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Real Analysis – II
MATH 72101
Spring 2022
Le Chen
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
1
§ 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
2
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
3
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
4
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.
5
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.
5
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.
5
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) :=
∫E
fdµ, for all E ∈ M.
6
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) :=
∫E
fdµ, for all E ∈ M.
6
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. �
7
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. �
7
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.
8
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 ;
and
E is null for ν ⇔ f = 0 µ-a.e. on E .
9
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. �
10
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. �
10
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. �
10
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. �
10
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.
11
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.
11
Proof (part 2 of Theorem 3.1-3). This is straight forward:
P \ P′ ⊂ P ⇒ P \ P′ is positive
and
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. �
12
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.
Set
P := ∪∞n=1Pn and N := X \ P
13
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.
14
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.
14
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.
14
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.
14
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).
15
Proof (continued). By property in (1), we can construct:
A1 ⊃ A2 ⊃ A3 · · · ⊃ An ⊃ · · ·
with
ν(A1) > 0 +1
n1
ν(A2) > ν(A1) +1
n2
ν(A3) > ν(A2) +1
n3
ν(A4) > ν(A3) +1
n4
... >...
where in each step the pair (Aj , nj) is chosen such that nj is the smallest integer thatsatisfies that inequality.
16
Proof (continued). set
A := ∩∞n=1An.
We claim that∑∞
j=11nj
< ∞. Indeed,
∞∑j=1
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
n.
Since∑
j1nj
converges, for some j we have
1
nj<
1
n⇔ n < nj .
But B ⊂ A implies that B ⊂ Aj−1. This contradicts how we pick up (Aj , nj). �
17
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.
18
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.
18
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.
18
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. �
19