Measure Theory

Advanced Measure Theory

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Khalid Naeem 1 Department of Mathematics and Statistics

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The University of Lahore, Pakistan

1 [email protected]

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K ha lid Na e em

Contents

1 Preliminaries 1.1

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Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 σ-algebra

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σ-algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.2

Borel σ-Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3

Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.4

Caratheodory’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2.1

3 Measurable Functions 3.1

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4 6 9 14

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Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4 Integration

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4.1

Integration of Non-negative Function . . . . . . . . . . . . . . . . . . . . . .

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4.2

The Monotone Convergence Theorem . . . . . . . . . . . . . . . . . . . . . .

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4.3

Fatou’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Integration of Complex Functions . . . . . . . . . . . . . . . . . . . . . . . .

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4.5

Dominated Convergence Theorem . . . . . . . . . . . . . . . . . . . . . . . .

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4.4

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Chapter 1 Preliminaries Preliminaries

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1.1

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Definition 1.1.1. A sequence in X is a mapping from the set N of natural numbers to X

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such that f : N → X.

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Definition 1.1.2. If f : N → X is a sequence and g : N → N is a mapping such that

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g(n1 ) < g(n2 ) for n1 < n2 , then the composition gof : N → X is called a subsequence. Definition 1.1.3. A mapping f : X → Y is called one-to-one, one-one, 1 − 1 or injective mapping if different elements of X have different images in Y i.e. if x1 , x2 ∈ X, then x1 6= x2 ⇒ f (x1 ) 6= f (x2 ) In other words f (x1 ) = f (x2 ) ⇒ x1 = x2 For example the mapping f : R → R defined by f (x) = x3 is injective whereas f (x) = x2 is not so. Definition 1.1.4. A mapping f : X → Y is called into if range of f is not whole of Y i.e. if Rf $ Y . For example the mapping f : R → R defined by f (x) = x2 is an into mapping.

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Definition 1.1.5. A mapping f : X → Y is called onto or surjective if range of f is whole of Y i.e. if Rf = Y . For example the mapping f : R → R defined by f (x) = x is an onto mapping. Definition 1.1.6. A mapping f : X → Y which is both one-one and onto is called a bijective mapping or simply a bijection. For example the mapping f : R → R defined by f (x) = x is a bijective mapping. Definition 1.1.7. A set A is called countable or denumerable if there exists a bijection between A and N. Definition 1.1.8. Let X be a non-empty set and ≤ be a relation on X. Then ≤ is called a partial ordering if the following conditions hold: 1) x ≤ x i.e. ≤ is reflexive

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2) If x ≤ y and y ≤ x, then x = y i.e. ≤ is anti-symmetric

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3) If x ≤ y and y ≤ z, then x ≤ z i.e. ≤ is transitive

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for all x, y, z ∈ X. The set X with the partial ordering ≤ is called a partially ordered set

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and is shortly written ”poset”. We express it by (X, ≤). Example 1.1.9. If X is a non-empty set, then P(X) is a partial ordering under the set inclusion ⊆. Definition 1.1.10. Let X be a non-empty set and ≤ be a relation on X. Then ≤ is called a linear ordering if the following conditions hold: 1) x ≤ x 2) Either x ≤ y or y ≤ x 3) If x ≤ y and y ≤ z, then x ≤ z for all x, y, z ∈ X. 2

Example 1.1.11. The sets N, Z and R are linearly ordered under the usual relation ≤. Definition 1.1.12. Let X be partially ordered with the partial ordering ≤, then an element x ∈ X is called maximum (minimum) if there is an element y ∈ X satisfying x ≤ y (x ≥ y) is x itself. Definition 1.1.13. Let X be a partially ordered set and E ⊂ X. An element x ∈ X is called an upper bound (lower bound ) of E if y ≤ x (x ≤ y), ∀y ∈ E. Remark.

1) The maximum and minimum elements of a set may or may not exist unless the ordering is linear.

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2) The upper and lower bounds of a partially ordered set exist.

case of linear order, they are same.

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3) The minimal and lower bounds of a partial ordering may not be equal. However, in

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Similar cases are for maximum and upper bound.

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Example 1.1.14. Consider the set N of natural numbers equipped with the order ≤. 1 is the minimal element and lower bound too, whereas neither maximal element nor upper bound exist.

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Chapter 2 σ-algebra

σ-algebra

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2.1

Definition 2.1.1. A collection of sets is called an algebra if it is closed under complement

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1) E c ∈ A, ∀E ∈ A,

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and finite union i.e. if

2) ∪ni=1 Ei ∈ A, ∀Ei ∈ A Example 2.1.2.

1) If X is a non-empty set, then P(X) and {φ, X} are algebras. 2) If X is an uncountable set, then the collection A = {E ⊆ X : E is countable or E c is countable} is an algebra. Definition 2.1.3. Let X be a non-empty set, then the collection A of subsets of X is called a σ-algebra if A is closed under complement and countable union. Thus A = {Ei ⊆ A : i = 1, 2, 3, ...} is σ-algebra if 4

1) E c ∈ A, ∀E ∈ A, 2) ∪∞ i=1 Ei ∈ A, ∀Ei ∈ A Example 2.1.4. Let X be a countable set and A = {E ⊆ X : E is countable or E c is countable}. For the 1st condition, there arise two cases: i) If E ∈ A is countable, then E c is uncountable and hence E c ∈ A ii) If E ∈ A is uncountable, then E c is countable and hence E c ∈ A. For the 2nd condition, there arise three cases: ∞ i) If all Ei ’s are countable, then so is ∪∞ i=1 Ei and hence ∪i=1 Ei ∈ A ∞ ii) If all Ei ’s are uncountable, then so is ∪∞ i=1 Ei and hence ∪i=1 Ei ∈ A

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iii) If some Ei ’s are countable and some uncountable, then ∪∞ i=1 Ei is uncountable and

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hence ∪∞ i=1 Ei ∈ A.

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Hence, A is a σ-algebra.

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Remark.

1) Both X and φ must belong to A. 2) Every σ-algebra is an algebra. However, the converse may not hold. For example, if P(X) is the collection of all subsets of an infinite set X which are finite or have finite complements. Then, P(X) is an algebra but not a σ-algebra. Definition 2.1.5. Let X be a non-empty set and ǫ ∈ P(X). Then there exists a σ-algebra which is intersection of all sets containing ǫ. This algebra is called σ-algebra generated by ǫ and is denoted by M(ǫ). Thus, M(ǫ) = ∩∞ i=1 {Ai : ǫ ⊂ Ai , Ai ∈ P(X)} Proposition 2.1.6. If ǫ ∈ M(F ), then M(ǫ) ⊂ M(F ). 5

Borel σ-Algebra

2.2

Definition 2.2.1. If X is any metric space/topological space, then the σ-algebra generated by open/closed sets is called a Borel σ-algebra and is denoted by BX . Remark.

1) Borel σ-algebra contains open sets, closed sets, Gδ sets and Fσ sets, where Gδ sets are the countable intersection of open sets and Fσ sets are countable union of closed sets. Notice that Fσ and Gδ sets are complements of each other. 2) Countable union of open sets is open.

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3) Countable intersection of closed sets is closed.

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4) Finite intersection of open sets is open.

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5) Finite union of closed sets is closed.   1 1 Example 2.2.2. If Cn = n , 1 + n , then ∩∞ n=1 Cn = (0, 1].

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Theorem 2.2.3. Let R be a topological space under usual topology, then Borel σ-algebra is generated by one of the following: 1) The open interval ǫ1 = {(a, b) : a < b}. 2) The closed interval ǫ2 = {[a, b] : a < b}. 3) ǫ3 = {(a, b] : a < b}, ǫ4 = {[a, b) : a < b}. 4) Open rays ǫ5 = {(−∞, a)}, ǫ6 = {(a, ∞)}. 5) Closed rays ǫ7 = {(−∞, a]}, ǫ8 = {[a, ∞)}. Proof:

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1) ǫ1 = {(a, b) : a < b}. Obviously M(ǫj ) ⊆ BR for j = 1, 2, 3, ..., 6. For converse, suppose that O ∈ BR . Then O may be written as a countable union of open intervals as O = ∪∞ n=1 In ∈ M(ǫ1 ) i.e. O ∈ M(ǫ1 ). Hence BR ⊆ M(ǫ1 ). Thus, BR = M(ǫ1 ). 2) ǫ2 = {[a, b] : a < b}. Let [a, b] ∈ BR , then   1 1 [a, b] = a+ ,b− n n ⇒ [a, b] ∈ M(ǫ2 ) ∪∞ n=1

∴ BR ⊆ M(ǫ2 ) ⇒ M(ǫ2 ) = BR 3) ǫ3 = {(a, b] : a < b}, ǫ4 = {[a, b) : a < b}. 

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(a, b] =

∪∞ n=1

⇒ (a, b] ∈ M(ǫ3 )

1 a, b − n



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⇒ BR ⊆ M(ǫ3 )

  1 [a, b) = a + ,b n ⇒ [a, b) ∈ M(ǫ4 ) ∪∞ n=1

⇒ BR ⊆ M(ǫ4 ) 4) ǫ5 = {(−∞, a)}, ǫ6 = {(a, ∞)}. Let A ∈ ǫ5 , then A is an open set and hence A ∈ BR . Thus, ǫ5 ⊂ BR and so, being σ(ǫ5 ) the σ-algebra generated by ǫ5 , we have σ(ǫ5 ) ⊂ BR . Notice that   1 ∞ ∈ σ(ǫ5 ) (−∞, a] = ∩n=1 − ∞, a + n so for any a, b ∈ R, we have (a, b) = (−∞, b) \ (−∞, a] ∈ σ(ǫ5 ) 7

Since all open sets in R may be expressed as a countable union of intervals (a, b), then being τ the set of open sets of R, we have τ ⊂ σ(ǫ5 ). Thus, σ(ǫ5 ) = BR . Now, consider ǫ6 = {(a, ∞)}. Let A ∈ ǫ6 , then A is an open set and hence A ∈ BR . Thus, ǫ6 ⊂ BR and so, being σ(ǫ6 ) the σ-algebra generated by ǫ6 , we have σ(ǫ6 ) ⊂ BR . Notice that ∩∞ n=1

[b, ∞) = so for any a, b ∈ R, we have

  1 b − , ∞ ∈ σ(ǫ6 ) n

(a, b) = (a, ∞) \ [b, ∞) ∈ σ(ǫ6 ) Since all open sets in R may be expressed as a countable union of intervals (a, b), then being τ the set of open sets of R, we have BR = σ(τ ) ⊂ σ(ǫ6 ). Thus, σ(ǫ6 ) = BR . 5) ǫ7 = {(−∞, a]}, ǫ8 = {[a, ∞)}. Let A ∈ ǫ7 , then A is a closed set and hence A ∈ BR . Notice that

1 − ∞, b − n



∈ σ(ǫ7 )

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So for any a, b ∈ R, we have



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(−∞, b) =

∪∞ n=1

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Thus, ǫ7 ⊂ BR and so, being σ(ǫ7 ) the σ-algebra generated by ǫ7 , we have σ(ǫ7 ) ⊂ BR .

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(a, b) = (−∞, b) \ (−∞, a] ∈ σ(ǫ7 ) Since all open sets in R may be expressed as a countable union of intervals (a, b), then being τ the set of open sets of R, we have τ ⊂ σ(ǫ7 ). Thus, BR = σ(τ ) ⊂ σ(ǫ7 ). So we conclude that σ(ǫ7 ) = BR . Now, consider ǫ8 = {(a, ∞)}. Let A ∈ ǫ8 , then A is a closed set and hence A ∈ BR . Thus, ǫ8 ⊂ BR and so, being σ(ǫ8 ) the σ-algebra generated by ǫ8 , we have σ(ǫ8 ) ⊂ BR . Notice that (a, ∞) = so for any a, b ∈ R, we have

∪∞ n=1



1 a+ ,∞ n



∈ σ(ǫ8 )

(a, b) = (a, ∞) \ [b, ∞) ∈ σ(ǫ8 ) Since all open sets in R may be expressed as a countable union of intervals (a, b), then being τ the set of open sets of R, we have BR = σ(τ ) ⊂ σ(ǫ8 ). Thus, σ(ǫ8 ) = BR . 8

Definition 2.2.4. Let X 6= φ and ǫ ⊆ P(X), then ǫ is called an elementary family of sets if 1) φ ∈ ǫ, 2) E, F ∈ ǫ ⇒ E ∪ F ∈ ǫ, and 3) For each E ∈ ǫ, E c ∈ ǫ may be written as a finite disjoint union of members of ǫ.

2.3

Measures

Definition 2.3.1. Let X 6= φ be equipped with a σ-algebra M. The mapping µ : M → [0, ∞] is called a measure if 1) µ(φ) = 0

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∞ ∞ 2) µ(∪∞ i=1 Ei ) ≤ Σi=1 µ(Ei ), where {Ei }i=1 ∈ M.

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Remark.

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The measure µ is called a countable measure.

1) If µ(∪ni=1 Ei ) ≤ Σni=1 µ(Ei ), then µ is called a finite measure. k−1 2) We can convert {Ei }∞ i=1 into disjoint sets as Fk = Ek \ ∪i=1 Ei so that ∪k Fk = ∪k Ek . ∞ Then condition (2) in definition 2.3.1 becomes µ(∪∞ i=1 Fi ) = Σi=1 µ(Fi ).

Definition 2.3.2. Let (X, M, µ) be a measure space. If for each E ∈ M with µ(E) = ∞, there exists F ⊆ E such that µ(F ) < ∞, then µ is called semi-finite measure. Proposition 2.3.3. If µ1 , µ2 , ..., µn are measures on (X, µ) and a1 , a2 , ..., an ∈ [0, ∞), then Σn1 ai µi is also a measure on (X, µ), where µ(x) = Σn1 ai µi (x). Proof:

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1) Since µ1 , µ2 , ..., µn are measures, so µ1 (φ) = 0, µ2(φ) = 0, ..., µn (φ) = 0 and hence µ(φ) = Σn1 ai µi (φ) = 0. 2) Let E1 , E2 , ... be a countable collection of disjoint elements of µ, then n ∞ µ(∪∞ 1 Ek ) = Σi=1 ai µ(∪1 Ek )

Since each µ(∪∞ 1 Ek ) is a measure, so n ∞ n ∞ n ∞ ∞ µ(∪∞ 1 Ek ) = Σi=1 ai µi (∪1 Ek ) = Σi=1 ai Σk=1 µi (Ek ) = Σi=1 Σk=1 ai µi (Ek ) = Σk=1 µ(Ek )

Hence, µ is a measure. Proposition 2.3.4. Let (X, M, µ) be a measure space and E ∈ M, then µE (A) = µ(A∩E), for A ∈ M, is also a measure.

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Proof:

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1) µE (φ) = µ(φ ∩ E) = µ(φ) = 0, because µ is a measure.

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2) Let A1 , A2 , ... ∈ M and Ai ∩ Aj = φ for i 6= j. Since µ is a measure, so we must have ∞ µ(∪∞ i=1 Ai ∩ E) = Σi=1 µ(Ai ∩ E) and hence ∞ ∞ ∞ µE (∪∞ i=1 Ai ) = µ(∪i=1 Ai ∩ E) = Σi=1 µ(Ai ∩ E) = Σi=1 µE (Ai )

Thus, µE is a measure. Definition 2.3.5. Let (X, M, µ) be a measure space. If µ(X) < ∞, then µ is called a finite measure. Definition 2.3.6. Let (X, M, µ) be a measure space and E1 , E2 , ... ∈ M with Ei ∩ Ej = φ, i 6= j, then X = ∪∞ i=1 Ei is called a σ-finite measure if µ(Ei ) < ∞ for each i = 1, 2, ... Definition 2.3.7. Let (X, M, µ) be a measure space. If for each E ∈ M with µ(E) = ∞, there exists F ⊆ E such that µ(F ) < ∞, then µ is called a semi-finite measure. 10

Remark. Every σ-finite measure is also a semi-finite measure. The converse, however, may not hold. Example 2.3.8. Let X 6= φ and f : X → [0, ∞] be a function that defines a measure given by µ(E) = Σx∈E f (x), M = P(X). If f (x) < ∞, ∀x ∈ E, then the measure µ defined on M is a semi-finite measure. If f (x) > 0, ∀x ∈ E is countable, then µ is a σ-finite measure. Example 2.3.9. Let X = R = ∪x∈R {x}, M = P(X) be the discrete σ-algebra and µ be the counting measure. Then µ is a semi-finite measure because each subset of R having measure ∞ has at least one subset which has finite number of elements. For example, if E = [0, 1], then µ(E) = ∞. There exists F = [ 21 , 21 ] ⊂ E with µ(F ) = 1. Theorem 2.3.10. Let (X, M, µ) be a measure space, then

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1) µ is monotone i.e. if E ⊂ F , then µ(E) < µ(F ).

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2) µ is sub-additive i.e. if E1 , E2 , ... ∈ M with µ(Ei ) < ∞, ∀i, then µ(∪∞ i=1 Ei ) <

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Σ∞ i=1 µ(Ei ).

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3) µ is continuous from below i.e. if E1 ⊂ E2 ⊂ ..., then µ(∪∞ i=1 Ei ) = limn→∞ µ(En ), where E1 , E2 , ... ∈ M with µ(Ei ) < ∞, ∀i. 4) µ is continuous from above i.e. if E1 ⊃ E2 ⊃ ..., then µ(∩∞ i=1 Ei ) = limn→∞ µ(En ), where E1 , E2 , ... ∈ M with µ(E1 ) < ∞. Proof:

1) Since E ⊂ F , so we may write F = E ∪ (F \ E). This in turn leads to µ(F ) = µ(E) + µ(F \ E) > µ(E) i.e. µ(F ) > µ(E). k−1 ∞ ∞ ∞ 2) Let Fk = Ek \ ∪i=1 Ei , so that ∪∞ k=1 Fk = ∪k=1 Ek . Thus, µ(∪k=1 Fk ) = µ(∪k=1 Ek ) = ∞ Σ∞ k=1 µ(Fk ) ≤ Σk=1 µ(Ek ), because Fk ⊆ Ek ⇒ µ(Fk ) ≤ µ(Ek ).

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∞ ∞ ∞ 3) Since E1 ⊂ E2 ⊂ ..., so ∪∞ k=1 Ek = ∪k=1 (Ek \ Ek−1 ). Thus, µ(∪k=1 Ek ) = µ(∪k=1 (Ek \ ∞ Ek−1 )) = Σ∞ k=1 µ(Ek \ Ek−1 ) = limn→∞ Σk=1 (µ(Ek ) − µ(Ek−1 )) = limn→∞ µ(En ).

4) Let Fk = E1 \Ek , then F1 ⊂ F2 ⊂ F3 ⊂ .... Thus, by (3) µ(∪∞ k=1 Fk ) = limn→∞ µ(Fn ) i.e. ∞ µ(∪∞ k=1 (E1 \ Ek )) = limn→∞ µ(E1 \ En ). This leads to µ(E1 \ ∩k=1 Ek ) = limn→∞ µ(E1 \

En ). Therefore, µ(E1 ) − µ(∩∞ k=1 Ek ) = limn→∞ (µ(E1 ) − µ(En )) which in turn yields µ(∩∞ k=1 Ek ) = limn→∞ µ(En ). Definition 2.3.11. Let X 6= φ and f : X → [0, ∞], M = P(X) and µ(E) = Σx∈E f (x). If f (x) = 1, ∀x ∈ E, then µ is called a counting measure. Definition 2.3.12. Let X 6= φ and f : X → [0, ∞], M = P(X). If for some x0 ∈ X ( 1, if x = x0 µ (x) = 0, if x 6= x0 then µ is called a dirac measure.

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Definition 2.3.13. Let (X, M, µ) be a dirac measure space. If for some E ∈ M, µ(E) = 0,

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then E is called a null set.

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Definition 2.3.14. A statement about a point x ∈ X is said to hold almost everywhere if

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it holds for all x except at the points which has measure 0. Definition 2.3.15. A measure µ is called a complete measure if the domain of µ includes all subsets of null sets. Example 2.3.16. Let X be an uncountable set and M be the σ-algebra of countable and co-countable sets. Then the function µ defined on M as ( 0, if E is countable µ (E) = 1, if E is uncountable then µ is a measure. Example 2.3.17. Let X be an infinite set and M = 2X . Define ( 0, if E is finite µ (E) = 1, if E is infinite then µ is a measure. 12

Definition 2.3.18. A non-negative extended real-valued set function µ∗ defined on 2X is called an outer measure if 1) µ∗ (φ) = 0; 2) µ∗ is monotone i.e. if A ⊆ B, then µ∗ (A) ≤ µ∗ (B); and ∞ ∗ 3) µ∗ is countably sub-additive i.e. µ∗ (∪∞ i=1 Ei ) ≤ Σi=1 µ (Ei ).

Proposition 2.3.19. Let ǫ ∈ P(X), φ, X ∈ ǫ and µ : E → [0, ∞]. Then for any A ∈ ǫ, ∞ µ∗ (A) = inf{Σ∞ i=1 µ(Ei ) : A ⊂ ∪i=1 Ei } is an outer measure.

Proof:

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1) Clearly µ∗ (φ) = 0.

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2) Let A ⊆ B. Since a cover of A is the smaller set than the cover of B, so the infimum over both covers will satisfy the inequality µ∗ (A) ≤ µ∗ (B).

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k ∞ 3) Let {Ai }∞ i=1 ∈ ǫ and Ai ⊂ ∪k=1 Ei , then

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k ∞ k µ∗ (Ai ) = inf{Σ∞ k=1 µ(Ei ) : Ai ⊆ ∪k=1 Ei }

Thus, there exists

ǫ 2

such that k ∗ Σ∞ k=1 µ(Ei ) ≤ µ (Ai ) +

ǫ 2i

∞ k Assume that A = ∪∞ i=1 Ai , then A ⊆ ∪i=1 Ei . Thus

k ∞ ∗ k ∞ ∗ µ∗ (A) ≤ µ∗ (∪∞ i=1 Ei ) = Σi,k=1 µ (Ei ) ≤ Σi=1 µ (Ai ) + ǫ

This in turn yields ∗ µ∗ (A) ≤ Σ∞ i=1 µ (Ai ) ∞ ∗ i.e. µ∗ (∪∞ i=1 Ai ) ≤ Σi=1 µ (Ai ).

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Definition 2.3.20. A set A ⊆ X is called outer measurable if for all E ⊆ X, µ∗ (A) = µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) The inequality µ∗ (A) ≤ µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) is obvious. Thus, in order to show that A is outer measurable, we just need to show that µ∗ (A) ≥ µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) Lemma 2.3.21. Let (X, M, µ) be a measure space and N = {N ⊂ M : µ(N) = 0} and M = {E ∪ F : F ⊂ Nfor some N ∈ N }, then the measure µ can be extended to µ whose σ-algebra is M.

Caratheodory’s Theorem

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2.4

Theorem 2.4.1. A collection M of all µ∗ -measurable sets is a σ-algebra and the restriction

Proof:

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of µ∗ to M is a complete measure.

1) Let A ∈ M. Since A is µ∗ -measurable, so for all E ⊆ X we have µ∗ (E) = µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) = µ∗ (E ∩ (Ac )c ) + µ∗ (E ∩ Ac ) = µ∗ (E ∩ Ac ) + µ∗ (E ∩ (Ac )c ) It shows that Ac is µ∗ -measurable i.e. Ac ∈ M. 2) Let A, B ∈ M. Then for all E ⊆ X, we have µ∗ (E) = µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) = µ∗ (E ∩ A ∩ B) + µ∗ (E ∩ A ∩ B c ) + µ∗ (E ∩ Ac ∩ B) + µ∗ (E ∩ Ac ∩ B c ) 14

Now, A ∪ B = (Ac ∩ B) ∪ (A ∩ B c ) ∪ (A ∩ B) So for all E ⊆ X we have E ∩ (A ∪ B) = (E ∩ (Ac ∩ B)) ∪ (E ∩ (A ∩ B c )) ∪ (E ∩ (A ∩ B)) ∴ µ∗ (E ∩ (A ∪ B)) ≤ µ∗ (E ∩ (Ac ∩ B)) + µ∗ (E ∩ (A ∩ B c )) + µ∗ (E ∩ (A ∩ B)) Hence, above equation becomes µ∗ (E) ≥ µ∗ (E ∩ (A ∪ B)) + µ∗ (E ∩ (A ∪ B)c ) which shows that A ∪ B is µ∗ -measurable i.e. A ∪ B ∈ M. ∞ 3) Let {Ai }∞ i=1 be a disjoint collection of measurable sets. Let A = ∪i=1 Ai and Bn =

∪ni=1 Ai . Since E ∩ Bn ⊂ X and An is measurable, so

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µ∗ (E ∩ Bn ) = µ∗ (E ∩ Bn ∩ An ) + µ∗ (E ∩ Bn ∩ Acn )

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n−1 = µ∗ (E ∩ An ) + µ∗ (E ∩ Bn−1 ){∵ Bn ∩ An = An and Bn ∩ Acn = Bn−1 = ∪i=1 Ai }

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= µ∗ (E ∩ An ) + µ∗ (E ∩ An−1 ) + µ∗ (E ∩ An−1 ) (By induction)

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= µ∗ (E ∩ An ) + µ∗ (E ∩ An−1 ) + ... + µ∗ (E ∩ A1 )

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= Σni=1 µ∗ (E ∩ An ) Notice that

 n−1  n−1 Bn ∩ (∪i=1 Ai ) ∪ (∪∞ = (∪ni=1 Ai ) ∩ (∪i=1 Ai ) ∪ (∪∞ i=n+1 Ai ) i=n+1 Ai ) n−1 = (∪i=1 Ai ) ∪ φ n−1 = ∪i=1 Ai

= Bn−1 Now, since Bn is measurable, so µ∗ (E) = µ∗ (E ∩ Bn ) + µ∗ (E ∩ Bnc ) ≥ µ∗ (E ∩ An ) + µ∗ (E ∩ Ac ) {∵ Bn ⊆ A ⇒ Ac ⊆ Bnc ∴ µ∗ (A ∩ Ac ) ≤ µ∗ (A ∩ Bnc )} 15

Now taking limit n → ∞, we obtain ∗ ∗ c µ∗ (E) ≥ Σ∞ n=1 µ (E ∩ An ) + µ (E ∩ A ) ∗ c ∗ ∞ ≥ Σ∞ n=1 µ (E ∩ (∪n=1 An )) + µ (E ∩ A ) ∗ ∗ c ≥ Σ∞ n=1 µ (E ∩ A) + µ (E ∩ A )

Thus, A = ∪∞ i=1 Ai ∈ M and hence M is a σ-algebra. Now, if µ∗ (A) = 0, then for any E ⊂ X we have µ∗ (E) ≤ µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) = µ∗ (E ∩ Ac ) ≤ µ∗ (E)

K

ha

lid

Na e

em

so that A ∈ M and hence µ∗ |M is a complete measure.

16

Chapter 3 Measurable Functions Measurable Functions

em

3.1

If for every open E ⊂ Y ,

Na e

Definition 3.1.1. Let X, Y be metric/topological spaces.

f −1 (E) ⊂ X is open, then f : X → Y is called a continuous function.

ha

lid

Definition 3.1.2. Let (X, M) and (Y, N ) be measurable spaces. Then the function f : that f is (M, N )-measurable.

K

X → Y is called measurable function if for every E ∈ N , we have f −1 (E) ∈ M. We say

Remark.

1) Every measurable function is continuous. The converse, however, may not hold. 2) Lebesgue σ-algebra is the completion of Borel σ-algebra. Theorem 3.1.3. Let (X, M) be a measurable space and f : X → R be a function. Then the following statements are equivalent: 1) f is M-measurable. 2) f −1 (a, ∞) ∈ M, ∀a ∈ R. 17

3) f −1 [a, ∞) ∈ M, ∀a ∈ R. 4) f −1 (−∞, a) ∈ M, ∀a ∈ R. 5) f −1 (−∞, a] ∈ M, ∀a ∈ R. Proof: (1) ⇒ (2) Let f be measurable, then for each E ∈ BR we have f −1 (E) ∈ M. In particular, if E = (a, ∞) ∈ BR then f −1 (a, ∞) ∈ M. (2) ⇒ (3) 1 Since [a, ∞) = ∩∞ n=1 (a − n , ∞), so we have

1 , ∞)) n 1 f −1 (a − , ∞) n

f −1 [a, ∞) = f −1 (∩∞ n=1 (a −

em

= ∩∞ n=1

Na e

−1 Since f −1 (a − n1 , ∞) ∈ M, ∀n, so ∩∞ (a − n1 , ∞) ∈ M and hence f −1 [a, ∞) ∈ M. n=1 f

ha

Since (−∞, a) = ([a, ∞))c , so

lid

(3) ⇒ (4)

K

f −1 (−∞, a) = f −1 ([a, ∞))c = (f −1 [a, ∞))c ∈ M because M is a σ-algebra, so f −1 [a, ∞) ∈ M ⇒ (f −1 [a, ∞))c ∈ M. (4) ⇒ (5) 1 Since (−∞, a] = ∩∞ n=1 (−∞, a + n ), so we have  ! 1 − ∞, a + f −1 (−∞, a] = f −1 ∩∞ n=1 n   1 −1 − ∞, a + = ∩∞ n=1 f n −1 Since f −1 (−∞, a + n1 ) ∈ M, ∀n, so ∩∞ (−∞, a + n1 ) ∈ M and hence f −1 (−∞, a] ∈ M. n=1 f

(5) ⇒ (1) Since (−∞, a] may be expressed as countable intersection of open sets as   1 ∞ (−∞, a] = ∩n=1 − ∞, a + n 18

so (−∞, a] is a Gδ set and hence (−∞, a] ∈ BR . But f −1 (−∞, a] ∈ M, so proved. Q Theorem 3.1.4. Let X1 , X2 , ..., Xn be metric spaces and X = ni=1 Xi be equipped with the N N product metric. Then ni=1 BXi ⊆ BX . If Xi ’s are separable, then ni=1 BXi = BX .

Theorem 3.1.5. Let (X, M) and (Yα , Nα ), α ∈ A, be measurable spaces. Assume further Q N that Y = α∈A Yα , N = α∈A Nα and πα : Y → Yα is a continuous map. Then f : X → Y is (M, N )-measurable if and only if fα = πα of is (M, Nα)-measurable for all α.

Definition 3.1.6. Let (X, M) be a measurable space and E ⊆ X. The characteristic or indicator function χE of E ⊆ X is defined as ( χE (X) =

1, 0,

if x ∈ E if x ∈ /E

Definition 3.1.7. A function f defined on X is called simple function if it can be written

em

as finite linear combination of characteristic function with complete complex coefficients i.e.

Na e

f = Σni=1 ai χEi

ha K

called simple function.

lid

where ai ∈ C and Ei = f −1 (ai ), ∀i = 1, 2, 3, ..., n. Thus, a function whose range is finite is

Theorem 3.1.8. Let (X, M) be a measure space. If f : X → [0, ∞] is measurable, then there is a sequence {φn } of simple functions such that 0 ≤ φ1 ≤ φ2 ≤ ... ≤ f , φn → f pointwise and uniformly on any set on which f is bounded. Remark. The space of all measurable functions whose range is set of positive real numbers, is denoted by L+ .

19

Chapter 4 Integration

Integration of Non-negative Function

em

4.1

Na e

Remark. The space of all measurable functions whose range is set of positive real numbers,

lid

is denoted by L+ .

K

ha

Definition 4.1.1. Let (X, M, µ) be a measurable space and f be a function defined on X. R Then X f dµ is defined as 1) If f is a simple function such that f = Σni=1 ai χEi , then R R R f dµ = Σni=1 ai χEi dµ i.e. f dµ = Σni=1 ai µ(Ei ) R because χE dµ = µ(E).

2) If f is not a simple function, then R R f = sup{ φ : 0 ≤ φ ≤ f, where φ is a simple function}.

Proposition 4.1.2. Let φ and ψ be simple functions. 1) If c ≥ 0, then 2)

R

(φ + ψ) =

R

R

cφ = c

φ+

R

ψ.

R

φ.

20

3) If φ ≤ ψ, then 4) The map A 7→

R

R

φ≤

A

R

ψ.

dµ is a measure on M.

Proof:

1) Since φ is a simple function, so it may be written as a finite linear combination of characteristic function as φ = Σni=1 ai χEi , where Ei ⊆ X and ∪ni=1 Ei = X Thus, cφ = Σni=1 cai χEi and hence Σni=1 cai χEi

em

cφ =

Z

= Σni=1 cai µEi

Na e

Z

K

ha

lid

= cΣni=1 ai µEi Z = c Σni=1 ai χEi Z = c φ

2) Since φ and ψ are simple functions, so both may be written as linear combinations of characteristic functions as φ = Σni=1 ai χEi , and ψ = Σm j=1 bj χFj , where Ei , Fj ⊆ X, and ∪ni=1 Ei = X = ∪m j=1 Fj . Now, φ + ψ = Σni=1 ai χEi + Σm j=1 bj χFj , therefore Z Z Z φ+ ψ = (Σni=1 ai χEi + Σm j=1 bj χFj ) Z Z n m = Σi=1 ai χEi + Σj=1 bj χFj = Σni=1 ai µ(Ei ) + Σm j=1 bj µ(Fj )

21

m Since ∪m j=1 (Ei ∩ Fj ) = Ei ∩ (cupj=1 Fj = Ei ∩ X = Ei , so

n Ei = ∪m j=1 (Ei ∩ Fj ) and Fj = ∪i=1 (Ei ∩ Fj )

Thus, Z

φ+

Z

m n ψ = Σni=1 ai Σm j=1 µ(Ei ∩ Fj ) + Σj=1 bj Σi=1 µ(Ei ∩ Fj )

= Σni=1 Σm j=1 (ai + bj ) µ(Ei ∩ Fj ) = Σn,m (ai + bj ) µ(Ei ∩ Fj ) Z i,j=1 = (φ + ψ)

3) If φ ≤ ψ, then obviously ai ≤ bj for i = 1, 2, ..., n, j = 1, 2, ..., m. Thus, Z Z φ = Σni=1 ai χEi = Σni=1 ai µ(Ei )

em

≤ Σni=1 bj Σm j=1 µ(Ei ∩ Fj )

Na e

n = Σm j=1 bj Σi=1 µ(Ei ∩ Fj )

lid

= Σm j=1 bj µ(Fj ) = ψ Z φ ≤ ψ

ha

K

⇒

Z

4) Let {Ak } be a disjoint sequence in M and A = ∪∞ k=1 Ak , then Z Z φ = Σi ai µ(A ∩ Ei ) = Σi,k ai µ(Ak ∩ Ei ) = Σk A

which shows that the map A 7→

φ

Ak

R

A

dµ is a measure on M.

Remark. The same results hold for non-simple functions.

4.2

The Monotone Convergence Theorem

Theorem 4.2.1. If {fn } is a sequence in L+ such that fj < fj+1 , ∀j, and f = limn→∞ fn (= R R supn fn ), then f = limn→∞ fn . 22

R Proof: Since { fn } is an increasing sequence of numbers, so its limit exits and is possibly R R R R equal to ∞. Furthermore, fn ≤ f for all n, so lim fn ≤ f . Hence, R R R limn→∞ fn = supn fn ≤ f R R i.e. limn→∞ fn ≤ f (1)

In order to prove the reverse inequality, let α ∈ (0, 1) and φ be a simple function

such that 0 ≤ φ ≤ f . Let En = {x : fn (x) ≥ αφ(x)}, then obviously {En } is an increasing sequence of measurable sets whose union is X. Now, Z Z Z αφ(x) fn ≥ fn ≥ En

En

∴ lim

n→∞

Z

fn ≥

Z

αφ(x) = α En

Z

φ(x)

This inequality holds for all 0 < α < 1, it remains true for α = 1 too. Also, Z Z Z Z lim fn ≥ φ ≥ sup φ = f limn→∞

R

fn ≥

R

f

(2)

Na e

i.e.

em

n→∞

lid

Combining (1) and (2), we get the desired result.

K

ha

Theorem 4.2.2. If {fn } is a finite or infinite sequence in L+ and f = Σn fn , then R Σn fn .

R

f =

Proof: We employ induction. Let f1 and f2 be two functions in L+ . Then by Theorem 3.1.8, there exist sequences of simple functions φi and ψj such that 0 ≤ φ1 ≤ φ2 ≤ ... ≤ f1 with limn→∞ φn = f1 , and 0 ≤ ψ1 ≤ ψ2 ≤ ... ≤ f2 with limn→∞ ψn = f2 . By Monotone Convergence Theorem, we have Z Z Z Z f1 = lim φn , f2 = lim ψn n→∞

n→∞

Moreover, 0 ≤ φ1 + ψ1 ≤ φ2 + ψ2 ≤ ... ≤ f1 + f2 so that Z

(f1 + f2 ) = lim

n→∞

Z

(φn + ψn ) = lim

n→∞

23

Z

φn + lim

n→∞

Z

ψn =

Z

f1 +

Z

f2

Now, assume that

so that lim

N →∞

Z

ΣN 1 fn

Z

ΣN 1 fn

⇒ Proposition 4.2.3. If f ∈ L+ , then

Z

R

ΣN 1

=

Z

= lim

N →∞

Σ∞ 1 fn

=

Σ∞ 1

fn

ΣN 1 Z

Z

fn

fn

f = 0 ⇔ f = 0 a.e.

Proof: There arise two cases:

Na e

em

Case-I: If f is a simple function, then we may write f = Σni=1 ai χEi . Now, Z f = 0 Z ⇔ Σni=1 ai χEi = 0 Z ⇔ Σni=1 ai χEi = 0

lid

⇔ Σni=1 ai µ(Ei ) = 0

ha

⇔ ai = 0 or µ(Ei ) = 0

K

⇔f = 0

Case-II: If f is not a simple function. Let f = 0 a.e. and 0 ≤ φ ≤ f , where φ is a simple R R R function. Then φ = 0 a.e. Thus, φ = 0. So sup φ = 0 and hence f = 0. R Conversely, suppose that f = 0. We have to show that f = 0 a.e. For this, suppose on the contrary that f 6= 0 a.e. So there exists En = {x : f (x) >

1 } n

such that

{x : f (x) > 0} = ∪∞ n=1 En . Notice that since En ’s are intervals and measure of an interval is its length, so it follows that µ(En ) > 0. Since 1 1 ≥ χEn n n

f (x) > so

1 f (x) ≥ n which leads to a contradiction. Z

Z

χEn >

24

1 µ(En ) > 0 n

Corollary 4.2.4. If {fn } ⊆ L+ , f ∈ L+ and fn (x) increases to f (x) a.e., then

R

f = lim fn .

Proof: If fn (x) increases to f (x) for x ∈ E where µ(E c ) = 0, then f − f χE = 0 a.e. and fn − fn χE = 0 a.e. So by Monotone Convergence Theorem, Z Z Z Z f = f χE = lim fn χE = lim fn

4.3

Fatou’s Lemma

Lemma 4.3.1. If {fn } is any sequence in L+ , then

R

lim inf fn ≤ lim inf

R

fn .

R R Proof: For each k ≥ 1, we have inf n≥k fn ≤ fj for j ≥ k. Thus, inf n≥k fn ≤ fj for j ≥ k R R and hence inf n≥k fn ≤ inf j≥k fj . Letting k → ∞ and employing Monotone Convergence Z

Z

em

Theorem, we get

inf fn ≤ lim inf

Z

fn R R Corollary 4.3.2. If {fn } ⊆ L+ , f ∈ L+ and fn → f a.e., then f ≤ lim inf fn . lim inf fn = lim

n≥k

lid

Na e

k→∞

ha

Proof: The result follows immediately from Fatou’s Lemma if fn → f everywhere, and

K

this can be achieved by modifying fn and f on a null set without affecting the integrals, by Proposition 4.2.3. Proposition 4.3.3. If f ∈ L+ and {x : f (x) > 0} is σ-finite.

4.4

R

f < ∞, then {x : f (x) = ∞} is a null set and

Integration of Complex Functions

We begin our discussion by defining positive and negative parts of a function f as follows: Definition 4.4.1. Let f : X → R∞ be an extended real-valued function, then the positive and negative parts of f are defined as f + = max f (x), 0 25




f − = max − f (x), 0 ∀x ∈ X, respectively.




Notice that • f = f + − f −. • If f is measurable, then so are f + and f − . • |f | = f + + f − . Definition 4.4.2. Let f : X → C be a complex-valued function, then polar decomposition of f is defined as f = sgn(f ).|f | where sgn (z) =

(

z , |z|

if z 6= 0

0,

if z = 0

is the signature function.

Na e

em

We may also write |f | = sgn(f ).f , where sgn(f ) denotes the conjugate of sgn(f ). R

|f | < ∞. In particR ular, a function f is integrable on E ∈ M, where M is a σ-algebra on X, if E |f | < ∞.

ha

lid

Definition 4.4.3. A complex-valued function f is called integrable if

K

Theorem 4.4.4. The set of integrable real-valued functions on X is a real vector space, and

the integral is a linear functional.

R Proof: Let S = {f : f is integrable such that |f | < ∞}. Let α, β ∈ R and f, g ∈ S, then Z Z Z |αf + βg| ≤ |α| |f | + |β| |g| < ∞ which shows that αf + βg is integrable i.e. αf + βg ∈ S. Hence, S is a subspace of R and hence a vector space over the field R of real numbers. Now, 1)

R

αf = α

R

f.

26

2) Let h = f + g, then h+ = f + + g + and h− = f − + g − so that h+ − h− = f + + g + − f − − g − ⇒ h+ + f − + g − Z Z − + f + g− ⇒ h Z Z Z + − ⇒ h + f + g− Z ⇒ (h+ − h− ) Z ⇒ h Z

= h− + f + + g + Z Z Z − + = h + f + g+ Z Z Z − + = h + f + g+ Z Z + − = (f − f ) + (g + − g − ) Z Z = f+ g

Remark. The space of complex-valued integrable functions is a complex vector space and the integral is is a complex linear functional on it. We denote this space by L1 (µ), L1 (X, µ), L1 (X) or simply by L1 .

Na e

em

R R Proposition 4.4.5. If f ∈ L1 , then | f | ≤ |f |.

lid

Proof: There arise three cases:

K

ha

R R Case-I: If f = 0 a.e., then | f | = 0 = |f |.

Case-II: If f is real-valued, then f = f + − f − , so that Z Z Z Z Z Z Z Z f = (f + − f − ) ≤ f + + f − ≤ |f + | + |f − | ≤ (f + + f − ) = |f | R

R R R R f = 6 0. Let α = sgn( f ), then | f | = α f = αf . R For our convenience, let’s take αf to be real such that Z Z Z Z Z f = Re αf = Re(αf ) ≤ |αf | = |f |

Case-III: If f is complex-valued and

4.5

Dominated Convergence Theorem

Theorem 4.5.1. Let {fn } be a sequence in L1 such that 27

1) fn → f a.e., 2) There exists a non-negative g ∈ L1 such that |fn | ≤ g a.e. for all n. Then f ∈ L1 and

R

f = limn→∞

R

fn .

Proof: We know that if {fn } is measurable and L1 is complete. Also fn → f a.e., so f is measurable and |f | ≤ g, so f ∈ L1 . For convenience, we assume that both fn and g are real-valued. So, |fn | ≤ g ⇒ −fn ≤ g,

fn ≤ g

⇒ fn + g ≥ 0,

g − fn ≥ 0

Thus, by Fatou’s Lemma ≤ lim inf

K

ha

(fn + g) Z Z ≤ lim inf fn + g Z Z ≤ lim inf fn + g Z ≤ lim inf fn

em

lid

lim inf(fn + g) Z ⇒ (f + g) Z Z ⇒ f+ g Z ⇒ f

Z

Na e

Z

and Z

Z

lim inf(g − fn ) ≤ lim inf (g − fn ) Z Z ⇒ (g − lim sup fn ) ≤ g − lim sup fn Z Z ⇒ − lim sup fn ≤ − lim sup fn Z Z ⇒ − f ≤ − lim sup fn Z Z ⇒ f ≥ lim sup fn Z

Combining above two inequalities, we have Z Z Z lim sup fn ≤ f ≤ lim inf fn 28

We know that lim fn exists if lim inf fn = lim sup fn = lim fn , so last inequality may be recast as lim

Z

Z

Z

fn ≤ f ≤ lim Z Z ⇒ lim fn = f

fn

Theorem 4.5.2. Suppose that {fj } is a sequence in L1 such that Σ∞ 1 R R ∞ R ∞ 1 ∞ Σ1 fj converges a.e. to a function in L , and Σ1 fj = Σ1 fj .

R

|fj | < ∞, then

Proof: Since

Therefore, g = Σ∞ 1

R

Z

Σ∞ 1

|fj | =

Σ∞ 1

Z

|fj | < ∞

|fj | is integrable. Hence, {x : g(x) > 0} is σ-finite and Σ∞ 1 |fj (x)| is

finite a.e. Thus, Σ∞ 1 fj (x) is convergent. Also, |Sn | = |Σn1 fj (x)| ≤ Σn1 |fj (x)| = g

em

Therefore,

Na e

|Sn | = |Σn1 fj (x)| ≤ g

K

ha

lid

So by Dominated Convergence Theorem, we Z lim Sn n→∞ Z ⇒ lim Σn1 fj n→∞ Z ⇒ Σ∞ 1 fj

have

Z

Σn1 fj Z n = lim Σ1 fj n→∞ Z ∞ fj = Σ1 =

29

lim

n→∞