Homework 3 Spring 2014 Solutions

Prévia do material em texto

Math 426: Homework 3
Mary Radcliffe
due 23 April 2014
In Bartle:
4A. If the simple function ϕ ∈ M+ has the representation ϕ = ∑mk=1 bkχFk ,
where bk ∈ R and Fk ∈ F , prove that
∫
ϕ dµ =
∑m
k=1 bkµ(Fk).
Proof. For 1 ≤ k ≤ m, let ψk = bkχFk + 0χX\Fk . Then ψk is in standard
form, so
∫
ψk dµ = bkµ(Fk). Moreover, ϕ =
∑m
k=1 ψk, and thus since
integration is linear, we have
∫
ϕ dµ =
∑∫
ψk dµ =
∑m
k=1 bkµ(Fk).
4B. The sum, scalar multiple, and product of simple functions are simple func-
tions.
Proof. Let ϕ =
∑m
k=1 akχEk and ψ =
∑n
j=1 bkχFk be two simple func-
tions, and let c ∈ R. Note that cϕ = ∑mk=1(cak)χEk is clearly simple.
Now, as the only possible values of ϕ are a1, a2, . . . , am and the only pos-
sible values of ψ are b1, b2, . . . , bn, we have that the only possible values of
ϕ+ψ are a1+b1, a1+b2, . . . , a1+bn, a2+b1, . . . , am+bn and the only pos-
sible values of ϕψ are a1b1, a1b2, . . . , a1bn, a2b1, . . . , ambn, and thus both
ϕ + ψ and ϕψ take only finitely many values. By definition, then, both
ϕ+ ψ and ϕψ are simple functions.
4G. Let X = Z+, F be the set of all subsets of X, and µ be counting measure.
If f is a nonnegative function on X, then f ∈ M+(X,F) and ∫ f dµ =∑∞
n=1 f(n).
Proof. Let fk = fχ[k], where [k] = {1, 2, . . . , k}. Then fk is a simple
function, as it can take at most k values, and fk can be written in the rep-
resentation fk =
∑k
i=1 f(i)χ{i}, so
∫
fk =
∑k
i=1 f(i)µ({i}) =
∑k
i=1 f(i).
Moreover, fk ≤ fk+1 for all k, and thus by the MCT, we have
∫
f dµ =
lim
∫
fk dµ = lim
∑k
i=1 f(i) =
∑∞
i=1 f(i).
4K. If (X,F , µ) is a finite measure space, and {fn} is a real-valued sequence
in M+(X,F) which converges uniformly to a function f , then f ∈ M+
and
∫
f dµ = lim
n→∞
∫
fn dµ.
Proof. By Cor. 2.10, f ∈M , and as fn ≥ 0 for all n, we also have f ≥ 0,
and thus f ∈M+. Let � > 0, and let N be sufficiently large that if n ≥ N ,
we have |fn(x) − f(x)| < � for all x ∈ X. Choose n ≥ N , and let ϕ be a
simple function with ϕ ≤ fn. Define ψ as
ψ(x) =
{
0 if ϕ(x) < �
ϕ(x)− � if ϕ(x) ≥ � .
1
https://www.coursehero.com/file/9770575/Homework-3-Spring-2014-Solutions/
Th
is s
tud
y r
eso
urc
e w
as
sha
red
 vi
a C
ou
rse
He
ro.
com
As ϕ takes only finitely many values, we have also that ψ takes only
finitely many values, and if ϕ =
∑n
i=1 aiχEi +
∑m
j=1 bjχFj , where ai ≥ �
for all i and bj < � for all j, then ψ =
∑n
i=1(ai − �)χEi . Moreover, as
|fn(x)− f(x)| < � for all x, we have ψ < f . Therefore,∫
f dµ ≥
∫
ψ dµ
=
n∑
i=1
(ai − �)χEi
=
n∑
i=1
aiχEi +
m∑
j=1
bjχFj −
m∑
j=1
bjχFj −
n∑
i=1
�χEi
≥
∫
ϕ dµ−
m∑
j=1
�χFj −
n∑
i=1
�χEi
=
∫
ϕ dµ− �µ(X)
4L. Let X be a finite closed interval [a, b] in R, and let F be the collection of
Borel sets in X, and λ be Lebesgue measure on F . If f is a nonnegative
continuous function on X, show that
∫
f dλ =
∫ b
a
f(x)dx.
Proof. Recall that we can define
∫ b
a
f(x)dx = limn→∞
∑2n−1
k=0 mn,k
b−a
2n ,
where
mn,k = min
a+(b−a) k2n≤x≤a+(b−a) k+12n
f(x).
That is, we take the limit of the step function defined on a partition that
divides [a, b] into 2n equal intervals. As f is continuous, all mn,k are finite,
and this limit is guaranteed to exist and equal the integral1 of f .
Define ϕn to be the simple function which takes value mn,k on [a + (b −
a) k2n , a+ (b− a)k+12n ] and 0 elsewhere. Note that the ϕn are an increasing
sequence of nonnegative functions with limit f . Therefore, by the MCT,∫
f dλ = limn→∞
∫
ϕn dλ = limn→∞
∑2n−1
k=0 mn,k
b−a
2n =
∫ b
a
f dx.
4M. Let X = [0,∞), F the set of Borel subsets of X, and λ be Lebesgue
measure on F . If f is a nonnegative continuous function on X, show that∫
f dλ = lim
b→∞
∫ b
0
f(x)dx.
Proof. Take {an} to be any nonnegative sequence that increases to∞. De-
fine fn = fχ[0,an]. Then the fn are an increasing sequence of nonnegative
functions with limit f , and thus by the MCT, we have∫
f dλ = lim
n→∞
∫
fn dλ = lim
n→∞
∫ an
0
f dx,
where we are able to pass from the Lebesgue integral to the Riemann
integral on the bounded interval [0, an] by the previous problem. As
this holds for any sequence tending to ∞, we must have that ∫ f dλ =
limb→∞
∫ b
0
f dx.
1If you took 425 with me last quarter, we did not define the Riemann integral this way.
However, this interpretation can be proven to be equivalent to ours. You should check that you
can prove it, using the It Doesn’t Matter Lemma (6.7) from Baby Rudin and an appropriately
defined refinement of an arbitrary partition.
2
https://www.coursehero.com/file/9770575/Homework-3-Spring-2014-Solutions/
Th
is s
tud
y r
eso
urc
e w
as
sha
red
 vi
a C
ou
rse
He
ro.
com
4S. If f ∈ M+(X,F) and ∫ f dµ < ∞, then for any � > 0 there exists a set
E ∈ F such that µ(E) <∞ and ∫ f dµ = ∫
E
f dµ+ �.
Proof. Let � > 0. For n ∈ Z+, define En = {x ∈ X | f(x) ≥ 1n}. Note
that as
∫
f dµ ≥ ∫
En
f dµ ≥ 1nµ(En), we have that µ(En) is finite for
all n. Moreover, if fn = fχEn then as the En are increasing sets, the fn
are increasing positive functions with limit f , and therefore by the MVT,∫
fndµ→
∫
f dµ. Therefore, ∃N such that ∣∣∫ fNdµ− ∫ f dµ∣∣ < �.
Therefore, taking E = EN yields the result.
4T. Suppose that {fn} ⊂ M+(X,F), that {fn} converges to f , and that∫
f dµ = lim
n→∞
∫
fn dµ <∞. Prove that
∫
E
f dµ = lim
n→∞
∫
E
fn dµ for all
E ∈ F .
Proof. Let us suppose, to the contrary, that there exists some E with∫
E
f dµ 6= limn→∞
∫
E
fn dµ. Note that by Fatou’s Lemma, this implies
that
∫
E
f dµ < lim inf
∫
E
fn dµ. Then we have the following.∫
f dµ =
∫
E
f dµ+
∫
X\E
f dµ
< lim inf
∫
E
fn dµ+ lim inf
∫
X\E
fn dµ
≤ lim inf
(∫
E
fn dµ+
∫
X\E
fn dµ
)
= lim
∫
X
fn dµ =
∫
f dµ,
where strict inequality follows in line 2 by Fatou’s Lemma and because
both lim inf
∫
E
fn dµ and lim inf
∫
X\E fn dµ are finite. This is a clear
contradiction, and thus we must have that
∫
E
f dµ = lim
∫
E
fn dµ for all
E ∈ F .
Also, complete the following:
1. (a) Show, by example, that if {fn} ⊂M+ is a sequence of functions with
fn(x) ≥ fn+1(x) (i.e., a decreasing sequence) with fn → f , that it
may not be true that
∫
fdµ = lim
∫
fndµ (and thus there is no MCT
for decreasing sequences).
Proof. Define fn = χ[n,∞]. Then fn is decreasing, and fn → 0 = f .
However,
∫
f dλ = 0 6=∞ = lim ∫ fn dλ.
(b) Let (X,F , µ) be a finite measure space. Prove that if {fn} ⊂M+(X,F)
is a decreasing sequence of uniformly bounded functions with fn → f ,
then
∫
fdµ = lim
∫
fndµ.
Proof. Let M be a uniform bound for fn. Define gn = M − fn.
Then gn is an increasing sequence with limit M − f , and thus by the
MCT,
∫
(M − f)dµ = lim ∫ (M − fn)dµ. Note that ∫ Mdµ is finite
since µ is finite, so the result follows by subtracting
∫
Mdµ from both
sides.
(c) Show that if {fn} ⊂ M+ is a decreasing sequence of functions with
fn → f and
∫
fndµ <∞ for some n, then lim
∫
fndµ =
∫
fdµ.
3
https://www.coursehero.com/file/9770575/Homework-3-Spring-2014-Solutions/
Th
is s
tud
y r
eso
urc
e w
as
sha
red
 vi
a C
ou
rse
He
ro.
com
Proof. Put gk = fn − fn+k−1 for all k ≥ 1. Then gk is an increasing
sequence of functions, with gk → fn−f . Thus, by the MCT, we have∫
(fn − f)dµ = limk→∞
∫
(fn − fn+k−1)dµ, and the result follows by
subtracting the (finite) integral
∫
fn dµ from both sides.
2. Prove that if X is a finite set under counting measure µ, then for every
sequence of function fn on X with fn → f , we have
∫
fndµ→
∫
fdµ.
Proof.WOLOG, we assume that X = {1, 2, . . . , N}. Then ∫ fn dµ =∑N
k=1 fn(k) and
∫
f dµ =
∑N
k=1 f(k). Taking a limit, we have lim
∫
fn dµ =
lim
n→∞
N∑
k=1
fn(k) =
N∑
k=1
lim
n→∞ fn(k) =
N∑
k=1
f(k), as desired.
4
https://www.coursehero.com/file/9770575/Homework-3-Spring-2014-Solutions/
Th
is s
tud
y r
eso
urc
e w
as
sha
red
 vi
a C
ou
rse
He
ro.
com
Powered by TCPDF (www.tcpdf.org)