S.: The reader should see the textbook, pp 391; we have the general discussion

b fxdx

(iv) the set of discontinuities of f on I has measure zero.

P.S.: The reader should see the textbook, pp 391; we have the general discussion.

7.27

Assume f has a derivative which is monotonic decreasing and satisfies f^x  m  0 for all x in a, b. Prove that



a

bcos fxdx  2m .

Hint: Multiply and divide the integrand by f^x and use Theorem 7.37(ii).

Proof: Since f^x  m  0, and _f¹_ is monotonic increasing ona, b, we consider



_a^bcos fxdx 



_a^{b cos f}_fx^xf^xdx

 1

f^b



_c^bcos fxf^xdx, by Theorem 7.37(ii)

 1

f^b



_fc^fbcos udu, by Change of Variable

 sin fb  sin fc

f^b

which implies that



a

bcos fxdx  2m .

7.28

Given a decreasing seq uence of real numbersGn such that Gn  0 as n  . Define a function f on 0, 1 in terms of Gn as follows: f0  1; if x is irrational, then fx  0; if x is rational m/n (in lowest terms), then fm/n  Gn.

Compute the oscillationfx at each x in 0, 1 and show that f  R on 0, 1.

Proof: Let x₀  Q^c  0, 1. Since limnGn  0, given   0, there exists a

positive integer K such that as n  K, we have |Gn|  . So, there exists a finite number

of positive integers n such that Gn  . Denote S  x : |fx|  , then #S  .

Choose a  0 such that x0  , x0   0, 1 does NOT contain all points of S. Note that fx0  0. Hence, we know that f is continuous at x0. That is,fx  0 for all

x  Q^c  0, 1.

Let x₀  0, then it is clear that f0  1 f0  0. So, f is not continuous at 0. Q  0, 1.

Let x₀  Q  0, 1, say x0  ^M_N (in lowest terms). SinceGn is monotonic decreasing, there exists a finite number of positive integers n such that Gn  GN.

Denote T  x : |fx|  GN, then #T  . Choose a   0 such that

x0  , x0    0, 1 does NOT contain all points of T. Let h  0, , then supfx  fy : x, y  x0  h, x0  h  0, 1  fx0  GN.

So,fx0  GN. That is, fx  fx for all x  Q  0, 1.

Remark: (1) If we have proved f is continuous on Q^c  0, 1, then f is automatically Riemann integrable on0, 1 since D Q  0, 1  Q, the set of discontinuities of f has measure zero.

(2) Here is a good exercise. We write it as a reference. Given a function f defined on

a, b, then the set of continuities of f on a, b is G set.

Proof: Let C denote the set of continuities of f ona, b, then C  x : _fx  0

 _k1^ x : fx  1/k

andx : _fx  1/k is open. We know that C is a G set.

Note: (i) We call S a G_ set if S  _n1^ On, where On is open for each n.

(ii) Given y  x  a, b : fx  1/k : I, then fy  1/k. Hence, there exists a d  0, such that

fBy, d  1/k, where By, d  a, b

For z  By, d, consider a smaller  so that Bz,   By, d. Hence,

fBz,   1/k which implies that

fz  1/k.

Hence, By, d  I. That is, y is an interior point of I. That is, I is open since every point of I is interior.

7.29

Let f be defined as in Exercise 7.28 with Gn  1/n. Let gx  1 if

0  x  1, g0  0. Show that the composite function h defined by hx  gfx is not Riemann-integrable on0, 1, although both f  R and g  R on 0, 1.

Proof: By Exercise 7.28, we know that

hx  0 if x  Q^c  0, 1

1 if x  Q  0, 1

which is discontinuous everywhere on0, 1. Hence, the function h (Dirichlet Function) is not Riemann-integrable on0, 1.

7.30 Use Lebesgue’s theorem to prove Theorem 7.49.

(a) If f is of bounded variation ona, b, then f  R on a, b.

Proof: Since f is of bounded variation ona, b, by Theorem 6.13 (Jordan Theorem), f  f1  f2, where f₁ and f₂ are increasing ona, b. Let Di denote the set of discontinuities of fi ona, b, i  1, 2. Hence, D, the set of discontinuities of f on a, b is

D  D1 D2.

Since |D1|  |D2|  0, we know that |D|  0. In addition, f is of bounded on a, b since f is bounded variation ona, b. So, by Theorem 7.48, f  R on a, b.

(b) If f  R on a, b, then f  R on c, d for every subinterval c, d  a, b, |f|  R and f²  R on a, b. Also, f  g  R on a, b whenever g  R on a, b.

Proof: (i) Let D_a,b and D_c,d denote the set of discontinuities of f ona, b and c, d, respectively. Then

D_c,d  D_a,b.

Since f  R on a, b, and use Theorem 7.48, |D_a,b|  0 which implies that |D_c,d|  0.

In addition, since f is bounded ona, b, f is automatically is bounded on c, d for every compact subintervalc, d. So, by Theorem 7.48, f  R on c, d.

(ii) Let D_f and D_|f| denote the set of discontinuities of f and |f| ona, b, respectively, then

D_|f|  Df.

Since f  R on a, b, and use Theorem 7.48, |Df|  0 which implies that |D|f||  0. In addition, since f is bounded ona, b, it is clear that |f| is bounded on a, b. So, by Theorem 7.48, |f|  R on a, b.

(iii) Let D_f and D_f2 denote the set of discontinuities of f and f² ona, b, respectively, then

D_f2  Df.

Since f  R on a, b, and use Theorem 7.48, |Df|  0 which implies that D_f²  0. In addition, since f is bounded ona, b, it is clear that f²is bounded ona, b. So, by Theorem 7.48, f²  R on a, b.

(iv) Let D_f and Dg denote the set of discontinuities of f and g ona, b, respectively.

Let D_fg denote the set of discontinuities of fg ona, b, then D_fg  Df  Dg

Since f, g  R on a, b, and use Theorem 7.48, |Df|  |Dg|  0 which implies that

|D_fg|  0. In addition, since f and g are bounded on a, b, it is clear that fg is bounded on

a, b. So, by Theorem 7.48, fg  R on a, b.

(c) If f  R and g  R on a, b, then f/g  R on a, b whenever g is bounded away from 0.

Proof: Let D_f and D_g denote the set of discontinuities of f and g ona, b, respectively.

Since g is bounded away from 0, we know that f/g is well-defined and f/g is also bounded ona, b. Consider Df/g, the set of discontinuities of f/g ona, b is

D_f/g  Df  Dg.

Since f  R and g  R on a, b, by Theorem 7.48, |Df|  |Dg|  0 which implies that

|D_f/g|  0. Since f/g is bounded on a, b with |Df/g|  0, f/g  R on a, b by Theorem 7.48.

Remark: The condition that the function g is bounded away from 0 CANNOT omit.

For example, say gx  x on 0, 1 and g0  1. Then it is clear that g  R on 0, 1, but

1/g  R on 0, 1. In addition, the reader should note that when we ask whether a function is Riemann-integrable or not, we always assume that f is BOUNDED on a COMPACT INTERVALa, b.

(d) If f and g are bounded functions having the same discontinuities ona, b, then f  R on a, b if, and only if, g  R on a, b.

Proof: () Suppose that f  R on a, b, then Df, the set of discontinuities of f ona, b

has measure zero by Theorem 7.48. From hypothesis, D_f  Dg, the set of discontinuities of g ona, b, we know that g  R on a, b by Theorem 7.48.

() If we change the roles of f and g, we have proved it.

(e)

^{Let g}  R on a, b and assume that m  gx  M for all x  a, b. If f is continuous onm, M, the composite function h defined by hx  fgx is

Riemann-integrable ona, b.

Proof: Note that h is bounded ona, b. Let Dh and Dgdenote the set of dsicontinuities of h and g ona, b, respectively. Then

D_h  Dg.

Since g  R on a, b, then |Dg|  0 by Theorem 7.48. Hence, |Dh|  0 which implies that h  R on a, b by Theorem 7.48.

Remark: (1) There has a more general theorem related with Riemann-Stieltjes Integral. We write it as a reference.

(Theorem)Suppose g  R on a, b, m  gx  M for all x  a, b. If f is continuous onm, M, the composite function h defined by hx  fgx  R on

a, b.

Proof: It suffices to consider the case that is increasing on a, b. If a  b, there is nothing to prove it. So, we assume thata  b. In addition, let

K  supxa,b|hx|. We claim that h satisfies Riemann condition with respect to  on

a, b. That is, given   0, we want to find a partition P such that UP, h,   LP, h,  



k1 n

Mkh  mkhk  .

Since f is uniformly continuous onm, M, for this   0, there is a _2K1^    0 such that as |x y|   where x, y  m, M, we have

|fx  fy|  

2b  a. 1

Since g  R on a, b, for this   0, there exists a partition P such that UP, g,   LP, g,  



k1 n

Mkg  mkgk  ². 2 Let P  A  B, where A  xj : M_kg  mkg   and

B  xj : M_kg  mkg  , then





B

k 



B

Mkh  mkhk  ² by (2) which implies that



B

k  .

So, we have

UA, h,   LA, h,  



A

Mkh  mkhk

 /2 by (1) and

UB, h,   LB, h,  



B

Mkh  mkhk

 2K



B

k

 2K

 /2.

It implies that

UP, h,   LP, h,   UA, h,   LA, h,   UB, h,   LB, h, 

 .

That is, we have proved that h satisfies the Riemann condition with respect to on a, b.

So, h  R on a, b.

Note: We mention that if we change the roles of f and g, then the conclusion does NOT hold. Since the counterexample is constructed by some conclusions that we will learn in Real Analysis, we do NOT give it a proof. Let C be the standard Cantor set in0, 1 and C^ the Cantor set with positive measure in0, 1. Use similar method on defining Cantor Lebesgue Function, then there is a continuous function f : 0, 1  0, 1 such that fC  C^. And Choose g  XC on0, 1. Then

h  g  f  X_C^ which is NOT Riemann integrable on0, 1.

(2) The reader should note the followings. Since these proofs use the exercise 7.30 and Theorem 7.49, we omit it.

(i) If f  R on a, b, then |f|  R on a, b, and f^r  R on a, b, where r  0, .

(ii) If |f|  R on a, b, it does NOT implies f  R on a, b. And if f²  R on a, b, it does NOT implies f  R on a, b. For example,

f  1 if x  Q  a, b

1  Q^c  a, b

(iii) If f³  R on a, b, then f  R on a, b.

7.31 Use Lebesgue’s theorem to prove that if f  R and g  R on a, b and if fx  m  0 for all x in a, b, then the function h defined by

hx  fx^gx

is Riemann-integrable ona, b.

Proof: Consider

hx  exph log f, then by Theorem 7.49,

f  R  log f  R  h log f  R  exph log f  h  R.

7.32

^{Let I}  0, 1 and let A1  I  ¹₃, ²₃  be the subset of I obtained by removing

those points which lie in the open middle third of I; that is, A₁  0, ¹₃   ²₃, 1. Let A2

be the subset of A₁ obtained by removing he open middle third of 0, ¹₃  and of ²₃, 1.

Continue this process and define A₃, A₄, . . . . The set C  _n1^ A_n is called the Cantor set.

Prove that

(a) C is compact set having measure zero.

Proof: Write C  _n1^ An. Note that every An is closed, so C is closed. Since A₁ 

is closed and bounded, A₁ is compact and C  A1, we know that C is compact by Theorem 3.39.

In addition, it is clear that |An|  ²₃ ⁿ for each n. Hence, |C|  limn|An|  0, which implies that C has a measure zero.

(b) x  C if, and only if, x 



_n1^{ an3n}, where each an is either 0 or 2.

Proof: ()Let x  C  _n1^ An, then x  Anfor all n. Consider the followings.

(i) Since x  A1  0, ¹₃   ²₃, 1, then it implies that a1  0 or 2.

(ii) Since x  A2  0, ¹₉   ²₉, ³₉   ⁶₉, ⁷₉   ⁸₉, ⁹₉ , then it implies that a2  0 or 2.

Inductively, we have an  0 or 2. So, x 



_n1^{ an3n}, where each an is either 0 or 2.

() If x 



_n1^{ a}n3^n, where each a_n is either 0 or 2, then it is clear that x  An for each n. Hence, x  C.

(c) C is uncountable.

Proof: Suppose that C is countable, write C  x1, x₂, . . .. We consider unique ternary expansion: if x 



_n1^{ an3n}, then x : a1, . . . , an, . . .. From this definition, by (b), we have

x_k  xk1, x_k2, . . . x_kk,. . . where each component is 0 or 2.

Choose y  y1, y₂, . . . where

yj  2 if x_jj  0 0 if x_jj  2.

By (b), y  C. It implies that y  xk for some k which contradicts to the choice of y.

Hence, C is uncountable.

Remark: (1) In fact, C  C^ means that C is a perfect set. Hence, C is uncountable.

The reader can see the book, Principles of Mathematical Analysis by Walter Rudin, pp 41-42.

(2) Let C  x : x 



_n1^{ an3n}, where each an is either 0 or 2 . Define a new function : C  0, 1 by

x 



n1

 an/2 2ⁿ ,

then it is clear that is 1-1 and onto. So, C is equivalent to 0, 1. That is, C is uncountable.

(d) Let fx  1 if x  C, fx  0 if x  C. Prove that f  R on 0, 1.

Proof: In order to show that f  R on 0, 1, it suffices to show that, by Theorem 7.48, f is continuous on0, 1  C since it implies that D  C, where D is the set of

discontinuities of f on0, 1.

Let x₀  0, 1  C, and note that C  C^, so there is a  0 such that

x0  , x0    C  , where x0  , x0   0, 1. Then given   0, there is a

  0 such that as x  x0  , x0  , we have

|fx  fx0|  0  .

Remark: (1) C  C^ :Given x  C  _n1^ An, and note that every endpoints of An

belong to C. So, x is an accumulation point of the set y : y is the endpoints of An . So, C  C^. In addition, C^  C since C is closed. Hence, C  C^.

(2) In fact, we have

f is continuous on0, 1  C and f is not continuous on C.

Proof: In (d), we have proved that f is continuous on0, 1  C, so it remains to show that f is not continuous on C. Let x₀  C, if f is continuous at x0, then given  1/2, there is a  0 such that as x  x0  , x0    0, 1, we have

|fx  fx0|  1/2

which is absurb since we can choose y  x0  , x0    0, 1 and y  C by the fact C does NOT contain an open interval since C has measure zero. So, we have proved that f is not continuous on C.

Note: In a metric space M, a set S M is called nonwhere dense if intclS  .

Hence, we know that C is a nonwhere dense set.

Supplement on Cantor set.

From the exercise 7.32, we have learned what the Cantor set is. We write some conclusions as a reference.

(1) The Cantor set C is compact and perfect.

(2) The Cantor set C is uncountable. In fact, #C  #R.

(3) The Cantor set C has measure zero.

(4) The Cantor set C is nonwhere dense.

(5) Every point x in C can be expressed as x 



_n1^{ an3n}, where each an is either 0 or 2.

(6) X_C : 0, 1  0, 1 the characteristic function of C on 0, 1 is Riemann integrable.

The reader should be noted that Cantor set C in the exercise is 1dimensional case. We can use the same method to construct a ndimensional Cantor set in the set

x1, . . . , xn : 0  xj  1, j  1, 2, . . n. In addition, there are many researches on Cantor set. For example, we will learn so called Space-Filling Curve on the textbook, Ch9, pp 224-225.

In addition, there is an important function called Cantor-Lebesgue Function related with Cantor set. The reader can see the book, Measure and Integral (An Introduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 35.

7.33

The exercise outlines a proof (due to Ivan Niven) that² is irrational. Let fx  xⁿ1  xⁿ/n!. Prove that:

(a) 0  fx  1/n! if 0  x  1.

Proof: It is clear.

(b) Each kth derivative f^k0 and f^k1 is an integer.

Proof: By Leibnitz Rule,

f^kx  1n!



j0 k

_j^kn   n  j  1x^nj 1^kjn   n  k  j  11  x^nkj which implies that

f^k0 

0 if k  n 1 if k  n

nk1^knn   2n  k  1 if k  n .

So, f^k0  Z for each k  N. Similarly, f^k1  Z for each k  Z.

Now assume that²  a/b, where a and b are positive integers, and let Fx  bⁿ



k0 n

1^kf^2kx^2n2k. Prove that:

(c) F0 and F1 are integers.

Proof: By (b), it is clear.

(d)²aⁿfx sin x  _dx^d F^x sin x  Fx cos x

Proof: Note that

F^x  ²Fx  bⁿ



k0 n

1^kf^2k2x^2n2k  ²bⁿ



k0 n

1^kf^2kx^2n2k

 bⁿ



k0 n1

1^kf^2k2x^2n2k  bⁿ1ⁿf^2n2x

 ²bⁿ



k1 n

1^kf^2kx^2n2k  ²bⁿfx²ⁿ

 bⁿ



k0 n1

1^kf^2k2x^2n2k  1^k1f^2k2x^2n2k

 bⁿ1ⁿf^2n2x  ²bⁿfx²ⁿ

 ²aⁿfx since f is a polynomial of degree 2n.

So,

dx d F^x sin x  Fx cos x

 sin xF^x  ²Fx

 ²aⁿfx sin x.

(e) F1  F0  aⁿ



₀^1fx sin xdx.

Proof: By (d), we have

²aⁿ



₀^1fx sin xdx  F^x sin x  Fx cos x|₀¹

 F^1 sin   F1 cos   F^0 sin 0  F0 cos 0

 F1  F0

which implies that

F1  F0  aⁿ



₀^1fx sin xdx.

(f) Use (a) in (e) to deduce that 0  F1  F0  1 if n is sufficiently large. This contradicts (c) and show that² (and hence) is irrational.

Proof: By (a), and sin x  0, 1 on 0, , we have

0  aⁿ



₀¹fx sin xdx  an!ⁿ



₀^{1sinxdx  2a}_{n! }^{n 0 as n  .}

So, as n is sufficiently large, we have, by (d),

0  F1  F0  1

which contradicts (c). So, we have proved that² (and hence) is irrational.

Remark: The reader should know that is a transcendental number. (Also, so is e). It is well-known that a transcendental number must be an irrational number.

In 1900, David Hilbert asked 23 problems, the 7th problem is that, if 0, 1 is an algebraic number and is an algebraic number but not rational, then is it true that ^is a transcendental number. The problem is completely solved by Israil Moiseevic Gelfand in 1934. There are many open problem now on algebraic and transcendental numbers. For example, It is an open problem: Is the Euler Constant

  lim 1  12 . . . 1n  logn a transcendental number.

7.34 Given a real-valued function, continuous on the interval a, b and having a finite bounded derivative^ ona, b. Let f be defined and bounded on a, b and assume that both integrals



_a^bfxdx and



_a^{bfxxdx}

exists. Prove that these integrals are equal. (It is not assumed that^ is continuous.) Proof: Since both integrals exist, given  0, there exists a partition

P  x0  a, . . . , xn  b such that

SP, f,  



a

bfxdx  /2 where

SP, f,  



j1 n

ftjj for t_j  xj1, x_j





j1 n

ftj^sjxj by Mean Value Theorem, where sj  xj1, xj * and

SP, f^ 



_a^{bfx}xdx  /2 where

SP, f^ 



j1 n

ftj^tjxj for tj  xj1, xj **

So, let t_j  sj, then we have

SP, f,   SP, f^.

Hence,



a

bfxdx 



a

bfx^xdx

 SP, f,  



_a^bfxdx  SP, f^ 



_a^{bfxxdx}

 .

So, we have proved that both integrals are equal.

7.35

Prove the following theorem, which implies that a function with a positive integral must itself be positive on some interval. Assume that f  R on a, b and that 0  fx  M on a, b, where M  0. Let I 



_a^bfxdx, let h  ¹₂I/M  b  a, and assume that I  0. Then the set T  x : fx  h contains a finite number of intervals, the sum of whose lengths is at least h.

Hint. Let P be a partition ofa, b such that every Riemann sum SP, f 



_k1^{n ft}kxk

satisfies SP, f  I/2. Split SP, f into two parts, SP, f 



_kA^



_kB^{, where}

A  k : xk1, x_k  T, and B  k : k  A.

If k  A, use the inequality ftk  M; if k  B, choose tk so that ftk  h. Deduce that



_kA^xk  h.

Proof: It is clear by Hint, so we omit the proof.

Remark: There is another proof about that a function with a positive integral must itself be positive on some interval.

Proof: Suppose NOT, it means that in every subinterval, there is a point p such that fp  0. So,

LP, f 



j1 n

m_jxj  0 since mj  0 for any partition P. Then it implies that

sup

P LP, f 



_a^{bfxdx  0}

which contradicts to a function with a positive integral. Hence, we have proved that a function with a positive integral must itself be positive on some interval.

Supplement on integration of vector-valued functions.

(Definition) Given f₁, . . . , fn real valued functions defined ona, b, and let

f  f1, . . . , fn : a, b  Rⁿ. If  on a, b. We say that f  R on a, b means that f_j  R on a, b for j  1, 2, . . . , n. If this is the case, we define



_a^{bfd }



_a^{bf1d, . . . ,}



_a^{bfnd .}

From the definition, the reader should find that the definition is NOT stranger for us. When we talk f  f1, . . . , fn  R on a, b, it suffices to consider each fj  R on a, b for j  1, 2, . . . , n.

For example, if f  R on a, b where   on a, b, then f  R on a, b.

Proof: Since f  R on a, b, we know that fj  R on a, b for j  1, 2, . . . , n.

Hence,



k1 n

f_j²  R on a, b

which implies that, by Remark (1) in Exercise 7.30,

f 



k1 n

f_j²  R on a, b

1/2

 R on a, b.

Remark: In the case above, we have



_a^{bfd }



_a^{bfd.}

Proof: Consider

y²  



_a^bf1d, . . . ,



_a^bfnd,



_a^bf1d, . . . ,



_a^{bfnd }





j1

n



_a^bfjd



_a^{bfjd ,}

which implies that, (let yj 



_a^bfjd, y y1, . . . , yn),

y² 



j1 n

yj



_a^bfjd





j1

n



a bf_jy_jd





_a^b



j1 n

f_jy_j d





_a^{bfyd}

 y



_a^bfd

which implies that

y 



a

bfd.

Note: The equality holds if, and only if, ft  kty.

In document Apostol Solution Ch7 Mathematical Analysis 2e (Page 37-47)