Math 426: 426: Homew Homework ork 4 Mary Radcliffe due 2 May 2014
In Bartle: Bartle: 5C. If f f L( L (X, , µ) and g and g is a -measurable -measurable real-valued function such that f ( f (x) = g( g (x) almost everywhere, then g then g L( L (X, , µ) and f dµ = dµ = g dµ. dµ.
∈ ∈
F
F
∈
F
− g(x)|. Then h Then h ∈ M + (X, F ), ), and h and h = 0 almost everywhere. Therefore, h is integrable and h dµ = dµ = 0. But this this implies implies − g) dµ ≤ that f −g is integrable, and thus g is integrable. Moreover, (f − − g) dµ = h dµ = dµ = 0, and thus (f − dµ = 0. The result immediately follows by Proof. Let h Let h((x) = f ( f (x)
|
linearity of integration.
5E. If f f L and g is a bounded, measurable function, then f g
∈ L. L .
∈ ∈
Proof. Let M Let M be be such that g M . M . Note that by linearity of integration, we have that M f L, L , and f g M f for all x all x.. Then by Corollary 5.4, f g is integrable.
| | ≤ | | ≤ | |
∈ ∈
5I. If f is f is a complex-valued function on X on X such such that Re f Re f and and Im f Im f belong belong to L(X, , µ), we say that f is f is integrable and define f dµ = dµ = Re f Re f dµ + dµ + i Im f Im f dµ. dµ. Let f b f bee a complex-v complex-valued alued measurable measurable function. function. Show Show that f is f is integrable if and only if f is integrable, in which case f dµ f dµ. dµ .
F | |
| | |
≤
Proof. Let us write f = f 1 + if + if 2 , so that f 1 = Re f Re f and f and f 2 = Im f Im f .. We have that f is f is measurable if and only if f 1 and f 2 are measurable, and 1/2 f = (f 1 )2 + (f (f 2 )2 . Thus Thus,, f f is measurable implies that f is also measurable.
||
| |
√
R, then Notice, Notice, if a, b a2 + b2 = (a + b)2 2ab (a + b)2 = a + b + b a + b . Theref Therefore ore,, if f is f is integrable, then f is a measurable function such that f f 1 + f 2 , and thus thus by Corollar Corollary y 5.4 5.4,, f is integrable.
|
∈ |≤| | ||
| | ≤ | | | |
−
| |
≤
||
Conversely, if f is integrable, note that f 1 < f and f 2 < f , so we have f have f 1 and f and f 2 are real-valued integrable functions by Corollary 5.4. But then by definition f definition f is also integrable.
| | |
| | | |
| | | |
For the inequality, write f dµ = dµ = re iθ , so that f dµ = r. r . Let g(x) = − iθ e f ( f (x), so that g (x) = g1 (x) + ig + ig 2 (x), and g dµ = e −iθ f ( f (x) dµ = − iθ e f ( f (x) dµ = dµ = r. r . But then then g 2 dµ = dµ = 0, so g2 = 0 almost everywhere. Moreover, f = e−iθ f = g = g1 almost everywhere.
| | |
Therefore, we have
f dµ = r = r =
|| | | | | |
g1 dµ = dµ =
1
g1 dµ
≤ |
g1 dµ = dµ =
|
| |
f dµ,
since g 1 is real valued and we may use Theorem 5.3.
5P. Let f n L(X, , µ), and suppose that f n converges to a function f . Show that if lim f n f dµ = 0, then f dµ = lim f n dµ.
∈
F | − |
{ } | |
Proof. Note that as f n
|
|
| − f | is integrable for n sufficiently large, we have f n − f is integrable, and thus by linearity f is integrable. Moreover, for all n, we have |f n | ≤ |f n − f | + |f |, so |f n |−|f | ≤ |f n − f |. Integrating and taking limits on both sides, we have lim |f n | dµ − |f | dµ ≤ lim |f n − f | dµ = 0, and thus lim |f n | dµ = |f | dµ
∞
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 ∞ and to obtain the formula 0 xn e−x dx = n!.
≥
≤
Proof. Now,
∞
b
e−tx dx =
lim
b→∞
0
=
lim
b→∞
=
lim
b→∞
1 , t
=
e−tx dx
0
b
− − 1 −tx e t x=0 1 −tb 1 e + t t
where the integral is obtained by the fundamental theorem of calculus. Let f (x, t) = e −tx for t [ 12 , 32 ]. Then ∂f xe−tx , so ∂f ∂t (x, t) = ∂t (x, t) 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 > 8ln4 is sufficient), we have that x < ex/4 . Define h(x) = g(x) if x 8ln4 and h(x) = e− 4 if x > 8ln4. 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
∈
x
−
x
−
1 t
= = = =
d dt d dt
∞
|≤ ∞ x
≤
≤
1 d = 2 t dt
|
e−tx dx
0
e−tx dλ(x)
[0,∞)
∂ −tx e dλ(x) [0,∞) ∂t
− −
xe−tx dλ(x)
[0,∞)
=
∞
xe−tx dλ(x).
0
Taking t = 1 and multiplying by
−1 yields the result for n = 1. ∞
Now, proceed by induction on n. Suppose it is known that 0 xn−1 e−tx dx = (n−1)! n−1 −tx e for t [ 12 , 23 ]. As above, we have t . Let f n (x, t) = x ∂f xn e−tx , so ∂f xn e− 2 . Again, for x sufficiently ∂t (x, t) = ∂t (x, t) n
n
−
|
|≤
n
2
∈
x
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
∞
n!
− tn+1
d = dt
(n
− 1)! tn
d dt
= = =
∞
n−1
x
e−tx dx
0
∂ n−1 −tx (x e )dλ(x) [0,∞) ∂t
−
∞
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 f (x, t) is -measurable for each t [a, b]. Suppose that for some t0 , t1 [a, b], the function x f (x, t0 ) is integrable on X , that ∂f ∂t (x, t1 ) exists,
F
× ∈
→
→ ∈
and that there exists an integrable function g on X such that g(x) for x X and t [a, b], t = t 1 . Then
∈
∈
d dt
f (x, t) dµ(x)
=
t=t1
f (x,t)−f (x,t1 ) t−t1
∂f (x, t1 ) dµ(x). ∂t
Proof. Choose tn to be any sequence with tn
≤
→ t1, tn = t1 (we may
f (x,t1 ) start at n = 2 to avoid confusion). Put hn (x) = f (x,tt )− , so by −t1 ∂f hypothesis hn (x) g(x) for all x, and hn (x) ∂t (x, t1 ). By the dominated convergence theorem, then, we have lim n→∞ X hn (x)dµ(x) = ∂f (x, t1 )dµ(x). On the other hand, X ∂t n
n
|
lim
n→∞
|≤
→
f (x, tn ) f (x, t1 ) dµ(x) n→∞ tn t1 1 = lim f (x, tn )dµ(x) n→∞ tn t1 d = f (x, t)dµ(x) , dt t=t1
hn (x)dµ(x) =
X
lim
− −
−
−
f (x, t1 )dµ(x)
as desired. Z+ , let f n 5T. Let f be a -measurable function on X to R. For n be the sequence of truncates of f . If f is integrable with respect to µ, then f dµ = lim f n dµ. Conversely, if sup f n dµ < , then f is integrable.
F
∈
|
Proof. Note that for all n, we have f n
| →
the DCT implies that f n dµ
|
{ }
∞
| ≤ f , and thus if f is integrable,
f dµ.
For the converse, note that f n is a monotonically increasing sequence of nonnegative functions with limit f , and thus the MCT implies that f n 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 the f n take on negative values. Let h M + (X, ), and suppose that h dµ < . If f n is a sequence in M (X, ) and h f n then liminf f n dµ liminf f n dµ.
∈
F F − ≤
3
∞ { } ≤
Proof. Note that f n + h
≥ 0. Thus we have
liminf f n dµ +
h dµ =
(lim inf f n + lim inf h) dµ
=
(lim inf(f n + h)) dµ
≤ lim inf
(f n + h) dµ
= liminf
f n dµ +
= liminf Subtracting
h dµ
f n dµ + h dµ.
h dµ yields the result.
Also, complete the following: 1. Compute the following limits. Justify each computational step using convergence theorems and/or calculus.
(b) (c)
∞
n 1+n2 x2 dx. −n ∞ limn→∞ 0 1 + nx log(2 + cos( nx ))dx n limn→∞ −n f 1 + nx2 g(x)dx, where g : R R is (Lebesgue) inteR is bounded, measurable, and continuous at grable and f : R
(a) limn→∞
0
→
→
1.
(a) We consider this as two integrals, over [0, 1] and [1, Note that on [0, 1], we have
Solution.
1
lim
n→∞
0
n dx = 1 + n2 x2
∞).
1
lim [arctan(nx)]0
n→∞
=
lim arctan(n) =
n→∞
π . 2
n 1 1 For the other part, notice that 1+nn2 x2 n2 x2 nx2 x2 , and ∞ 1 moreover, 1 x2 = 1 < , so x12 L([1, ), , λ). Therefore, by the dominated convergence theorem, we have
lim
n→∞
∞
1
∞
≤
∈
n dx = 1 + n2 x2
∞ L
≤
≤
n dλ(x) n→∞ [1,∞) 1 + n2 x2 n = lim dλ(x) 2 2 [1,∞) n→∞ 1 + n x lim
=
0dλ(x) = 0.
[1,∞)
∞
n π 1+n2 x2 dx = 2 . e−x , and as cos( nx ) 1, we have log(2 + −n 1 + nx log(2+cos( nx )) (log 3)e−x , which
≤ ≤
Therefore, we obtain limn→∞
0
(b) Note that (1 + nx )−n cos( nx )) log 3. Thus, is clearly integrable. Therefore, by the dominated convergence theorem, we have lim
n→∞
∞
0
x −n x 1+ log 2 + cos n n
≤
dx =
≤
∞
0
=
∞
lim
n→∞
= log(3)
x −n x 1+ log 2 + cos n n
e−x log(3)dx
0
4
dx
(c) Let us rewrite the integral as n
x lim f 1 + 2 g(x)dx = lim n→∞ −n n→∞ n
x f 1 + 2 χ[−n,n] g(x) dλ(x). n R
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
|
|
|
n
|
∈
| | ∈
x lim f 1 + 2 g(x)dx = n→∞ −n n
| |∈
≤
x χ[−n,n] g(x) dλ(x) n→∞ R n2 x = lim f 1 + 2 χ[−n,n] g(x) dλ(x) →∞ n n R
lim
=
f 1 +
f (1)χR g(x)dλ(x)
R
= f (1)
g(x)dλ(x),
R
where the limit exists because f is continuous at 1.
t
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 (t).
−
(b) Show that h(t) + g(t) =
π 4 for
1 e−t2 (1+x2 ) dx. 1+x2 0
Show that h (t) =
all t.
(c) Use the previous parts to conclude that lim f (t) =
√ π 2
t→∞
∞
2
e−x dx =
0
. 2
(a) Note that h (t) = 2f (t)f (t) = 2e−t f (t), by the fundamental theorem of calculus. 2 2 − 2 (1+ 2 ) ∂ e− (1+ ) For g (t), let f (x, t) = e 1+x2 , and note that ∂t = 1+x2
Solution.
t
2
2
2
x
2 2
2
t
x
−2te−t (1+x ) = −2te−t e−t x . Put η(t) = 2te−t . Note that η (t) = (2 − 4t2 )e−t , and thus by the first derivative test, η has a global maximum of 2/e at t = 1/2, and a global minimum of − 2/e √ at t = − 12. Therefore, −2te−t ≤ 2/e =: C for all t. Thus, 2
≤
2 2
∂f ∂t
− − −
2
C e−t x C , which is clearly integrable on [0, 1]. Therefore, by Theorem 5.9, d dt
1
0
2
≤ 2
e−t (1+x ) dx 1 + x2
1
=
0
1
=
2
2
e−t (1+x 1 + x2
∂ ∂t
2
2te−t
(1+x2 )
)
dx
dx
0
1
2
2te−t
=
2
2e−t
=
0 t
e−t
2 2
x
dx
2
e−u du
0
=
2
−2e−t f (t) = −h(t),
as desired. 5
using the substitution u = tx
(b) As (h + g) = 0 for all t, we have that h + g is constant. Note that 1 1 π h(0) = 0. Moreover, g(0) = 0 1+x 2 dx = arctan(1) = 4 . Therefore, (h + g)(0) = π4 , and as h + g is constant, the result follows.
1 (c) Note that f (x, t) 1+x2 for all x, t, and thus by the dominated convergence theorem, we have
|
|≤
1
lim g(t) =
t→∞
0
1
=
2
2
e−t (1+x ) lim dx t→∞ 1 + x2 0 dx = 0.
0
But then limt→∞ h(t) =
π 4,
and thus limt→∞ f (t) =
π/4 =
√ π 2
.
Extras: 5A. If f L(X, , µ) and a > 0, show that the set x X f (x) > a has finite measure. In addition, the set x X f (x) = 0 has σ-finite measure.
∈
F
{ ∈ | | { ∈ | }
Proof. As f
|
}
| | ∈ L, we have that |f | dµ = A < ∞ . Let E a = { x ∈ X | |f (x)| > a}. Then we have |f | = |f |χE + |f |χE ≥ aχ E + |f |χE , and thus
A =
| |
f dµ
c a
a
≥ aµ(E a) +
≤ A/a. Moreover, {x ∈ X | f (x) = 0} = ∪n∈
| |
f dµ
E ac
c a
a
≥ aµ(E a).
Therefore, µ(E a )
Z+
E n , and thus the set has σ-finite
measure.
5B. If f is a -measurable real-valued function and if f (x) = 0 for µ-almost all x X , then f L(X, , µ) and f dµ = 0.
∈
F
∈ F Proof. Note that | f | ∈ M + is µ-almost 0, and thus |f | dµ = 0. This implies that |f | ∈ L, and thus f ∈ L. Moreover, f dµ ≤ |f | dµ = 0,
and thus f dµ = 0.
5D. If f L(X, , µ) and > 0, then there exists a measurable simple function ϕ such that f ϕ dµ < .
∈
F | − |
Proof. As f + and f − are M + , we have measurable simple functions ϕ+
and ϕ − such that ϕ + f + , ϕ − f − , and (f + ϕ+ ) dµ < 2 , ϕ− ) dµ < 2 . Let ϕ = ϕ+ ϕ− . Then we have ϕ is simple, and
| − | |
− ϕ−)|
≤
f ϕ dµ =
−
≤
f + ϕ+ (f − ϕ− ) dµ
− − − |
− ≤ | − f +
ϕ+ + (f −
| |
(f −
−
5F. If f belongs to L, it does not follow that f 2 belongs to L. 1 Proof. Let f = √ on [0, 1]. Then we have x
√ 1 2 x|0 = 2. f 2 ∈ / L.
However, f 2 =
1 x does
6
[0,1]
f dλ =
1 1 √ x dx 0
=
not have a finite integral on [0, 1], so
dµ < .
5G. Suppose that f L(X, , µ), and that its indefinite integral is λ(E ) = f dµ for E . Show that λ(E ) 0 for all E 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 .
∈ ∈ F
≥
F ∈ ∈
≥
∈ F
Proof. Write f = f +
− f −. Suppose λ(E ) >≥ 0 for all E ∈ F . Let > 0, and put 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∈ E 1/n has measure 0, and therefore f ≥ 0 µ-almost
Z+
everywhere.
On the other hand, if f 0 µ-almost everywhere, then for all E , 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 f (x) > .
| |
|
}
∈ F
{x ∈
5H. Suppose that f 1 , f 2 L(X, , µ), and let λ1 , λ2 be their indefinite integrals. Show that λ 1 (E ) = λ2 (E ) for all E if and only if f 1 (x) = f 2 (x) for almost all x X .
∈
F
∈ F
∈
Proof. Note that λ 1 (E ) = λ 2 (E ) for all E if and only if (λ1
− λ2)(E ) = 0 − f 2 = 0 µ-almost everywhere (by problem
for all E , if and only if f 1 G).
7
