Math 426: Homework 4

Mary Radcliffe

due 2 May 2014

In Bartle:

5C. If f L(X,F , ) and g is a F-measurable real-valued function such thatf(x) = g(x) almost everywhere, then g L(X,F , ) and f d = g d.Proof. Let h(x) = |f(x) g(x)|. Then h M+(X,F), and h = 0 almosteverywhere. Therefore, h is integrable and

h d = 0. But this implies

that fg is integrable, and thus g is integrable. Moreover, (f g) d h d = 0, and thus

(f g) d = 0. The result immediately follows by

linearity of integration.

5E. If f L and g is a bounded, measurable function, then fg L.

Proof. Let M be such that |g| M . Note that by linearity of integration,we have that Mf L, and |fg| |Mf | for all x. Then by Corollary 5.4,fg is integrable.

5I. If f is a complex-valued function on X such that Re f and Im f belong toL(X,F , ), we say that f is integrable and define f d = Re f d +i

Im f d. Let f be a complex-valued measurable function. Show thatf is integrable if and only if |f | is integrable, in which case f d |f | d.Proof. Let us write f = f1 + if2, so that f1 = Re f and f2 = Im f . Wehave that f is measurable if and only if f1 and f2 are measurable, and

|f | = ((f1)2 + (f2)2)1/2. Thus, f is measurable implies that |f | is alsomeasurable.

Notice, if a, b R, then a2 + b2 = (a+ b)2 2ab (a+ b)2 =|a + b| |a| + |b|. Therefore, if f is integrable, then |f | is a measurablefunction such that |f | |f1| + |f2|, and thus by Corollary 5.4, |f | isintegrable.

Conversely, if |f | is integrable, note that |f1| < |f | and |f2| < |f |, so wehave f1 and f2 are real-valued integrable functions by Corollary 5.4. Butthen by definition f is also integrable.

For the inequality, writef d = rei, so that

f d = r. Let g(x) =eif(x), so that g(x) = g1(x) + ig2(x), and

g d =

eif(x) d =

eif(x) d = r. But then

g2 d = 0, so g2 = 0 almost everywhere.

Moreover, |f | = |ei||f | = |g| = |g1| almost everywhere.Therefore, we have f d = r = g1 d = g1 d |g1| d = |f | d,

1

since g1 is real valued and we may use Theorem 5.3.

5P. Let fn L(X,F , ), and suppose that {fn} converges to a function f .Show that if lim

|fn f | d = 0, then |f | d = lim |fn| d.Proof. Note that as |fn f | is integrable for n sufficiently large, we havefn f is integrable, and thus by linearity f is integrable. Moreover, forall n, we have |fn| |fnf |+ |f |, so |fn| |f | |fnf |. Integrating andtaking limits on both sides, we have lim

|fn| d |f | d lim |fn f | d = 0, and thus lim |fn| d = |f | d

5Q. If t > 0, then0etxdx = 1t . Moreover, if t a > 0, then etx eax.

Use this and Exercise 4M to justify differentiating under the integral signand to obtain the formula

0xnexdx = n!.

Proof. Now, 0

etxdx = limb

b0

etxdx

= limb

1tetx

bx=0

= limb

(1tetb +

1

t

)=

1

t,

where the integral is obtained by the fundamental theorem of calculus.

Let f(x, t) = etx for t [ 12 , 32 ]. Then ft (x, t) = xetx, so |ft (x, t)| xe

x2 for all t. Let g(x) = xe

x2 . We claim that g is integrable on [0,).

This can be seen in several ways. Note that g is measurable. Moreover,note that if x is sufficiently large (in fact, x > 8 ln 4 is sufficient), wehave that x < ex/4. Define h(x) = g(x) if x 8 ln 4 and h(x) = e x4 ifx > 8 ln 4. Then g h, and h is clearly integrable, and thus by Corollary5.4, g is also integrable. But then we have, by Theorem 5.9, that

1t2

=d

dt

(1

t

)=

d

dt

( 0

etxdx)

=d

dt

([0,)

etxd(x)

)

=

[0,)

t

(etx

)d(x)

=

[0,)

xetxd(x)

= 0

xetxd(x).

Taking t = 1 and multiplying by 1 yields the result for n = 1.Now, proceed by induction on n. Suppose it is known that

0xn1etxdx =

(n1)!tn . Let fn(x, t) = x

n1etx for t [ 12 , 32 ]. As above, we havefnt (x, t) = xnetx, so |fnt (x, t)| xne

x2 . Again, for x sufficiently

2

large, we have that xn < ex/4, so as above, we have that xnex2 is inte-

grable on [0,). Then as above, we have

n!tn+1

=d

dt

((n 1)!tn

)=

d

dt

( 0

xn1etxdx)

=

[0,)

t(xn1etx)d(x)

= 0

xnetxdx.

Taking t = 1 and multiplying by 1 yields the result for n.

5R. Suppose that f is defined on X [a, b] to R and that the function x 7f(x, t) is F-measurable for each t [a, b]. Suppose that for some t0, t1 [a, b], the function x 7 f(x, t0) is integrable on X, that ft (x, t1) exists,and that there exists an integrable function g onX such that

f(x,t)f(x,t1)tt1 g(x) for x X and t [a, b], t 6= t1. Then[

d

dt

f(x, t) d(x)

]t=t1

=

f

t(x, t1) d(x).

Proof. Choose tn to be any sequence with tn t1, tn 6= t1 (we maystart at n = 2 to avoid confusion). Put hn(x) =

f(x,tn)f(x,t1)tnt1 , so by

hypothesis |hn(x)| g(x) for all x, and hn(x) ft (x, t1). By thedominated convergence theorem, then, we have limn

Xhn(x)d(x) =

Xft (x, t1)d(x). On the other hand,

limn

X

hn(x)d(x) = limn

f(x, tn) f(x, t1)

tn t1 d(x)

= limn

1

tn t1

(f(x, tn)d(x)

f(x, t1)d(x)

)=

[d

dt

f(x, t)d(x)

]t=t1

,

as desired.

5T. Let f be a F-measurable function on X to R. For n Z+, let {fn}be the sequence of truncates of f . If f is integrable with respect to ,then

f d = lim

fn d. Conversely, if sup

|fn| d < , then f isintegrable.

Proof. Note that for all n, we have |fn| f , and thus if f is integrable,the DCT implies that

fn d

f d.

For the converse, note that |fn| is a monotonically increasing sequenceof nonnegative functions with limit |f |, and thus the MCT implies that |fn| d |f | d. Thus, by hypothesis, |f | d < , and therefore|f | and thus f are integrable.

4O. Fatous Lemma has an extension to a case where the fn take on negativevalues. Let h M+(X,F), and suppose that h d < . If {fn} is asequence in M(X,F) and h fn then

lim inf fn d lim inf

fn d.

3

Proof. Note that fn + h 0. Thus we havelim inf fn d+

h d =

(lim inf fn + lim inf h) d

=

(lim inf(fn + h)) d

lim inf

(fn + h) d

= lim inf

(fn d+

h d

)= lim inf

fn d+

h d.

Subtractingh d yields the result.

Also, complete the following:

1. Compute the following limits. Justify each computational step using con-vergence theorems and/or calculus.

(a) limn0

n1+n2x2 dx.

(b) limn0

(1 + xn

)nlog(2 + cos( xn ))dx

(c) limn nn f

(1 + xn2

)g(x)dx, where g : R R is (Lebesgue) inte-

grable and f : R R is bounded, measurable, and continuous at1.

Solution. (a) We consider this as two integrals, over [0, 1] and [1,).Note that on [0, 1], we have

limn

10

n

1 + n2x2dx = lim

n [arctan(nx)]10

= limn arctan(n) =

pi

2.

For the other part, notice that n1+n2x2 nn2x2 1nx2 1x2 , andmoreover,

1

1x2 = 1 < , so 1x2 L([1,),L, ). Therefore, by

the dominated convergence theorem, we have

limn

1

n

1 + n2x2dx = lim

n

[1,)

n

1 + n2x2d(x)

=

[1,)

limn

(n

1 + n2x2

)d(x)

=

[1,)

0d(x) = 0.

Therefore, we obtain limn0

n1+n2x2 dx =

pi2 .

(b) Note that (1 + xn )n ex, and as cos( xn ) 1, we have log(2 +

cos( xn )) log 3. Thus,(1 + xn

)nlog(2+cos( xn )) (log 3)ex, which

is clearly integrable. Therefore, by the dominated convergence theo-rem, we have

limn

0

(1 +

x

n

)nlog(

2 + cos(xn

))dx =

0

limn

((1 +

x

n

)nlog(

2 + cos(xn

)))dx

=

0

ex log(3)dx

= log(3)

4

(c) Let us rewrite the integral as

limn

nn

f(

1 +x

n2

)g(x)dx = lim

n

Rf(

1 +x

n2

)[n,n]g(x) d(x).

LetM be such that |f(x)| < M for all x. Then f (1 + xn2 )[n,n]g(x) M |g(x)|, and since g L, we also have |g| L and thus M |g| L.Therefore, by the dominated convergence theorem, we have

limn

nn

f(

1 +x

n2

)g(x)dx = lim

n

Rf(

1 +x

n2

)[n,n]g(x) d(x)

=

R

limn

(f(

1 +x

n2

)[n,n]g(x)

)d(x)

=

Rf(1)Rg(x)d(x)

= f(1)

Rg(x)d(x),

where the limit exists because f is continuous at 1.

2. Let f(t) = t0ex

2

dx. We will use DCT to evaluate limt f(t), evenwithout being able to evaluate the indefinite integral.

(a) Put h(t) = f(t)2 and g(t) = 10et

2(1+x2)

1+x2 dx. Show that h(t) =

g(t).(b) Show that h(t) + g(t) = pi4 for all t.

(c) Use the previous parts to conclude that limt f(t) =

0

ex2

dx =pi

2.

Solution. (a) Note that h(t) = 2f(t)f (t) = 2et2

f(t), by the funda-mental theorem of calculus.

For g(t), let f(x, t) = et2(1+x2)1+x2 , and note that

t

[et

2(1+x2)

1+x2

]=

2tet2(1+x2) = 2tet2et2x2 . Put (t) = 2tet2 . Note that (t) =(2 4t2)et2 , and thus by the first derivative test, has a globalmaximum of

2/e at t =

1/2, and a global minimum of 2/e

at t = 12. Therefore,2tet2 2/e =: C for all t. Thus,ft Cet2x2 C, which is clearly integrable on [0, 1]. Therefore,

by Theorem 5.9,

d

dt

[ 10

et2(1+x2)

1 + x2dx

]=

10

t

(et

2(1+x2)

1 + x2

)dx

=

10

2tet2(1+x2)dx

= 2tet2 10

et2x2dx

= 2et2 t0

eu2

du using the substitution u = tx

= 2et2f(t) = h(t),as desired.

5

(b) As (h + g) = 0 for all t, we have that h + g is constant. Note thath(0) = 0. Moreover, g(0) =

10

11+x2 dx = arctan(1) =

pi4 . Therefore,

(h+ g)(0) = pi4 , and as h+ g is constant, the result follows.

(c) Note that |f(x, t)| 11+x2 for all x, t, and thus by the dominatedconvergence theorem, we have

limt g(t) =

10

limt

et2(1+x2)

1 + x2dx

=

10

0 dx = 0.

But then limt h(t) = pi4 , and thus limt f(t) =pi/4 =

pi2 .

Extras:

5A. If f L(X,F , ) and a > 0, show that the set {x X | |f(x)| > a}has finite measure. In addition, the set {x X | f(x) 6= 0} has -finitemeasure.

Proof. As |f | L, we have that |f | d = A < . Let Ea = {x X | |f(x)| > a}. Then we have |f | = |f |Ea + |f |Eca aEa + |f |Eca ,and thus

A =

|f | d a(Ea) +

Eca

|f | d a(Ea).

Therefore, (Ea) A/a.Moreover, {x X | f(x) 6= 0} = nZ+En, and thus the set has -finitemeasure.

5B. If f is a F-measurable real-valued function and if f(x) = 0 for -almostall x X, then f L(X,F , ) and f d = 0.Proof. Note that |f | M+ is -almost 0, and thus |f | d = 0. Thisimplies that |f | L, and thus f L. Moreover, f d |f | d = 0,and thus

f d = 0.

5D. If f L(X,F , ) and > 0, then there exists a measurable simple function such that

|f | d < .Proof. As f+ and f are M+, we have measurable simple functions +

and such that + f+, f, and (f+ +) d < 2 , (f ) d < 2 . Let =

+ . Then we have is simple, and|f| d =

|f++(f)| d

(|f+ +|+ |(f )|) d < .

5F. If f belongs to L, it does not follow that f2 belongs to L.

Proof. Let f = 1x

on [0, 1]. Then we have[0,1]

f d = 10

1xdx =

2x|10 = 2. However, f2 = 1x does not have a finite integral on [0, 1], so

f2 / L.

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5G. Suppose that f L(X,F , ), and that its indefinite integral is (E) =Ef d for E F . Show that (E) 0 for all E F if and only if

f(x) 0 for almost all x X. Moreover, (E) = 0 for all E if and onlyif f(x) = 0 for almost all x X.

Proof. Write f = f+ f. Suppose (E) > 0 for all E F . Let > 0, and put E = {x X | f(x) < }. As f is measurable, E F .Moreover,

Ef d (E) 0, and thus (E) = 0. But then {x

X | f(x) < 0} = nZ+E1/n has measure 0, and therefore f 0 -almosteverywhere.

On the other hand, if f 0 -almost everywhere, then for all E F ,fE 0 -almost everywhere, and thus by 5B,

fE d 0 for all

E F .For the second piece, we repeat the above argument with E = {x X | |f(x)| > }.

5H. Suppose that f1, f2 L(X,F , ), and let 1, 2 be their indefinite inte-grals. Show that 1(E) = 2(E) for all E F if and only if f1(x) = f2(x)for almost all x X.

Proof. Note that 1(E) = 2(E) for all E if and only if (1 2)(E) = 0for all E, if and only if f1 f2 = 0 -almost everywhere (by problemG).

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