Post on 27-Feb-2022
Real Analysis – II
MATH 72101
Spring 2022
Le Chen
lzc0090@auburn.edu
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.
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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.
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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.
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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.
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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.
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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. �
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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. �
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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.
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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 .
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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. �
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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. �
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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. �
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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. �
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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.
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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.
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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. �
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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
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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.
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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.
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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.
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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.
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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).
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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.
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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). �
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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.
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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.
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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.
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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. �
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