Measure Theory and Probability

Measure Theory and Probability

Module VI: Continuity Theorem

Dr. Arindom Chakraborty

October 31, 2015

Objectives

I Concept of continuity for a sequence of sets

I Continuity theorem of measure and Probability

Objectives

I Concept of continuity for a sequence of sets

I Continuity theorem of measure and Probability

Statement

I Let λbe an additive set function defined on (Ω,F). Then

1. λisσ−additive =⇒forAn∈ F ↑n,

limλ(An) =λ(limAn) =λ(∪^∞_n=1An) =λ( ¯A) (λis continuous from below)

2. λisσ−additive =⇒forA_n∈ F ↓n, λ(A₁)<∞ limλ(A_n) =λ(limA_n) =λ(∩^∞_n=1A_n) =λ(A) (λis continuous from above)

λis continuous≡λis continuous from above and below 3. λis continuous from below⇒λis σ−additive

4. λis finite and continuous atφ⇒λisσ−additive

Statement

I Let λbe an additive set function defined on (Ω,F). Then 1. λisσ−additive =⇒forAn∈ F ↑n,

limλ(An) =λ(limAn) =λ(∪^∞_n=1An) =λ( ¯A) (λis continuous from below)

2. λisσ−additive =⇒forA_n∈ F ↓n, λ(A₁)<∞ limλ(A_n) =λ(limA_n) =λ(∩^∞_n=1A_n) =λ(A) (λis continuous from above)

λis continuous≡λis continuous from above and below 3. λis continuous from below⇒λis σ−additive

4. λis finite and continuous atφ⇒λisσ−additive

Statement

I Let λbe an additive set function defined on (Ω,F). Then 1. λisσ−additive =⇒forAn∈ F ↑n,

limλ(An) =λ(limAn) =λ(∪^∞_n=1An) =λ( ¯A) (λis continuous from below)

2. λisσ−additive =⇒forA_n∈ F ↓n, λ(A₁)<∞ limλ(A_n) =λ(limA_n) =λ(∩^∞_n=1A_n) =λ(A)

(λis continuous from above)

λis continuous≡λis continuous from above and below 3. λis continuous from below⇒λis σ−additive

4. λis finite and continuous atφ⇒λisσ−additive

Statement

I Let λbe an additive set function defined on (Ω,F). Then 1. λisσ−additive =⇒forAn∈ F ↑n,

limλ(An) =λ(limAn) =λ(∪^∞_n=1An) =λ( ¯A) (λis continuous from below)

2. λisσ−additive =⇒forA_n∈ F ↓n, λ(A₁)<∞ limλ(A_n) =λ(limA_n) =λ(∩^∞_n=1A_n) =λ(A) (λis continuous from above)

λis continuous≡λis continuous from above and below 3. λis continuous from below⇒λis σ−additive

4. λis finite and continuous atφ⇒λisσ−additive

Statement

I Let λbe an additive set function defined on (Ω,F). Then 1. λisσ−additive =⇒forAn∈ F ↑n,

limλ(An) =λ(limAn) =λ(∪^∞_n=1An) =λ( ¯A) (λis continuous from below)

2. λisσ−additive =⇒forA_n∈ F ↓n, λ(A₁)<∞ limλ(A_n) =λ(limA_n) =λ(∩^∞_n=1A_n) =λ(A) (λis continuous from above)

λis continuous≡λis continuous from above and below

3. λis continuous from below⇒λis σ−additive 4. λis finite and continuous atφ⇒λisσ−additive

Statement

I Let λbe an additive set function defined on (Ω,F). Then 1. λisσ−additive =⇒forAn∈ F ↑n,

limλ(An) =λ(limAn) =λ(∪^∞_n=1An) =λ( ¯A) (λis continuous from below)

2. λisσ−additive =⇒forA_n∈ F ↓n, λ(A₁)<∞ limλ(A_n) =λ(limA_n) =λ(∩^∞_n=1A_n) =λ(A) (λis continuous from above)

λis continuous≡λis continuous from above and below 3. λis continuous from below⇒λis σ−additive

4. λis finite and continuous atφ⇒λisσ−additive

Statement

I Let λbe an additive set function defined on (Ω,F). Then 1. λisσ−additive =⇒forAn∈ F ↑n,

limλ(An) =λ(limAn) =λ(∪^∞_n=1An) =λ( ¯A) (λis continuous from below)

2. λisσ−additive =⇒forA_n∈ F ↓n, λ(A₁)<∞ limλ(A_n) =λ(limA_n) =λ(∩^∞_n=1A_n) =λ(A) (λis continuous from above)

λis continuous≡λis continuous from above and below 3. λis continuous from below⇒λis σ−additive

4. λis finite and continuous atφ⇒λisσ−additive

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞ n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞ n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞

n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞

n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞

n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞

n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞

n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n]

λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞

n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An)

λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 1. LetA_n↑n

LetBi = A^C₁A^C₂...A^C_i−1Ai, then ∪^∞_i=1Ai=∪^∞_i=1Bi

λ(∪^∞_i=1A_i) = λ(∪^∞_i=1B_i) =λ(∪^∞_i=1A^C₁A^C₂...A^C_i−1A_i)

=

∞

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞

n

X

i=1

λ(A^C₁A^C₂...A^C_i−1Ai)

= lim

n→∞λ(∪ⁿ_i=1A^C₁A^C₂...A^C_i−1Ai)[λisσ−additive]

= lim

n→∞λ(∪ⁿ_i=1A_i) = lim

n→∞λ(A_n) 2. LetA_n↓ndefineB_n=A₁−A_n,↑n

λ(∪^∞_i=1Bi) = lim

n→∞λ(Bn)[ by previous result ]

= lim

n→∞λ(A₁−A_n)[A₁= (A₁−A_n)∪A_n] λ(∪^∞_i=1(A1−Ai)) = λ(A1)− lim

n→∞λ(An) λ(A1)−λ(∩^∞_i=1Ai) = λ(A1)− lim

n→∞λ(An)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn=∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn) = lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞ n

X

i=1

λ(A_i)[λis additive ] =

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0 LetA_i ∩A_j =φ. DefineBn= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn =∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn)

= lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞ n

X

i=1

λ(A_i)[λis additive ] =

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0 LetA_i ∩A_j =φ. DefineBn= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn =∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn) = lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞ n

X

i=1

λ(A_i)[λis additive ] =

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0 LetA_i ∩A_j =φ. DefineBn= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn =∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn) = lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞

n

X

i=1

λ(A_i)[λis additive ]

=

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0 LetA_i ∩A_j =φ. DefineBn= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn =∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn) = lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞

n

X

i=1

λ(A_i)[λis additive ] =

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0

LetA_i ∩A_j =φ. DefineBn= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn =∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn) = lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞

n

X

i=1

λ(A_i)[λis additive ] =

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0 LetA_i ∩A_j =φ. DefineB_n= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn =∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn) = lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞

n

X

i=1

λ(A_i)[λis additive ] =

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0 LetA_i ∩A_j =φ. DefineB_n= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)

Proof

I 3. Let A1,A2, ...,An∈ F be a sequence of pairwise disjoint sets. Define

Bn=∪ⁿ_i=1Ai ↑n. Then

limn→∞Bn=∪^∞_n=1Bn =∪^∞_i=1Ai

λ(∪^∞_i=1Ai) = λ( lim

n→∞Bn) = lim

n→∞λ(Bn)[λcontinuous from below ]

= lim

n→∞λ(∪ⁿ_i=1B_i) = lim

n→∞

n

X

i=1

λ(A_i)[λis additive ] =

∞

X

i=1

λ(A_i)

I 4. Let B_n↓φ. Then

limn→∞λ(Bn) =λ(limn→∞Bn) =λ(φ) = 0 LetA_i ∩A_j =φ. DefineB_n= ¯A−(∪ⁿ_i=1A_i)↓n

n→∞lim λ(Bn) =λ( lim

n→∞Bn) = 0

⇔ lim

n→∞λ( ¯A−(∪ⁿ_i=1A_i)) = λ( lim

n→∞( ¯A−(∪ⁿ_i=1A_i)))

⇔ lim

n→∞[λ( ¯A)−λ(∪ⁿ_i=1A_i)] = 0⇔λ( ¯A) = lim

n→∞ n

X

i=1

λ(A_i)