59.6. 0,1 LAWS 1875

and soP(A∩BC)= P(A)−P(A∩B) = P(A)(1−P(B)) = P(A)P

(BC) .

and so BC ∈ G . If {Bi}∞

i=1 are disjoint sets in G ,

P(A∩∪∞k=1Bk) =

∞

∑k=1

P(A∩Bk) = P(A)∞

∑k=1

P(Bk)

= P(A)P(∪∞k=1Bk)

and so ∪∞k=1Bk ∈ G . Therefore by the Lemma on π systems Lemma 12.12.3 on Page 329,

it follows G = σ (K ) = σ(∪∞

k=n+1Fk).

Thus for any B ∈ σ(∪∞

k=n+1Fk)= Tn+1,P(A∩B) = P(A)P(B). However, A is in all

of these Tn+1 and so P(A∩A) = P(A) = P(A)2 so P(A) equals either 0 or 1.What sorts of things are tail events of independent σ algebras?

Theorem 59.6.5 Let {Xk} be a sequence of independent random variables having valuesin Z a Banach space. Then

A≡ {ω : {Xk (ω)} converges}

is a tail event of the independent σ algebras {σ (Xk)} . So is

B≡

{ω :

{∞

∑k=1

Xk (ω)

}converges

}.

Proof: Since Z is complete, A is the same as the set where {Xk (ω)} is a Cauchysequence. This set is

∩∞n=1∩∞

p=1∪∞m=p∩l,k≥m {ω : ||Xk (ω)−Xl (ω)||< 1/n}

Note that

∪∞m=p∩l,k≥m {ω : ||Xk (ω)−Xl (ω)||< 1/n} ∈ σ

(∪∞

j=pσ (X j))

for every p is the set where ultimately any pair of Xk,Xl are closer together than 1/n,

∩∞p=1∪∞

m=p∩l,k≥m {ω : ||Xk (ω)−Xl (ω)||< 1/n}

is a tail event. The set where {Xk (ω)} is a Cauchy sequence is the intersection of all theseand is therefore, also a tail event.

Now consider B. This set is the same as the set where the partial sums are Cauchysequences. Let Sn ≡ ∑

nk=1 Xk. The set where the sum converges is then

∩∞n=1∩∞

p=2∪∞m=p∩l,k≥m {ω : ||Sk (ω)−Sl (ω)||< 1/n}

Say k < l and consider for m≥ p

{ω : ||Sk (ω)−Sl (ω)||< 1/n, k ≥ m}