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Math 426: Homework 4 Mary Radcliffe due 2 May 2014

In Bartle: 5C. If f ∈ L(X, F, µ) and g is a F-measurable real-valued function such R R that f (x) = g(x) almost everywhere, then g ∈ L(X, F, µ) and f dµ = g dµ. + Proof. Let h(x) = |f (x) − g(x)|. Then h ∈ M R (X, F), and h = 0 almost everywhere. Therefore, h is integrable and h dµ = 0. But R this implies (f − g) dµ ≤ that f −g is integrable, and thus g is integrable. Moreover, R R h dµ = 0, and thus (f − g) dµ = 0. The result immediately follows by linearity of integration.

5E. If f ∈ L and g is a bounded, measurable function, then f g ∈ L. Proof. Let M be such that |g| ≤ M . Note that by linearity of integration, we have that M f ∈ L, and |f g| ≤ |M f | for all x. Then by Corollary 5.4, f g is integrable. 5I. If f is a complex-valued function on X such that ReR f and ImR f belong to L(X, R F, µ), we say that f is integrable and define f dµ = Re f dµ + i Im f dµ. Let f be a complex-valued measurable function. Show that R f dµ ≤ Rf is integrable if and only if |f | is integrable, in which case |f | dµ. Proof. Let us write f = f1 + if2 , so that f1 = Re f and f2 = Im f . We have that f is measurable if and only if f1 and f2 are measurable, and
1/2 |f | = (f1 )2 + (f2 )2 . Thus, f is measurable implies that |f | is also measurable. p p √ 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 measurable function such that |f | ≤ |f1 | + |f2 |, and thus by Corollary 5.4, |f | is integrable. Conversely, if |f | is integrable, note that |f1 | < |f | and |f2 | < |f |, so we have f1 and f2 are real-valued integrable functions by Corollary 5.4. But then by definition f is also integrable. R R For the inequality, write f dµ = reiθ , so thatR f dµ R= r. Let g(x) = e−iθ fR(x), so that g(x) = g1 (x)R + ig2 (x), and g dµ = e−iθ f (x) dµ = e−iθ f (x) dµ = r. But then g2 dµ = 0, so g2 = 0 almost everywhere. Moreover, |f | = |e−iθ ||f | = |g| = |g1 | almost everywhere. Therefore, we have Z Z Z Z Z f dµ = r = g1 dµ = g1 dµ ≤ |g1 | dµ = |f | dµ, 1

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

5P. Let fn ∈ L(X, F, R µ), and suppose that R{fn } convergesR to a function f . Show that if lim |fn − f | dµ = 0, then |f | dµ = lim |fn | dµ. Proof. Note that as |fn − f | is integrable for n sufficiently large, we have fn − f is integrable, and thus by linearity f is integrable. Moreover, for all n, we have |fn | ≤ |fn − f | + |f |, so |fnR| − |f | ≤ |fnR− f |. Integrating R and taking limits on both sides, we have lim |f | dµ − |f | dµ ≤ lim |fn − n R R f | dµ = 0, and thus lim |fn | dµ = |f | dµ R∞ 5Q. If t > 0, then 0 e−tx dx = 1t . Moreover, if t ≥ a > 0, then e−tx ≤ e−ax . Use this and Exercise 4M to justify differentiating under the integral sign R∞ and to obtain the formula 0 xn e−x dx = n!. Proof. Now, Z

∞

e−tx dx

Z =

lim

b→∞

0

b

e−tx dx

0

b 1 −tx lim − e b→∞ t x=0   1 −tb 1 lim − e + b→∞ t t 1 , t

= = =

where the integral is obtained by the fundamental theorem of calculus. −tx , so | ∂f Let f (x, t) = e−tx for t ∈ [ 21 , 32 ]. Then ∂f ∂t (x, t) = −xe ∂t (x, t)| ≤ x −x − xe 2 for all t. Let g(x) = xe 2 . 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), we x have that x < ex/4 . Define h(x) = g(x) if x ≤ 8 ln 4 and h(x) = e− 4 if x > 8 ln 4. Then g ≤ h, and h is clearly integrable, and thus by Corollary 5.4, g is also integrable. But then we have, by Theorem 5.9, that   Z ∞  1 d 1 d −tx − 2 = = e dx t dt t dt 0 ! Z d −tx e dλ(x) = dt [0,∞) Z ∂ −tx
= e dλ(x) [0,∞) ∂t Z = −xe−tx dλ(x) [0,∞) Z ∞

= −

xe−tx dλ(x).

0

Taking t = 1 and multiplying by −1 yields the result for n = 1. R∞ Now, proceed by induction on n. Suppose it is known that 0 xn−1 e−tx dx = (n−1)! Let fn (x, t) = xn−1 e−tx for t ∈ [ 21 , 32 ]. As above, we have tn . ∂fn n −tx n −x n 2. , so | ∂f Again, for x sufficiently ∂t (x, t) = −x e ∂t (x, t)| ≤ x e

2

x

large, we have that xn < ex/4 , so as above, we have that xn e− 2 is integrable on [0, ∞). Then as above, we have   Z ∞  d (n − 1)! d n! n−1 −tx = x e dx − n+1 = t dt tn dt 0 Z ∂ n−1 −tx = (x e )dλ(x) [0,∞) ∂t Z ∞ xn e−tx dx. = − 0

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 7→ f (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 ∂f ∂t (x, t1 ) exists, (x,t1 ) and that there exists an integrable function g on X such that f (x,t)−f ≤ t−t1 g(x) for x ∈ X and t ∈ [a, b], t 6= t1 . Then  Z  Z d ∂f = f (x, t) dµ(x) (x, t1 ) dµ(x). dt ∂t t=t1 Proof. Choose tn to be any sequence with tn → t1 , tn 6= t1 (we may (x,t1 ) start at n = 2 to avoid confusion). Put hn (x) = f (x,tntn)−f , so by −t1 ∂f hypothesis |hn (x)| ≤ g(x) for all x, and hn (x) → ∂tR(x, t1 ). By the dominated convergence theorem, then, we have limn→∞ X hn (x)dµ(x) = R ∂f (x, t1 )dµ(x). On the other hand, X ∂t Z Z f (x, tn ) − f (x, t1 ) lim hn (x)dµ(x) = lim dµ(x) n→∞ X n→∞ tn − t1 Z  Z 1 = lim f (x, tn )dµ(x) − f (x, t1 )dµ(x) n→∞ tn − t1  Z  d , = f (x, t)dµ(x) dt t=t1

as desired. 5T. Let f be a F-measurable function on X to R. For n ∈ Z+ , let {fn } be theR sequence of Rtruncates of f . If f is integrable with respect to µ, R then f dµ = lim fn dµ. Conversely, if sup |fn | dµ < ∞, then f is integrable. Proof. Note that for all R n, we have R |fn | ≤ f , and thus if f is integrable, the DCT implies that fn dµ → f dµ. For the converse, note that |fn | is a monotonically increasing sequence of R nonnegative R functions with limit |f |, and Rthus the MCT implies that |fn | dµ → |f | dµ. Thus, by hypothesis, |f | dµ < ∞, and therefore |f | and thus f are integrable. 4O. Fatou’s Lemma has an extension to a case where R the fn take on negative values. Let h ∈ M + (X, F), and supposeR that h dµ < ∞. If {f R n } is a sequence in M (X, F) and −h ≤ fn then lim inf fn dµ ≤ lim inf fn dµ. 3

Proof. Note that fn + h ≥ 0. Thus we have Z Z Z lim inf fn dµ + h dµ = (lim inf fn + lim inf h) dµ Z = (lim inf(fn + h)) dµ Z ≤ lim inf (fn + h) dµ  Z Z fn dµ + h dµ = lim inf Z Z = lim inf fn dµ + h dµ. Subtracting

R

h dµ yields the result.

Also, complete the following: 1. Compute the following limits. Justify each computational step using convergence theorems and/or calculus. R∞ (a) limn→∞ 0 1+nn2 x2 dx.
−n R∞ (b) limn→∞ 0 1 + nx log(2 + cos( nx ))dx
Rn (c) limn→∞ −n f 1 + nx2 g(x)dx, where g : R → R is (Lebesgue) integrable and f : R → R is bounded, measurable, and continuous at 1. Solution. (a) We consider this as two integrals, over [0, 1] and [1, ∞). Note that on [0, 1], we have Z 1 n 1 dx = lim [arctan(nx)]0 lim n→∞ n→∞ 0 1 + n2 x2 π = lim arctan(n) = . n→∞ 2 For the other part, notice that 1+nn2 x2 ≤ n2nx2 ≤ nx1 2 ≤ x12 , and R∞ moreover, 1 x12 = 1 < ∞, so x12 ∈ L([1, ∞), L, λ). Therefore, by the dominated convergence theorem, we have Z ∞ Z n n lim dx = lim dλ(x) 2 2 2 2 n→∞ 1 n→∞ 1+n x [1,∞) 1 + n x   Z n dλ(x) = lim 2 2 [1,∞) n→∞ 1 + n x Z = 0dλ(x) = 0. [1,∞)

R∞

n π 1+n2 x2 dx = 2 . x −n ≤ e−x , and as cos( nx ) ≤ 1, n)
−n Thus, 1 + nx log(2+cos( nx )) ≤

Therefore, we obtain limn→∞

0

(b) Note that (1 + we have log(2 + x cos( n )) ≤ log 3. (log 3)e−x , which is clearly integrable. Therefore, by the dominated convergence theorem, we have  Z ∞ Z ∞   x    x  x −n x −n lim 1+ log 2 + cos dx = lim 1+ log 2 + cos dx n→∞ 0 n→∞ n n n n Z0 ∞ = e−x log(3)dx 0

= 4

log(3)

(c) Let us rewrite the integral as Z n  Z  x x f 1 + 2 g(x)dx = lim lim f 1 + 2 χ[−n,n] g(x) dλ(x). n→∞ −n n→∞ R n n
Let M be such that |f (x)| < M for all x. Then f 1 + nx2 χ[−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 Z  Z n  x x f 1 + 2 g(x)dx = lim f 1 + 2 χ[−n,n] g(x) dλ(x) lim n→∞ R n→∞ −n n n Z    x = lim f 1 + 2 χ[−n,n] g(x) dλ(x) n→∞ n ZR = f (1)χR g(x)dλ(x) R Z = f (1) g(x)dλ(x), R

where the limit exists because f is continuous at 1.

Rt 2 2. Let f (t) = 0 e−x dx. We will use DCT to evaluate limt→∞ f (t), even without being able to evaluate the indefinite integral. (a) Put h(t) = f (t)2 and g(t) = −g 0 (t). (b) Show that h(t) + g(t) =

π 4

R1 0

2

e−t (1+x 1+x2

2)

dx. Show that h0 (t) =

for all t. Z

(c) Use the previous parts to conclude that lim f (t) = t→∞ √ π . 2

∞

2

e−x dx =

0

2

Solution. (a) Note that h0 (t) = 2f (t)f 0 (t) = 2e−t f (t), by the fundamental theorem of calculus. h −t2 (1+x2 ) i −t2 (1+x2 ) ∂ e For g 0 (t), let f (x, t) = e 1+x2 , and note that ∂t = 1+x2 2

2

2

2

2

2

−2te−t (1+x ) = −2te−t e−t x . Put η(t) = 2te−t . Note that η 0 (t) = 2 (2 − 4t2 )e−t ,pand thus bypthe first derivative test, η has a global p a global minimum of − 2/e maximum of 2/e at t = 1/2, and p √ 2 at t = − 12. Therefore, −2te−t ≤ 2/e =: C for all t. Thus, 2 2 ∂f ∂t ≤ Ce−t x ≤ C, which is clearly integrable on [0, 1]. Therefore, by Theorem 5.9, "Z # ! Z 1 2 2 1 −t2 (1+x2 ) d e ∂ e−t (1+x ) dx = dx dt 0 1 + x2 1 + x2 0 ∂t Z 1 2 2 = −2te−t (1+x ) dx 0

= =

2

−2te−t −2e

−t2

Z

Z

0 t

1

e−t

2

x2

dx

2

e−u du

0

=

2

−2e−t f (t) = −h0 (t),

as desired. 5

using the substitution u = tx

(b) As (h + g)0 = 0 for all t, we have that h + g is constant. Note that R1 1 π h(0) = 0. Moreover, g(0) = 0 1+x 2 dx = arctan(1) = 4 . Therefore, π (h + g)(0) = 4 , and as h + g is constant, the result follows. 1 (c) Note that |f (x, t)| ≤ 1+x 2 for all x, t, and thus by the dominated convergence theorem, we have Z 1 2 2 e−t (1+x ) dx lim g(t) = lim t→∞ 1 + x2 0 t→∞ Z 1 = 0 dx = 0. 0

But then limt→∞ h(t) =

π 4,

and thus limt→∞ f (t) =

p

√

π/4 =

π 2 .

Extras: 5A. If f ∈ L(X, F, µ) 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 σ-finite measure. R Proof. As |f | ∈ L, we have that |f | dµ = A < ∞. Let Ea = {x ∈ X | |f (x)| > a}. Then we have |f | = |f |χEa + |f |χEac ≥ aχEa + |f |χEac , and thus Z Z A = |f | dµ ≥ aµ(Ea ) + |f | dµ ≥ aµ(Ea ). Eac

Therefore, µ(Ea ) ≤ A/a. Moreover, {x ∈ X | f (x) 6= 0} = ∪n∈Z+ En , and thus the set has σ-finite measure. 5B. If f is a F-measurable real-valued function and if f (x) = 0 for µ-almost R all x ∈ X, then f ∈ L(X, F, µ) and f dµ = 0. R Proof. Note that |f | ∈ M + is µ-almost 0, and thus |f | dµ R R = 0. This f dµ ≤ |f | dµ = 0, implies that |f | ∈ L, and thus f ∈ L. Moreover, R and thus f dµ = 0. 5D. If f ∈ L(X, F, R µ) and  > 0, then there exists a measurable simple function ϕ such that |f − ϕ| dµ < . Proof. As f + and f − are M + , we have measurable simple functions ϕ+ R R and ϕ− such that ϕ+ ≤ f + , ϕ− ≤ f − , and (f + − ϕ+ ) dµ < 2 , (f − − ϕ− ) dµ < 2 . Let ϕ = ϕ+ − ϕ− . Then we have ϕ is simple, and Z Z Z
|f −ϕ| dµ = |f + −ϕ+ −(f − −ϕ− )| dµ ≤ |f + − ϕ+ | + |(f − − ϕ− )| dµ < .

5F. If f belongs to L, it does not follow that f 2 belongs to L. R R1 Proof. Let f = √1x on [0, 1]. Then we have [0,1] f dλ = 0 √1x dx = √ 1 2 x|0 = 2. However, f 2 = x1 does not have a finite integral on [0, 1], so f2 ∈ / L. 6

5G. Suppose that f ∈ L(X, F, µ), and that its indefinite integral is λ(E) = R f dµ for E ∈ F. Show that λ(E) ≥ 0 for all E ∈ F if and only if E f (x) ≥ 0 for almost all x ∈ X. Moreover, λ(E) = 0 for all E if and only if f (x) = 0 for almost all x ∈ X. Proof. Write f = f + − f − . Suppose λ(E) >≥ 0 for all E ∈ F. Let  > 0, and Rput E = {x ∈ X | f (x) < −}. As f is measurable, E ∈ F. Moreover, E f dµ ≤ −µ(E ) ≥ 0, and thus µ(E ) = 0. But then {x ∈ X | f (x) < 0} = ∪n∈Z+ E1/n has measure 0, and therefore f ≥ 0 µ-almost everywhere. On the other hand, if f ≥ 0 µ-almost everywhere,R then for all E ∈ F, f χE ≥ 0 µ-almost everywhere, and thus by 5B, f χE dµ ≥ 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, µ), and let λ1 , λ2 be their indefinite integrals. Show that λ1 (E) = λ2 (E) for all E ∈ F if and only if f1 (x) = f2 (x) for almost all x ∈ X. Proof. Note that λ1 (E) = λ2 (E) for all E if and only if (λ1 − λ2 )(E) = 0 for all E, if and only if f1 − f2 = 0 µ-almost everywhere (by problem G).

7