1874 CHAPTER 59. BASIC PROBABILITY

The Kolmogorov zero one law follows next. It has to do with something called a tailevent.

Definition 59.6.2 Let {Fn} be a sequence of σ algebras. Then Tn ≡ σ(∪∞

k=nFk)

wherethis means the smallest σ algebra which contains each Fk for k ≥ n. Then a tail event is aset which is in the σ algebra, T ≡∩∞

n=1 Tn.

As usual, (Ω,F ,P) is the underlying probability space such that all σ algebras arecontained in F .

Lemma 59.6.3 Suppose {Fn}∞

n=1 are independent σ algebras and suppose A is a tail eventand Aki ∈Fki , i = 1, · · · ,m are given sets. Then

P(Ak1 ∩·· ·∩Akm ∩A

)= P

(Ak1 ∩·· ·∩Akm

)P(A)

Proof: Let K be the π system consisting of finite intersections of the form

Bm1 ∩Bm2 ∩·· ·∩Bm j

where mi ∈Fki for ki > max{k1, · · · ,km} ≡ N. Thus σ (K ) = σ(∪∞

i=N+1Fi). Now let

G ≡{

B ∈ σ (K ) : P(Ak1 ∩·· ·∩Akm ∩B

)= P

(Ak1 ∩·· ·∩Akm

)P(B)

}Then clearly K ⊆ G . It is also true that G is closed with respect to complements andcountable disjoint unions. By the lemma on π systems, G = σ (K ) = σ

(∪∞

i=N+1Fi).

Since A is in σ(∪∞

i=N+1Fi)

due to the assumption that it is a tail event, it follows that

P(Ak1 ∩·· ·∩Akm ∩A

)= P

(Ak1 ∩·· ·∩Akm

)P(A)

Theorem 59.6.4 Suppose the σ algebras, {Fn}∞

n=1 are independent and suppose A is atail event. Then P(A) either equals 0 or 1.

Proof: Let A ∈T . I want to show that P(A) = P(A)2. Let K denote sets of the formAk1 ∩·· ·∩Akm for some m, Ak j ∈Fk j where each k j > n. Thus K is a π system and

σ (K ) = σ(∪∞

k=n+1Fk)≡Tn+1

LetG ≡

{B ∈Tn+1 ≡ σ

(∪∞

k=n+1Fk)

: P(A∩B) = P(A)P(B)}

Thus K ⊆ G because

P(Ak1 ∩·· ·∩Akm ∩A

)= P

(Ak1 ∩·· ·∩Akm

)P(A)

by Lemma 59.6.3. However, G is closed with respect to countable disjoint unions andcomplements. Here is why. If B ∈ G ,

P(A∩BC)+P(A∩B) = P(A)