real analysis – ii math 7210based on g. b. folland's real

Real Analysis – II

MATH 72101

Spring 2022

Le Chen

[email protected]

Last updated on

February 8, 2022

Auburn UniversityAuburn AL

1Based on G. B. Folland’s Real Analysis, Modern Techniques and Their Applications, 2nd Ed.0

Chapter 3. Signed measures and differentiation

§ 3.1 Signed measures

§ 3.2 The Lebesgue-Radon-Nikodym theorem

§ 3.3 Complex measures

§ 3.4 Differentiation on Euclidean space

§ 3.5 Functions of bounded variation

Definition 3.1-1 Let (X ,M) be a measurable space. A signed measure on (X ,M)is a function ν : M → [−∞,+∞] such that

1. ν(∅) = 0;

2. ν assumes at most one of the values ±∞;

3. if {Ej} is a sequence of disjoint sets in M, then

ν (∪∞n=1En) =

∞∑n=1

ν (En) ,

where the latter sum converges absolutely if ν (∪∞n=1En) is finite.

1. ν(∅) = 0;

ν (∪∞n=1En) =

∞∑n=1

ν (En) ,

1. ν(∅) = 0;

ν (∪∞n=1En) =

∞∑n=1

ν (En) ,

Two examples

Example 3.1-1 Suppose that µ1 and µ2 are measures on M and at least one ofthem is finite. Then µ = µ1 − µ2 is a signed measure.

Example 3.1-2 Suppose that µ is a (positive) measure on M andf : X → [−∞,∞] is a measurable function such that

either∫

f+dµ or∫

f−dµ

is finite. Then f induces one signed measure ν defined as

ν(E) :=

fdµ, for all E ∈ M.

Two examples

Example 3.1-1 Suppose that µ1 and µ2 are measures on M and at least one ofthem is finite. Then µ = µ1 − µ2 is a signed measure.

Example 3.1-2 Suppose that µ is a (positive) measure on M andf : X → [−∞,∞] is a measurable function such that

either∫

f+dµ or∫

f−dµ

is finite. Then f induces one signed measure ν defined as

ν(E) :=

fdµ, for all E ∈ M.

Proposition 3.1-1 (Continuity) Let ν be a signed measure on (X ,M). Then wehave

1. if {En} is an increasing sequence in M, then

ν (∪∞n=1En) = lim

n→∞ν (En) .

2. if {En} is a decreasing sequence in M and ν(E1) < ∞, then

ν (∩∞n=1En) = lim

n→∞ν (En) .

Proof. Homework. �

Proposition 3.1-1 (Continuity) Let ν be a signed measure on (X ,M). Then wehave

1. if {En} is an increasing sequence in M, then

ν (∪∞n=1En) = lim

n→∞ν (En) .

2. if {En} is a decreasing sequence in M and ν(E1) < ∞, then

ν (∩∞n=1En) = lim

n→∞ν (En) .

Proof. Homework. �

Definition 3.1-2 Let ν be a signed measure on (X ,M). A set E ∈ M is calledpositive for ν if for all F ∈ M such that F ⊂ E , one has

ν(F ) ≥ 0.

Similarly, a set E ∈ M is called negative for ν if for all F ∈ M such that F ⊂ E ,one has

ν(F ) ≤ 0;

and a set E ∈ M is called null for ν if it is both positive and negative, orequivalently, if for all F ∈ M such that F ⊂ E , one has

ν(F ) = 0.

Remark 3.1-1 In Example 3.1-2, if ν is induced by a measurable function f on(X ,M), namely, ν(E) =

∫E fdµ for E ∈ M, then

E is positive for ν ⇔ f ≥ 0 µ-a.e. on E .

Similarly,

E is negative for ν ⇔ f ≤ 0 µ-a.e. on E ;

E is null for ν ⇔ f = 0 µ-a.e. on E .

Lemma 3.1-2 It holds that

1. any measurable subset of a positive set is positive;

2. the union of any countable family of positive sets is positive.

Proof. Part 1 is clear from the definition.

Part 2 can be proved using Proposition 3.1-1. �

Theorem 3.1-3 (The Hahn Decomposition Theorem) If ν is a signed measureon (X ,M), then

1. there exist a positive set P and a negative set N for ν such that P ∪ N = X andP ∩ N = ∅.

2. if P′ and N ′ is another such pair, then both P∆P′ and N∆N ′ are null for ν.

Definition 3.1-3 P and N in above theorem is called a Hahn decomposition for ν.

Theorem 3.1-3 (The Hahn Decomposition Theorem) If ν is a signed measureon (X ,M), then

1. there exist a positive set P and a negative set N for ν such that P ∪ N = X andP ∩ N = ∅.

2. if P′ and N ′ is another such pair, then both P∆P′ and N∆N ′ are null for ν.

Definition 3.1-3 P and N in above theorem is called a Hahn decomposition for ν.

Proof (part 2 of Theorem 3.1-3). This is straight forward:

P \ P′ ⊂ P ⇒ P \ P′ is positive

P \ P′ ⊂ N ′ ⇒ P \ P′ is negative.

Hence, P \ P′ is null. Therefore,

P∆P′ =(P \ P′) ∪ (

P′ \ P)

is null for ν.

Similarly, N∆N ′ is null for ν. �

Proof (part 1 of Theorem 3.1-3). By definition, we may assume that ν does notassume the value ∞.

Construction of P and N:

m := sup {ν (En) : En is positive for ν} .

Then there exists a sequence {Pn} of positive sets such that

limn→∞

µ(Pn) = m.

P := ∪∞n=1Pn and N := X \ P

Proof (continued). We can see that

1. P is a positive set for ν thanks to part 2 of Lemma 3.1-2;

2. ν(P) = m < ∞ thanks to the continuity proposition 3.1-1.

It remains to show that N is a negative set for ν. We will prove this by contradiction.

Now let us assume that N is not a negative set.

Proof (continued). Since N is not negative, there exists A ⊂ N such thatν(A) > 0.

We first exclude the possibility of A being a positive set. Indeed, if so, then A ∪ P isalso positive and hence,

ν(A ∪ P) = ν(A) + ν(P) > ν (P) = m,

which contradicts the definition of m.

Now we can claim that N satisfies the property that

A ⊂ N, ν(A) > 0 ⇒ ∃B ⊂ A, ν(B) > ν(A). (1)

Indeed, because A cannot be positive, there exists C ⊂ A with ν(C) < 0. SetB = A \ C. It is then clear that

ν(B) = ν(A)− ν(C) > ν(A).

Proof (continued). By property in (1), we can construct:

A1 ⊃ A2 ⊃ A3 · · · ⊃ An ⊃ · · ·

ν(A1) > 0 +1

ν(A2) > ν(A1) +1

ν(A3) > ν(A2) +1

ν(A4) > ν(A3) +1

... >...

where in each step the pair (Aj , nj) is chosen such that nj is the smallest integer thatsatisfies that inequality.

Proof (continued). set

A := ∩∞n=1An.

We claim that∑∞

j=11nj

< ∞. Indeed,

∞∑j=1

nj< ν(A) < ∞.

For this A, apply the property (1) again, there exists B ⊂ A such that for someinteger n,

ν(B) > ν(A) +1

Since∑

converges, for some j we have

n⇔ n < nj .

But B ⊂ A implies that B ⊂ Aj−1. This contradicts how we pick up (Aj , nj). �

Definition 3.1-4 Let µ and ν be two signed measures on (X ,M). They aremutually singular, denoted as

µ ⊥ ν,

if there exist E ,F ∈ M such that

1. E ∩ F = ∅ and E ∪ F = X ;

2. µ(E) = 0;

3. ν(F ) = 0.

µ ⊥ ν,

1. E ∩ F = ∅ and E ∪ F = X ;

2. µ(E) = 0;

3. ν(F ) = 0.

µ ⊥ ν,

1. E ∩ F = ∅ and E ∪ F = X ;

2. µ(E) = 0;

3. ν(F ) = 0.

Theorem 3.1-4 (The Jordan decomposition theorem) If ν is a signedmeasure, there exist unique positive measures ν+ and ν− such that

ν = ν+ − ν− and ν+ ⊥ ν−.

Proof. Read the textbook. �

real analysis – ii math 7210based on g. b. folland's real

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