SOLUTIONS OF HW8
MINGFENG ZHAO
April 20, 2011
1. [Problem 11, in Page 92] Let µ be a positive measure on X. A collection of functions {fα }α∈A ⊂
L1 (X, µ) is called uniformly integrable if for every > 0, there exists δ > 0 such that whenever
µ(E) < δ, we have
Z
fα dµ < ,
for all α ∈ A
E
Show that
a. Any finite subset of L1 (X, µ) is uniformly integrable.
1
1
1
b. If {fn }∞
n=1 is a sequence in L (X, µ) that converges in the L metric to f ∈ L (X, µ), then
{fn }∞
n=1 is uniformly integrable.
Proof. a. Let f1 , f2 , · · · , fn ∈ L1 (X, µ). For each fi , 1 ≤ i ≤ n, by Corollary 3.6, in Page 89, we know
for any > 0, there exists δi > 0, 1 ≤ i ≤ n such that whenever µ(E) < δi , we have
Z
fi dµ < ,
1 ≤ i ≤ n.
E
For the above > 0, take δ = min {δ1 , δ2 , · · · , δn }. Whenever µ(E) < δ, then we know µ(E) < δ ≤
δi for all 1 ≤ i ≤ n, which implies
Z
fi dµ < for all 1 ≤ i ≤ n.
E
Therefore, {f1 , f2 , · · · , fn } is uniformly integrable.
1
1
1
b. Let {fn }∞
n=1 be a sequence in L (X, µ) that converges in the L metric to f ∈ L (X, µ), that is,
Z
|fn (x) − f (x)|dµ = 0
lim
n→∞
X
1
2
MINGFENG ZHAO
So for any > 0, there exists N ≥ 1 such that whenever n ≥ N + 1, we have
Z
|fn (x) − f (x)|dµ <
X
.
2
Since f ∈ L1 (X, µ), so by the Corollary 3.6, in Page 89, we know for the above > 0, there exists
δN > 0 such that whenever µ(E) < δN , we have
Z
f (x)dµ < .
2
E
So for all n ≥ N + 1, we have
Z
Z
f (x)dµ
= (fn (x) − f (x))dµ +
Z
fn (x)dµ
E
E
E
Z
Z
≤ (fn (x) − f (x))dµ + f (x)dµ
E
E
Z
≤
E
<
Z
|fn (x) − f (x)|dµ + f (x)dµ
E
+
2 2
= .
For f1 , f2 , · · · , fN , which is a subset in L1 (X, µ), by the result of the part a., we know for the above
0
> 0, there exists δN
> 0 such that
Z
fn dµ < for all 1 ≤ n ≤ N .
E
0
Now for the above > 0, take δ = min {δN
, δN } > 0, we know whenever µ(E) < δ, we have
Z
fn dµ < for all n ≥ 1.
E
Therefore, {fn }∞
n=1 is uniformly integrable.
2. [Problem 17, in Page 93] Let (X, M, µ) be a measure space, N be a sub σ-algebra of M, and ν =
µ|M , such that (X, N , ν) is a σ-finite measure space. If f ∈ L1 (X, µ), then there exists g ∈ L1 (X, ν)
SOLUTIONS OF HW8
(thus g is N -measurable) such that
R
E
f dµ =
R
E
3
gdν for all E ∈ N , and g 0 is another such function,
then g = g 0 ν-a.e.
Proof. Let f ∈ L1 (X, µ), define the singed measure µ induced by f on M given by for any E ∈ M,
we have
Z
µ(E) =
f dµ.
E
And let ν = µ|N , we know that ν is a signed measure on N . So for E ∈ N , such that ν(E) = 0,
then we have µ(E) = 0, so µ(E) =
R
E
f dµ = 0, which implies that ν ν. Since ν is σ-finite, then
by the Lebesgue-Radon-Nikodym theorem 3.8, in Page 90, we know that there exists an extended
ν-integrable function g : X → R such that dν = gdν. Therefore, for all E ∈ N , we have
Z
Z
f dµ = µ(E) = ν(E) =
E
gdν.
E
If g 0 ∈ L1 (X, ν) is another function such that for all E ∈ N , we have
Z
Z
f dµ =
E
gdν.
E
This implies that dν = g 0 dν. By the Lebesgue-Radon-Nikodym theorem 3.8, in Page 90, we know
g = g 0 ν-a.e.
Remark 1. Here, I assume (X, N , ν) is a σ-finite measure space, of course, this implies that (X, M, µ)
is a σ-finite measure space. But the book does not assume this, and the book just assumes (X, M, ν)
is a σ-finite measure space. But I think this is not enough, here I give a counterexample.
For example X = R, M = BR , and µ =Lebesgue measure on BR . Let N = {φ, R} be the trivial
σ-subalgebra of M, and ν = µ|N . Consider the function f (x) =
Z
Z
∞
f (x)dx =
R
−∞
x2
1
x2 +1
1
dx = π.
+1
on R. We know
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MINGFENG ZHAO
On the other hand, we know any N -measurable simple function g on R must be of the form
g(x) ≡ c
for some c ∈ R. So
R
R
for all x ∈ R
g(x)dν = c · ν(R), which implies that if g ∈ L1 (X, ν), then c = 0. Hence for
all ν-integrable function g, we know g ≡ 0 on R. So we can not find ν-integrable function g such that
Z
Z
f dµ =
E
gdν.
E
for all E ∈ N .
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu