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REAL ANALYSIS II HOMEWORK 1
˙
CIHANBAHRAN
The questions are from Folland’s text.
Section 3.1
1. Prove Proposition 3.1.
Proposition 1. Let ν be a signed measure on (X,M). If {Ej} is an increasing sequence
S
in M, then ν( ∞E ) = lim ν(E ). If {E } is a decreasing sequence in M and ν(E )
1T j j→∞ j j 1
is finite, then ν( ∞E ) = lim ν(E ).
1 j j→∞ j
S
Proof. Assume that {E } is an increasing sequence in M. Write E = ∞E,E =∅
j r 1 j 0
and for every j ≥ 1 define F = E E . Then
j j j−1
• {Fj} is a disjoint sequence in M,
S
• E = ∞Fj, and
1
• EN =SNFj for every N ≥ 1.
1
Therefore Ñ é
∞ N
ν(E) = Xν(F ) = lim Xν(F) = lim ν(E ).
j N→∞ j N→∞ N
j=1 j=1
Now assume that {E } is a decreasing sequence in M with |ν(E )| < ∞. Write E =
j 1
T∞E. Then{ErE }isanincreasingsequenceinMsuchthatS∞(E rE )=E rE.
1 j 1 j 1 1 j 1
Therefore by the first part
ν(E rE)= lim ν(E rE ).
1 j→∞ 1 j
Note that for every j, ν(E ) = ν(E )+ν(E rE ) is finite, so ν(E ) is finite. Similarly
1 j 1 j j
ν(E) is finite. Thus the above equality can be rewritten as
ν(E )−ν(E)= lim (ν(E )−ν(E )) .
1 j→∞ 1 j
As ν(E ) is finite, we get
1
ν(E) = lim ν(Ej).
j→∞
2. If ν is a signed measure, E is ν-null iff |ν|(E) = 0. Also, if ν and µ are signed
measures, ν ⊥ µ iff |ν| ⊥ µ iff ν+ ⊥ µ and ν− ⊥ µ.
Assume |ν|(E) = 0. Then given F ⊆ E, since |ν| is a measure we have |ν|(F) = 0.
Thus ν+(F) = 0 = ν−(E) = 0 and so ν(F) = 0.
1
REAL ANALYSIS II HOMEWORK 1 2
Conversely, assume that E is ν-null. Since ν+ ⊥ ν− there exists disjoint sets A,B ⊆ X
such that A∪B = X and ν+ is null on B and ν− is null on A. Observe that
0 = ν(E ∩A) = ν+(E ∩A)−ν−(E ∩A)=ν+(E∩A)=ν+(ErB)=ν+(E)
and
0 = ν(E ∩B) = ν+(E ∩B)−ν−(E∩B)=−ν−(E∩B)=−ν−(ErA)=ν−(E).
Thus |ν|(E) = ν+(E)+ν−(E) = 0.
To keep track of the equivalences, let’s write
(1) ν ⊥ µ.
(2) |ν| ⊥ µ.
(3) ν+ ⊥ µ and ν− ⊥ µ.
(1) ⇒ (2): There exists A,B ⊆ X such that X is the disjoint union of A and B, and
ν is null on B and µ is null on A. By above, |ν|(B) = 0. Being a positive measure, so
|ν| is null on B. Hence |ν| ⊥ µ.
(2) ⇒ (3): There exists A,B ⊆ X such that X is the disjoint union of A and B, and
|ν| is null on B and µ is null on A. In particular ν+ and ν− are null on B, so ν+ ⊥ µ
and ν− ⊥ µ.
(3) ⇒ (1): There exists A,B ⊆ X such that X is the disjoint union of A and B, and
ν+ is null on B and µ is null on A. And there exists C,D ⊆ X such that X is the
disjoint union of C and D, and ν− is null on D and µ is null on C. So µ is null on
A∪C. Moreover, given E ⊆ Xr(A∪C) = B∩D, we have ν+(E) = 0 and ν−(E) = 0
so |ν|(E) = 0. Thus by above, ν is null on B ∩ D. Thus ν ⊥ µ.
3. Let ν be a signed measure on (X,M).
a. L1(ν) = L1(|ν|).
b. If f ∈ L1(ν), |R fdν| ≤ R |f|d|ν|.
c. If E ∈ M, |ν|(E) = sup{|RE fdν| : |f| ≤ 1}.
a. By definition L1(ν) = L1(ν+) ∩ L1(ν−). Also since |ν| = ν+ + ν−, we also have
L1(|ν|) = L1(ν+)∩L1(ν−).
b. We have
Z Z Z
+ −
fdν = fdν − fdν
Z Z
+ −
≤ fdν + fdν
≤Z |f|dν++Z |f|dν− = Z |f|d|ν|.
c. If |ν|(E) = ∞, then either ν+(E) or ν−(E) is ∞, so in any case ∞ = |ν(E)| =
| RE 1dν|. Thus the equality holds.
Next, assume |ν|(E) < ∞. Then every ν-measurable f with |f| ≤ 1 satisfies fχE ∈
L1(|ν|) = L1(ν). Thus by (b),
Z Z Z Z
fdν = fχ dν ≤ |fχ |d|ν| = |f|χ d|ν| ≤ |ν|(E).
E E E
E
Thus |ν|(E) is an upper bound for the given set, hence we get the “≥” part of the
desired inequality. For the other direction, let X = P ∪N be a Hahn decomposition of
REAL ANALYSIS II HOMEWORK 1 3
ν and consider f = χ −χ ∈L1(ν). Since P ∩N = ∅, |f| ≤ 1 and we have
P∩E N∩E
Z Z Z
fdν = fdν − fdν
E P∩E N∩E
=|ν(P ∩E)+ν(N ∩E)|
+ −
= ν (E)+ν (E)
=|ν|(E)
which yields “≤”. R
6. Suppose ν(E) = Efdµ where µ is a positive measure and f is an extended
µ-integrable function. Describe the Hahn decompositions of ν and the positive,
negative, and total variations of ν.
Write (X,M,µ) for the measure space. Let P = {x ∈ X : f(x) ≥ 0} and N = {x ∈
X : f(x) < 0}. Clearly X is the disjoint union of P and N. Moreover for every
E∈Mcontained in P we have ν(E) = R fdµ ≥ 0 so P is ν-positive and similarly N
E
is ν-negative. Thus X = P ∪N is a Hahn decomposition of ν.
Therefore for every E ∈ M Z Z
ν+(E) = ν(P ∩E) = fdµ= f+dµ
P∩E E
and Z Z Z
ν−(E) = −ν(N ∩E) = − fdµ=− −f−dµ= f−dµ.
N∩E E E
Finally,
|ν|(E) = ν+(E)+ν−(E) = Z (f+ +f−)dµ.
E
Section 3.2
8. ν ≪ µ iff |ν| ≪ µ iff ν+ ≪ µ and ν− ≪ µ.
Assume ν ≪ µ. Let X = P ∪N be a Hahn decomposition. Then given E ∈ M with
µ(E) = 0, we have µ(P ∩E) = 0 so ν+(E) = ν(P ∩E) = 0 and similarly ν−(E) = 0.
Thus ν+ ≪ µ and ν− ≪ µ.
Assume ν+ ≪ µ and ν− ≪ µ. Then given E ∈ M with µ(E) = 0, we have |ν|(E) =
ν+(E)+ν−(E)=0+0=0. Thus |ν|≪µ.
Assume |ν| ≪ µ. Then given E ∈ M with µ(E) = 0, we have ν+(E)+ν−(E) = 0 so
ν+(E) = ν−(E) = 0 since ν+, ν− are positive measures. Thus ν(E) = 0 − 0 = 0, so
ν ≪µ.
9. Suppose {νj} is a sequence of positive measures, If νj ⊥ µ for all j, then
P P
∞ν ⊥µandif ν ≪µfor all j, then ∞ν ≪µ.
1 j j 1 j
P
Write ν = ∞ν . Assume ν ⊥ µ for all j. So for every j there exists A ,B ∈ M such
1 j j j j
that
• Aj ∩Bj = ∅,
• Aj ∪Bj = X,
REAL ANALYSIS II HOMEWORK 1 4
• ν is null on B ,
j j
• µ is null on A .
j
Let A = S∞Aj and B = Tj Bj. Then
1 1
• A∩B=∅,
• A∪B=X,
• µ is null on A, since µ is null on every Aj,
• ν is null on B.
Thus ν ⊥ µ.
Now assume ν ≪ µ for every j. So given E ∈ M with µ(E) = 0, we have ν (E) = 0
j j
for every j and hence
∞
ν(E) = Xνj(E) = 0,
1
hence ν ≪ µ.
10. Theorem 3.5 may fail when ν is not finite. (Consider dν(x) = dx/x and
dµ(x) = dx on (0,1), or ν = counting measure and µ(E) = P 2−n on N.)
n∈E
Consider the measure space (N,P(N),µ) where
µ:P(N)→R
X −n
E7→n∈E2 .
Clearly µ is a measure. Now if we let ν to be the counting measure, then if µ(E) = 0
for some E ⊆ N, that means E = ∅ hence ν(E) = 0. Thus we have ν ≪ µ.
However, suppose the conclusion of Theorem 3.5 holds for our µ and ν. So there exists
P −n
δ > 0 such that ν(E) < 1 whenever µ(E) < δ. Since n∈N2 converges, there exists
N∈Nsuchthat X
2−n < δ.
n≥N
So picking E = {n ∈ N : n ≥ N}, we have µ(E) < δ. However ν(E) = ∞.
11. Let µ be a positive measure. A collection of functions {f } ⊆L1(µ)iscalled
α α∈A R
uniformly integrable if for every ε > 0 there exists δ > 0 such that | f dµ| < ε
E α
for all α ∈ A whenever µ(E) < δ.
1
a. Any finite subset of L (µ) is uniformly integrable.
b. If {f } is a sequence in L1(µ) that converges in the L1 metric to f ∈ L1(µ),
n
then {f } is uniformly integrable.
n
Write (X,M,µ) for the measure space at hand. Let’s first show why singletons are
uniformly integrable. Given f ∈ L1(µ) the map
ν : M → R
E7→Z fdµ
E
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