [0, 1], it deduces from Problem 16.11 that Z

−x·y

e

Q

dx =

Z

1

−x1

e

0



dx1 × · · · ×

Z

1

0

  e − 1 n , e−xn dxn = e

completing the proof of the problem.



Problem 16.13 ⋆ Let a < A and b < B. Suppose that f (x, y) = [a, A] × [b, B] and

I=

Z

∂2 F (x, y) is continuous on Q = ∂x∂y

f (x, y) d(x, y). Q

Show that I = F (A, B) − F (a, B) − F (A, b) + F (a, b).

16.3. Integration on Rn

183

Proof. Since f is continuous on Q, it satisfies all the hypotheses of Fubini’s Theorem. Therefore, we have Z B Z A  f (x, y) dx dy. (16.22) I= a

b

By the Second Fundamental Theorem of Calculus, we know that Z A Z A A ∂2 ∂ ∂ ∂ f (x, y) dx = F (x, y) dx = F (x, y) = F (A, y) − F (a, y). ∂x∂y ∂y ∂y ∂y a a a

(16.23)

Substituting the result (16.23) into the integral (16.22), we get Z B Z B Z Bh i ∂ ∂ ∂ ∂ F (A, y) − F (a, y) dy = F (A, y) dy − F (a, y) dy. I= ∂y ∂y b ∂y b ∂y b

Applying the Second Fundamental Theorem of Calculus again, we obtain immediately that B B I = F (A, y) − F (a, y) = F (A, B) − F (A, b) − F (a, B) + F (a, b). b

b

This completes the proof of the problem.



Problem 16.14 ⋆ Suppose that S = [0, 1] × [0, 1] and f (x, y) = Prove that

  x + y − 1, if x + y ≤ 1;  Z

0,

(16.24)

otherwise.

1 f d(x, y) = − . 6 S

Proof. Since f is continuous on S, it satisfies all the requirements of Fubini’s Theorem. Thus we have Z 1 Z 1 Z  f (x, y) dy dx. (16.25) f d(x, y) = S

0

0

By the definition (16.24), the integral (16.25) reduces to Z 1  Z 1−x Z  (x + y − 1) dy dx f d(x, y) = 0 0 S Z 1  1−x y2 xy + = dx −y 2 0 0 Z 1h i (1 − x)2 x(1 − x) + = − (1 − x) dx 2 0 Z 1 1 (1 − x)2 dx =− 2 0 1 =− , 6 completing the proof of the problem.



Chapter 16. Integral Calculus of Functions of Several Variables

184

Problem 16.15 ⋆

⋆ Denote S = [0, 1] × [0, 1]. Define f : S → R by f (x, y) =

  0, 

if at least one of x or y is irrational; (16.26)

1 n,

if x, y ∈ Q and x =

m n,

where m and n are relatively prime and n > 0. Prove that Z Z 1Z 1 Z 1  f (x, y) d(x, y) = 0, f (x, y) dx dy = f (x, y) dx = 0

0

0

S

but f (x, y) ∈ / R on [0, 1] for every rational x. Proof. On ([0, 1] \ Q) × [0, 1], we have f (x, y) is continuous and zero. Given ǫ > 0. Let ǫ ǫ , qk + 2k+1 ). Then we have {q1 , q2 , . . .} = Q ∩ [0, 1] and Ik = (qk − 2k+1 Q ∩ [0, 1] ⊆

∞ [

Ik

and

k=1

∞ X

µ(Ik ) =

k=1

∞ X ǫ < ǫ. 2k k=1

By the definition, µ(Q ∩ [0, 1]) = 0 and Theorem 16.10 (The Lebesgue’s Integrability Condition) implies that f (x, y) ∈ R on S. If P is a partition of S into p rectangles I1 , I2 , . . . , Ip . Since each rectangle must contain a point with irrational coordinates, the definition (16.26) shows that L(P, f ) = 0 and then Z f (x, y) d(x, y) = 0

(16.27)

S

by Theorem 16.8 (The Riemann Integrability Condition).

Next, if y is irrational, then f (x, y) = 0 for all x ∈ [0, 1]. Therefore, Remark 16.2(b) shows that Z 1

f (x, y) dx = 0.

(16.28)

0

If y ∈ Q ∩ [0, 1], then we know from [18, Exercise 18, p. 100] that the functiond f (x, y) is continuous at every irrational x in [0, 1]. Thus it follows from [25, Theorem 9.4, p. 159] that f (x, y) ∈ R on [0, 1] for every rational y ∈ Q ∩ [0, 1]. By Theorem 14.6(a), we see that Z Z Z 1 Z f (x, y) dm. (16.29) f (x, y) dm + f (x, y) dm = R f (x, y) dx = 0

[0,1]

Q∩[0,1]

[0,1]\Q

Since m(Q ∩ [0, 1]) = 0 and f (x, y) = 0 on [0, 1] \ Q by the definition (16.26), Theorem 14.3(f) ensures that the two Lebesgue integrals on the right-hand side of the equation (16.29) are zero. In other words, we have Z 1

f (x, y) dx = 0

(16.30)

0

if y ∈ Q ∩ [0, 1]. Now we combine the results (16.28) and (16.30) to conclude that Z 1 f (x, y) dx = 0 0

d

It is, in fact, the Riemann function, see [25, Problems 7.5 and 9.8, pp. 105, 106, 166]

(16.31)

16.3. Integration on Rn

185 for every y ∈ [0, 1] and so

Z

0

1Z 1 0

 f (x, y) dx dy = 0.

(16.32)

Hence our desired results follows immediately from the results (16.27), (16.31) and (16.32). However, suppose that x ∈ Q ∩ [0, 1]. Then we have x = m n , where m and n are relatively prime and n > 0. Thus we get from the definition (16.26) that   0, if y ∈ [0, 1] \ Q; (16.33) f (x, y) =  1 n , if y ∈ Q ∩ [0, 1].

Obviously, the function (16.33) is nowhere continuous because it is a multiple of the Dirichlet / R on [0, 1] for every fixed x ∈ Q∩[0, 1]. This completes function D(y).e Consequently, f (x, y) ∈ the proof of the problem.  Remark 16.7 Problem 16.15 shows that the condition f (·, y) ∈ R on [c, d] for every x ∈ [a, b] in Fubini’s Theorem cannot be relaxed. In fact, one can find counterexamples to show that the other hypotheses cannot be omitted.

Problem 16.16 ⋆ Suppose that E = {(x, y, z) ∈ R3 | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x and 0 ≤ z ≤ 1 − x − y} and f (x, y, z) = x. Prove that Z 1 . f dx = 24 E Proof. We notice that if F = {(x, y) ∈ R2 | 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 − x}, φ(x, y) = 0 and ϕ(x, y) = 1 − x − y, then F is clearly a closed Jordan measurable subset of R2 (in fact, F is the area bounded by the straight lines y = 1 − x, x = 0 and y = 0) and both φ and ϕ are continuous on F . Thus Theorem 16.18 implies that Z  Z 1−x−y ZZZ Z  x dz d(x, y). (16.34) f (x, y, z) dx dy dz = f dx = E

F

E

0

Similarly, if I = [0, 1], φ(x) = 0 and ϕ(x) = 1 − x, then I is a closed Jordan measurable subset of R and both φ(x) = 0 and ϕ(x) = 1 − x are continuous on I. Therefore, we apply Theorem 16.18 again to the integral on the right-hand side of (16.34) to obtain Z  Z 1−x−y Z  x dz d(x, y) f dx = 0 F E Z 1 Z 1−x Z 1−x−y = x dz dy dx 0 0 0 Z 1 Z 1−x (x − x2 − xy) dy dx = 0

e

0

Read [25, Problems 7.2 and 7.11, pp. 103, 104, 110, 111]

Chapter 16. Integral Calculus of Functions of Several Variables Z 1 1 = (x − 2x2 + x3 ) dx 2 0 1 , = 24 completing the proof of the problem.

186



16.4 Applications of the Mean Value Theorem Problem 16.17 ⋆

⋆ Suppose that E ⊆ Rn , f ∈ R on E and Z f dx = 0.

(16.35)

E

Let F = {x ∈ E | f (x) < 0} and J(F ) = 0. Prove that there corresponds a set S with µ(S) = 0 and f (x) = 0 for every x ∈ E \ S. Proof. Since J(F ) = 0, F is Jordan measurable and the particular case of the Mean Value Theorem for Multiple Integrals implies that Z f dx = 0. (16.36) F

Besides, it follows from Theorem 16.14 that E \ F is also Jordan measurable. By the conditions (16.35) and (16.36), and Theorem 16.17(c) (Properties of Integration on E), we obtain Z Z Z f dx = 0, f dx − f dx = E\F

E

F

i.e., f ∈ R on E \ F . Next, if A is the set of the discontinuities of f in E \ F , then we deduce from Theorem 16.16 that µ(A) = 0. Take p ∈ E \ (F ∪ A ∪ ∂E). Since p ∈ / A, f is continuous at p. If f (p) 6= 0, then f (p) > 0. Since p ∈ / ∂E, the continuity of f at p shows that there exists a δ > 0 such that Nδ (p) ⊆ E \ F and f (p) (16.37) f (x) > 2 for all x ∈ Nδ (p). By the definition, if x ∈ E\F , then we have f (x) ≥ 0. Note that (E\F )\Nδ (p) is Jordan measurable so that Z f dx ≥ 0 (16.38) (E\F )\Nδ (p)

by Theorem 16.17(b) (Properties of Integration on E). In addition, we know from the inequalities (16.37) and (16.38) that Z Z Z Z f (p) f dx > µ(Nδ (p)) · f dx ≥ >0 f dx + f dx = 2 Nδ (p) (E\F )\Nδ (p) Nδ (p) E\F

187

16.4. Applications of the Mean Value Theorem

which contradicts the hypothesis (16.35). In conclusion, we must have f (p) = 0 and thus f (x) = 0

(16.39)

on E \ (F ∪ A ∪ ∂E).

Finally, we claim that the set S = F ∪ A ∪ ∂E

satisfies the requirements. By the result (16.39), it suffices to prove that µ(S) = 0. Recall that J(F ) = 0, so µ(F ) = 0 by Problem 16.2. By the properties of the boundary of a set,f we have ∂E ⊆ ∂F ∪ ∂(E \ F )

and ∂(E \ F ) = E \ F ∩ E \ (E \ F ) = E \ F ∩ F = F .

Consequently, we have ∂E ⊆ ∂F ∪ F . By Theorem 16.13, we have J(∂F ) = 0, so µ(∂F ) = 0 by Problem 16.2. By Problem 16.1, we know that J(F ) = J ∗ (F ) = J ∗ (F ) = J(F ) = 0 which implies µ(F ) = 0 by Problem 16.2 again. Using Remark 16.2(a) twice, we get µ(∂E) = 0 and µ(S) = 0 which proves the claim. Hence we complete the analysis of the problem.



Problem 16.18 ⋆ Let E be a Jordan measurable set of Rn , f ∈ R on E and g : E → R be bounded. If F ⊆ E satisfies J(F ) = 0 and g(x) = f (x) on E \ F , prove that g ∈ R on E and Z Z g dx. f dx = E

E

Proof. Since f = g on E \ F , it follows from Theorem 16.17(c) (Properties of Integration on E) and Remark 16.5(a) that Z Z Z Z Z Z Z f dx. f dx = f dx + g dx + 0 = g dx = g dx + g dx = E

E\F

E\F

F

This completes the proof of the problem.

E\F

F

E



Problem 16.19 ⋆ ⋆ Suppose that E is an open connected Jordan measurable subset of Rn , f ∈ R on E and f is continuous on E. Prove that there is a point p ∈ E such that Z f dx = f (p)J(E). (16.40) E

f

Refer to [2, Exercises 3.51 and 3.52, p. 69].

Chapter 16. Integral Calculus of Functions of Several Variables

188

Proof. Since f ∈ R on E, f is bounded on E. Therefore, both M = sup{f (x) | x ∈ E} and m = inf{f (x) | x ∈ E} are finite and m ≤ f (x) ≤ M on E. By the Mean Value Theorem for Multiple Integrals, we have Z f dx ≤ M J(E) mJ(E) ≤ E

which means that there exists a λ ∈ [m, M ] such that Z

f dx = λJ(E).

(16.41)

E

If λ = m, then we must have m = M . Otherwise, one can find a point a ∈ E such that m < f (a) < M.

(16.42)

Since f is continuous on E, the Sign-preserving Propertyg implies that there exists a δ > 0 such that f (x) > m on Nδ (a). However, Theorem 16.17 (Properties of Integration on E) and Remark 16.6 give Z

f dx =

Z

f dx +

Nδ (a)

E

>

Z

m dx + Z

f dx E\Nδ (a)

Z

m dx

E\Nδ (a)

Nδ (a)

=m

Z

dx

E

= mJ(E)

(16.43)

which contradicts the formula (16.41). In other words, f is a constant function and in fact, f (x) = m on E. Hence the formula (16.40) holds trivially. Next, if λ = M , then we also have m = M . Otherwise, the inequality (16.42) holds for another point b ∈ E. Thus the previous analysis can be repeated to obtain the same contradiction (16.43). Now we may suppose that m < λ < M. Since f is continuous on E and E is connected, f (E) is a connectedh subset of [m, M ]. In fact, we conclude from [25, Theorem 4.16, p. 31] that f (E) is an interval. By the definitions of M and m, we certainly have (m, M ) ⊆ f (E) ⊆ [m, M ]. These observations show that one can find a point p ∈ E such that f (p) = λ. This completes the proof of the problem.

g h

In fact, we require its generalized version with Rn as the domain in [25, p. 112]. See [25, Theorem 7.12, p. 100].



189

16.5. Applications of the Change of Variables Theorem

16.5 Applications of the Change of Variables Theorem Problem 16.20 ⋆ Evaluate

Z

D

sin

p

x2 + y 2 d(x, y),

where D = {(x, y) | π 2 ≤ x2 + y 2 ≤ 4π 2 }.

Proof. Let E = {(r, θ) | π ≤ r ≤ 2π and 0 < θ < 2π} = [π, 2π] × (0, 2π). Then E is obviously Jordan measurable because J(∂E) = 0 and φ(E) = D, where φ(r, θ) = (r cos θ, r sin θ) π 5π and D is an annulus. Next, we note that if V = ( π2 , 5π 2 ) × (− 2 , 2 ), then V is an open set containing E = [π, 2π] × [0, 2π] and does not intersect the θ-axis. See Figure 16.3 for detials.

Figure 16.3: The mapping φ : E → D. p Finally, the function f (x, y) = sin x2 + y 2 is continuous on R2 and f (r cos θ, r sin θ) = sin r, so we must have f ∈ R on D and f (r cos θ, r sin θ) ∈ R on E. Hence we may apply the formula (16.9) to get Z Z p 2 2 sin x + y dx dy = f (r cos θ, r sin θ)r dr dθ D E Z 2π Z 2π r sin r dr dθ = 0 π Z 2π h 2π Z 2π i cos r dr dθ r cos r − =− π π 0 Z 2π 3π dθ =− 0

Chapter 16. Integral Calculus of Functions of Several Variables

190

= −6π 2 . This completes the analysis of the problem.



Problem 16.21 ⋆ Evaluate

Z

(x + y)3 dx dy,

R

where R is the parallelogram with vertices (1, 0), (3, 1), (2, 2) and (0, 1).

Proof. Consider f (x, y) = (x + y)3 and φ : R2 → R2 defined by  2u + v u − v  φ(u, v) = . , 3 3

Then it is continuously differentiable and injective on R2 . In addition, if E is the parallelogram with vertices (1, −2), (4, −2), (4, 1) and (1, 1), then it is easily shown that φ(E) = D, see Figure 16.4 below:

Figure 16.4: The mapping φ : E → R. Since f is continuous on R2 , we have f ∈ R on R by Theorem 16.16. Since f (φ(u, v)) = u3 which is continuous on R2 , we have f ◦ φ ∈ R on E by Theorem 16.16. Finally, since E is clearly a Jordan measurable set, E ⊆ R2 and  2 1   3 det Jφ ((u, v)) = det   1 3

3   = − 1 6= 0,  3 1 − 3

we obtain from the formula (16.8) that Z Z Z Z Z 1 1 4 3 255 3 1 3 3 u · du dv = , (x + y) dx dy = u du dv = (x + y) dx dy = 3 3 −2 1 3 E φ(E) R

191

16.5. Applications of the Change of Variables Theorem

completing the proof of the problem.



Problem 16.22 ⋆ Evaluate

Z

e−x

2 −y 2

d(x, y),

D

where D = {(x, y) ∈ R2 | x2 + y 2 ≤ 1}.

2

2

Proof. Let f (x, y) = e−x −y . Since it is continuous on D, f ∈ R on D by Theorem 16.16. Let E = {(r, θ) | 0 ≤ r ≤ 1 and 0 ≤ θ < 2π} = [0, 1] × [0, 2π). Then E is Jordan measurable because J(∂E) = 0 and φ(E) = D, where φ(r, θ) = (r cos θ, r sin θ). Using the polar coordinates (with the aid of Remark 16.4(b)), we have Z

−x2 −y 2

e

d(x, y) =

D

Z

Z

e−x

2 −y 2

d(x, y)

φ(E) 2

re−r dr dθ E Z 2π Z 1 2 = re−r dr dθ 0 0 Z 2π 1 =− (e−1 − 1) dθ 2 0 = π(1 − e−1 ). =

This ends the proof of the problem.



Problem 16.23 ⋆ Suppose that E is a Jordan measurable subset of Rn and for every x ∈ E, we have −x ∈ E. Let f be an odd function such that f ∈ R on E. Prove that Z f dx = 0. E

Proof. Define φ(x) = −x. Then φ is continuously differentiable and injective on Rn . In addition, we have φ(E) = E so that f ∈ R on φ(E). Therefore, we get f ◦ φ ∈ R on E. Finally, since E ⊆ Rn and   −1 0 · · · 0  0 −1 · · · 0    det Jφ (x) = det  .  = (−1)n 6= 0, .. . .  .. . . 0  0

0

· · · −1

Chapter 16. Integral Calculus of Functions of Several Variables

192

the integral in question satisfies all the requirements of the Change of Variables Theorem. Hence we deduce from the formula (16.8) and the fact f (φ(x)) = f (−x) = −f (x) that Z Z Z Z f (x) dx. (16.44) f (φ(x)) · | det Jφ (x)| dx = − f (y) dy = f (y) dy = E

E

E

φ(E)

Since the variable is dummy, the formula (16.45) implies that Z f dx = 0 E

which is our desired result. This completes the proof of the problem.



Problem 16.24 ⋆

⋆ Show that

ZZ

0