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Folland: Real Analysis, Chapter 2Sebastien Picard

Problem 2.3

If {fn} is a sequence of measurable functions on X, then {x : lim fn(x) exists} is a measurable set.

Solution:Define h = limsupfn, g = liminffn. By Proposition 2.7, h, g are measurable. Let

E =n=1

g1(n, ) h1(n, )

E =n=1

g1(, n) h1(, n)

.

It is clear that both E and E are measurable sets. Next, define

w(x) =

29 if g = h = g(x) h(x) else

Then w is a measurable function by Exercise 2. Therefore

E1 = (w1(, 0) w1(0, ))c

is measurable. Hence the set {x : limfn exists} is measurable since it is equal to

{x X : g(x) = h(x)} = E1 E E.

Problem 2.9Letf : [0, 1] [0, 1] be the Cantor function (1.5), and let g(x) = f(x) + x.

a. g is a bijection from [0, 1] to [0, 2] and h = g1 is continuous from [0, 2] to [0, 1].b. If C is the Cantor set, m(g(C)) = 1.c. By Exercise 29 of Chapter 1, g(C) contains a Lebesgue nonmeasurable setA. LetB = g1(A).

Then B is Lebesgue measurable but not Borel.d. There exists a Lebesgue measurable function F and a continuous function G onR such that

F G is not Lebesgue measurable.

Solution:(a) We know that f is an increasing function, and therefore f(x) + x is a strictly increasing function.Also, since f is continuous, then f(x) + x is continuous. Since g(0) = 0 and g(1) = 2, and g iscontinuous and strictly increasing on [0, 1], then g is a bijection from [0, 1] to [0, 2]. It follows thatg1 exists.

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Since g is continuous map from a compact set to a Hausdorff space, then g1 is continuous:indeed, g(K) is compact for any compact set K, and compacts sets are closed in a Hausdorff space,so (g1)1(C) is closed for any closed set C in the domain of g.

(b) Let

[0, 1] Cc =n=1

En

where En are countably many disjoint intervals. Since f is constant on a given En, then g(En) isa translate of En, so m(g(En)) = m(En). Therefore

m(g([0, 1] Cc)) = m(m=1

g(En)) = m(m=1

En) = m([0, 1] Cc) = 1

.

Since 2 = m([0, 2]) = m(g(C)) + m(g([0, 1] Cc)), we have m(g(C)) = 1.

(c) Suppose B is a Borel set. Since g1 is continuous, (g1)1(B) = A is Borel measurable. But Ais not even Lebesgue measurable, which is a contradiction.

However, B C. Since C is a null set, B is Lebesgue measurable by completeness of Lebesguemeasure.

(d) Define G and F as follows:

G(x) =

g1(x) if x [0, 2]x if x 01 if x 2

F(x) =

x if x B69 else

G is continuous by the Pasting Lemma from elementary topology. F is measurable since F1(a, )is either empty, the whole real line, or a subset of B (which is measurable since B has measure zero).

F G is not Lebesgue measurable since G1(F1(1, )) = G1(B) = A.

Problem 2.14If f L+, let (E) =

E

f d for E M. Then is a measure on M, and for any g L+,gd =

fgd. (First suppose that g is simple.)

Solution:We will show that is a measure on M. It is clear that () = 0. Let {Ak} be a countable

sequence of disjoint sets, and define A = k=1Ak.

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(Ak) =

A

f d =

A

(k=1

Akf)d =k=1

A

Akf d =k=1

(Ak).

The summation can be taken out from the integral by Theorem 2.15.Let g L+ be a simple function. Then g =

ni=1 aiAi where Ai are measurable sets. Therefore

gd =ni=1

ai(Ai) =ni=1

ai

f Aid =

ni=1

aiAif d =

fgd.

Now take any g L+. By Theorem 2.10 there exists a sequence {n} of simple functions thatconverges to g pointwise such that 0 1 2 g. Since {n} and {f n} are monotoneincreasing, we can apply the monotone convergence theorem twice to obtain

gd =

limnd = lim

nd = lim

nf d =

fgd.

Problem 2.16Iff L+ and

f < , for every > 0 there existsE M such that(E) < and

E

f > (

f).

Solution:Let f L+ such that

f < . Let > 0. By definition of

f, there exists a simple function

=ni=1 aiEi such that 0 f and

f 0. It will be shown that there exists a y (x , x + ) such thatg(y) > M. This will prove that g is unbounded on each interval and that g is discontinuous at everypoint.

There exists a rational number rn (x , x + ). Furthermore, there exists a y (x , x + )such that

0 < y rn M.

This proof does not fail after redefining g on any Lebesgue null set, since one can still find anirrational y with the desired properties.

(c) Since g < a.e., it follows that g2 < a.e. However,

g2

n=1

22nf2(x rn)dx =

n=122n

10

f2(z)dz 1

4

10

dz

z> .

Problem 2.27Letfn(x) = ae

nax benbx where 0 < a < b.a.

1

0

|fn(x)|dx = .b.

1

0 fn(x)dx = 0.

c.

1 fn L1([0, ), m), and

0

1 fn(x)dx = log(b/a).

Solution:

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(a) Since fn is the difference of two exponential functions, we can find a point c R such that fn < 0on (0, c) and fn > 0 on (c, ). In order to find this point c, we solve

benbc = aenac

log(b/a) = nc(b a)

c = log(b/a) 1n(b a)

We can now split up the integral in order to integrate |fn|:

0

|fn| =

co

aenax benbx

dx +

c

aenax benbx

dx

=enax

n

enbx

n

c0+enbx

n

enax

n

c

=2

nenac enbc

= 2n

e

aba

log(b/a) eb

balog(b/a)

.

Therefore0

|fn| is proportional to (1/n), hence

1

0

|fn| = .

(b)

0

fn(x)dx =

o

aenaxdx

0

benbxdx

= enax

n

0

+ enbx

n

0

=1

n(1 + 1) = 0.

Therefore

1

0

fn(x)dx = 0.

(c)

n=1

fn =n=1

aenax benbx

= a n=0

enax 1

b n=0

enbx 1

= a 1

1 eax 1

b

11 ebx

1

=a

1 eax

b

1 ebx+ (b a).

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We can see thatn=1 fn is positive on (0, ), so to show that

n=1 fn L

1 we must compute0

n=1 fn:

o

n=1 fn =

0 a

1 eax

b

1 ebx+ (b a)dx

=

log(1 eax) log(1 ebx) + (b a)x

0

= log(1 eax)(e(ba)x)

1 ebx

0

= limz0

limx

log(1 eax)(1 ebz)ebxeax

(1 ebx)(1 eaz)ebzeaz

= limz0

limx

log(1 ebz eax + eax+bz)ebxeaz

(1 eaz ebx + ebx+az)ebzeax

= limz0

limx

logeax(eax ebzax 1 + ebz)ebxeaz

ebx

(ebx

eazbx

1 + eaz

)ebz

eax

= limz0

limx

log(eax ebzax 1 + ebz)eaz

(ebx eazbx 1 + eaz)ebz

= limz0

log(ebz 1)eaz

(eaz 1)ebz

= limz0

log1 ebz

1 eaz

= log(b/a).

The last step follows from lHopitals rule:

limz0

1 ebz

1 eaz= limz0

bebz

aeaz=

b

a.

Problem 2.28Compute the following limits and justify the calculations:

a. limn0

(1 + (x/n))n sin(x/n)dx.

b. limn10

(1 + nx2)(1 + x2)ndx.c. limn

0n sin(x/n)[x(1 + x2)]1dx.

d. limna n(1 + n

2

x2

)1

dx. (The answer depends on whether a > 0, a = 0, or a < 0. Howdoes this accord with the various convergence theorems?

Solution:(a) Denote

fn =sin(x/n)

(1 + (x/n))n.

First, notice that by the binomial theorem, when n 2, for all x [0, ) we have

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0

dx

1 + x2= arctan(x)

0

= /2 < .

By the Dominated Convergence Theorem we can conclude

limn0 fndx =

0

limn

sin(x/n)

x/n (1 + x

2

)

1dx =

0

dx

1 + x2 = /2.

(d) In this case we can evaluate the integral directly:

limn

a

n

1 + n2x2dx = limn

na

dy

1 + y2= limn

arctan(y)na

= /2 limn

arctan(na).

Hence

limn

a

n

1 + n2x2dx =

0 a > 0/2 a = 0

a < 0

This agrees with the fact that one cannot apply the Dominated Convergence Theorem unless a > 0since there is no way to bound fn(0).

Problem 2.32Suppose (X) < . If f and g are complex-valued measurable functions on X, define

(f, g) =

|f g|1 + |f g|

d.

Then is a metric on the space of measurable functions if we identify functions that are equal a.e.,and fn f with respect to this metric iff fn f in measure.

Solution:We will first show that is a metric on the space of measurable functions if we identify functions

that are equal a.e.It is clear that (f, g) = (g, f) and that (f, g) 0. We can also see that

(f, g) = 0

|f g|

1 + |f g|d = 0

|f g|

1 + |f g|= 0 a.e. (by Proposition 2.16)

f = g a.e.

It only remains to show the triangle inequality. Let x,y,z X. First, suppose |x z| |x y|and |z y| |x y|. Then we have

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|x y|

1 + |x y|

|x z|

1 + |x y|+

|z y|

1 + |x y|

|x z|

1 + |x z|+

|z y|

1 + |z y|.

On the other hand, suppose |x z| |x y|. Then

|x y|

1 + |x y|

|x y|

1 + |x y| +

|z y|

1 + |z y|

|x z|

1 + |x z| +

|z y|

1 + |z y| .

The above argument can be repeated when |z y| |x y|. Hence for all x, y, z X,

|x y|

1 + |x y|

|x z|

1 + |x z|+

|z y|

1 + |z y|.

Using basic properties of the integral (Proposition 2.13), we can conclude that (f, g) (f, h) +(h, g) for any measurable functions f , g , h. This completes the proof that is a metric.

Suppose fn f with respect to . Let En, = {x : |fn(x) f(x)| }.

|fn f|1 + |fn f|

d En,

|fn f|

1 + |fn f|d

1 + (En,

).

It follows that

(En,) 1 +

|f g|

1 + |f g|d 0.

Conversely, suppose fn f in measure. Let > 0. Choose N N such that for all integersn > N, we have

x : |fn(x) f(x)|

2(X)

N:

(fn, f) =

|f g|

1 + |f g|d =

A

|f g|

1 + |f g|d +

X\A

|f g|

1 + |f g|d

(A) +

2(X)(X\A)

< /2 + /2 = .

Problem 2.40In Egoroffs theorem, the hypothesis (X) < can be replaced by |fn| g for all n, whereg L1().

Solution:Suppose f1, f2, . . . and f are measurable complex-valued functions on X such that fn f a.e.

and |fn| g for all n, where g L1(). We will follow the proof of Theorem 2.33 and make someminor adjustments.

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Without loss of generality, assume that fn f everywhere on X. For k, n N, let

En(k) =m=n

{x : |fm(x) f(x)| k1}.

For fixed k, En(k) decreases as n increases and n=1En(k) = . To apply continuity of measure

from above, we need (E1) < . Since |fn f| 2|g|, we observe that

E1(k) A(k) := {x : 2|g(x)| k1}.

We can use the fact that

> 2

X

|g|

A(k)

2|g| k1(A(k)),

in order to conclude

(E1(k)) (A(k)) < .

Therefore, by continuity of measure from above, (En(k)) 0 as n . Given > 0 and k N,there exists a positive integer nk such that (Enk(k)) < 2

k.If we define E = k=1Enk(k), then (E) < and fn f uniformly on E

c.

Problem 2.49Prove Theorem 2.39 by using Theorem 2.37 and Proposition 2.12 together with the following lemmas.

a. If E M N and (E) = 0, then (Ex) = (Ey) = 0 for a.e. x and y.b. If f is L-measurable and f = 0 -a.e., then fx and f

y are integrable for a.e. x and y, and

fx d = fy d = 0 for a.e. x and y. (Here the completeness of and is needed.)Solution:(a) Suppose E M N and (E) = 0. Define f = E. Then fx = Ex and f

y = Ey . ApplyFubinis theorem:

0 =

f d( ) =

fxd(y)

d(x) =

fyd(x)

d(y).

It follows that

Exd = 0 -a.e. and (Ex) = 0 -a.e., and similarly

Eyd = 0 -a.e. and(Ey) = 0 -a.e..

(b) Suppose f is L-measurable and f = 0 -a.e.. Define

A = {(x, y) M N : f(x, y) = 0}.

Then A E for some E M N such that (E) = 0. By part (a), (Ex) = 0 and (Ey) = 0

for a.e. x and y. Since Ax Ex and Ay Ey, we have (Ax) = 0 and (Ay) = 0. Therefore|fx|d =

Ax|fx|d = 0 for -a.e. x and

|fy|d =

Ay |fy|d = 0 for -a.e. y.

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We now prove Theorem 2.39. Suppose f is L-measurable and either (a) f 0 or (b) f L1(). ByProposition 2.12, there exists a M N-measurable function g such that f = g -almost everywhere.By Proposition 2.34, gx is N-measurable and g

y is M-measurable. Define h = g f. Then h = 0-a.e.. By lemma (b), hx is N-measurable for a.e. x and h

y is M-measurable for almost every y,hence fx is N-measurable for a.e. x and f

y is M-measurable for almost every yIn case (b), g L1( ), so by lemma (b), hx = gx fx is integrable for a.e. x. By Fubinis

theorem, gx is integrable for a.e. x, hence fx is integrable for a.e. x. Similarly fy is integrable for a.e.y.

By applying lemma (b) on the function h, we can see that for a.e. x we have

(gx fx)d = 0hence

gxd =

fxd. Similarly for a.e. y we have

gyd =

fyd. In case (a), g L+(M N),

so by Tonellis theorem x

gxd =

fxd is measurable, and y

gyd =

fyd is measurable.In case (b), g L1(M N), so by Fubinis theorem x

gxd =

fxd is integrable, and

y

gyd =

fyd is integrable.Since

gd =

f d, after applying Tonellis theorem (case a) or Fubinis theorem (case b) and

using the fact that

gyd =

fyd and

gxd =

fxd for almost every x and y, we obtain

f d =

f(x, y)dd =

f(x, y)dd.

Problem 2.55Let E = [0, 1] [0, 1]. Investigate the existence and equality of

E

f dm2,10

10

f(x, y) dx dy, and10

10

f(x, y) dy dx for the following f.a. f(x, y) = (x2 y2)(x2 + y2)2.b. f(x, y) = (1 xy)a (a > 0).c. f(x, y) = (x 1/2)3 if 0 < y < |x 1/2|, f(x, y) = 0 otherwise.

Solution:(a) First, we evaluate 1

0

x2 y2

(x2 + y2)2dx.

Using the substitution x = y tan , dx = y sec2 d, we obtain

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10

x2 y2

(x2 + y2)2dx =

arctan(1/y)0

(tan2 1)y3 sec2 d

(y2(tan2 + 1))2

= arctan(1/y)

0

(tan2 1)d

y sec2

=1

y

arctan(1/y)0

(sin2 cos2 )d

=1

y

arctan(1/y)0

(1 2cos2 )d

=1

y

arctan(1/y)0

cos2d

=1

2ysin(2 arctan(1/y))

=

1

y sin(arctan(1/y)) cos(arctan(1/y))

=1

y

sin(arctan(1/y))

cos(arctan(1/y))

cos(arctan(1/y))1

cos(arctan(1/y))

=1

y

tan(arctan(1/y))

1 + tan2(arctan(1/y))

=1

1 + y2

Therefore,

10

10

x2

y2

(x2 + y2)2dxdy =

10

dy1 + y2

= /4.

We observe that10

x2 y2

(x2 + y2)2dx =

10

y2 x2

(x2 + y2)2dx =

10

x2 y2

(x2 + y2)2dy.

Hence 10

10

x2 y2

(x2 + y2)2dydx =

10

dx

1 + x2= /4.

By Fubinis theorem,Ef dm2 is not defined.

(b) Since f is non-negative on [0, 1] [0, 1], f L+(E), so by Tonellis theoremEf dm

2 =10

10

f(x, y)dxdy =10

10

f(x, y)dydx. The integral may be infinite for some values of a... I haventhad time to do this computation yet.

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(c) First, we compute

10

10

f(x, y)dydx =

10

|x0.5|0

(x 1

2)3dydx =

10

(x 12)dx

|x 12 |3

.

The function(x 1

2)dx

|x1

2 |3

is not integrable on [0, 1]:10

|x 12

|dx

|x 12

|3=

10

dx

|x 12

|2= .

Therefore, the integral10

10 f(x, y)dydx does not exist, and hence

Ef dm

2 does not exist.

However,10

10

f(x, y)dxdy = 0:

10

f(x, y)dx =

10

{yy}dzz3

=

y1/2

dz

z3+

1/2y

dz

z3

= 2 +1

2y2

1

2y2+ 2 = 0.

Problem 2.57

Show that0 e

sx

x

1

sin x dx = arctan(s

1

) for s > 0 by integrating e

sxy

sin x with respect to x andy. (It may be useful to recall that tan(/2 ) = (tan )1. Cf. Exercise 31d.)

Solution:We will investigate the integral

0

1

esxy sin xdydx.

We want to apply Fubinis theorem, so first we verify that

0

1

|esxy sin x|dydx

0

1

esxyxdydx =

0

esx

s

dx < .

Next, we observe that 0

1

esxy sin xdydx =1

s

0

esxx1 sin xdx.

This integral can be computed by switching the order of integration. The first step is to useintegration by parts to compute

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I =

0

esxy sin xdx =

0

(sy)esxy cos xdx esxy cos x0

=

0

(sy)esxy cos xdx + 1

=

0

(s2y2)esxy sin xdx syesyx sin x0

+ 1

=

0

(s2y2)esxy sin xdx + 1.

I = s2y2I + 1.

Therefore

I =1

1 + s2y2.

Thus 1

0

esxy sin xdxdy =

1

dy

1 + s2y2=

1

s

s

dz

1 + z2=

1

s

2

arctan(s)

.

It is given that

tan(

2 ) = (tan )1

which implies

tan

2 arctan(s)

=

1

s .

Therefore

2 arctan(s) = arctan(s1).

By Fubinis theorem, we conclude

1

s

0

esxx1 sin xdx =1

sarctan(s1).

0

esxx1 sin xdx = arctan(s1).

Problem 2.58Show that

esxx1 sin2 x dx = 1

4log(1 + 4s2) for s > 0 by integrating esx sin2xy with respect to x

and y.

Solution:We will investigate the integral

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0

10

esx sin(2xy)dydx.

We want to apply Fubinis theorem, so first we verify that

0

10 |e

sx

sin(2xy)|dydx 0

10 e

sxy

2xydydx =0 e

sx

xdx < .

Next, we observe that0

10

esx sin(2xy)dydx =

0

esx

x

12

1

2cos2x

=

0

esx sin2 xdx

x.

This integral can be computed by switching the order of integration. The first step is to useintegration by parts to compute

I =

0

esx sin(2xy)dx =

0

esx

s

2y cos(2yx)dx esx

s

sin(2yx)

0

=

0

esx

s2y cos(2yx)dx

=

0

(2y)2esx

s2sin(2yx)dx

2y

s2esx cos(2yx)

0

=

0

(2y)2esx

s2sin(2yx)dx +

2y

s2.

I =4y2I

s2+

2y

s2.

I =2y

s2 + 4y2.

Therefore

10

0

esx sin(2xy)dxdy =

10

2y

s2 + 4y2dy =

1

4log(s2 + 4)

1

4log(s2) =

1

4log(1 + 4s2).

By Fubinis theorem, this proves that

0

esx sin2 xdx

x=

1

4log(1 + 4s2).

Problem 2.60(x)(y)/(x + y) =

10 tx1(1 t)y1dt for x, y > 0. (Recall that was defined in 2.3. Write

(x)(y) as a double integral and use the argument of the exponential as a new variable of integra-tion.)

Solution:

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By definition, we have

(x)(y) =

0

0

tx1sy1etsdsdt.

We perform the change of variables s = u uv, t = uv. The Jacobian is

(s, t)(u, v)

= su

tv

sv

tu

= (1 v)u (uv) = u.

The change of variables formula for multiple integrals yields

(x)(y) =

10

0

(uv)x1(u uv)y1euududv =

10

0

vx1(1 v)y1euux+y1dudv.

By definition of (x + y) and Fubinis theorem, this can be rewritten as

(x)(y) = 1

0

vx1(1 v)y1dv(x + y).This proves that

(x)(y)

(x + y)=

10

tx1(1 t)y1dt.

Problem 2.63The technique used to prove Proposition 2.54 can also be used to integrate any polynomial over Sn1.In fact, suppose f(x) =

n1 xjj (j N {0}) is a monomial. Then

f d = 0 if any j is odd, and

if all js are even, f d =

2(1) (n)

(1 + + n, where j =

j + 1

2.

Solution:By Theorem 2.49, we know that

Rn

e|x|2

nj=1

xjj dx =

0

Sn1

er2

nj=1

xj|xj|

jrjrn1ddr.

First we compute the right-hand side:

0

Sn1

er2

nj=1

xj|xj |

jrjrn1ddr =

Sn1

f d

0

er2

rn1+jdr

=

Sn1

f d

0

es

2s j

2 sn12 s

1

2 ds

=

Sn1

f d

1

2( j + 1

2).

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If any i is odd, then

ex2

i xii dxi = 0 by symmetry, hence then left-hand side is zero:Rn

e|x|2

nj=1

xjj dx =

nj=1

ex2

j xjj dxj = 0.

Therefore Sn1

f d = 0.

If every i is even, then by symmetry e

x2i xii dxi = 20 e

x2i xii dxi. The left-hand side isthen

Rn

e|x|2

nj=1

xjj dx = 2

n

nj=1

0

ex2

j xjj dxj =

nj=1

0

essj

2 1

2 ds =

nj=1

(j + 1

2).

By combining the identities of the left-hand side and right-hand side, we obtain

Sn1

f d = 2(1) (n)(1 + + n)

where j =j+1

2.