26.5. BANACH SPACE VALUED RANDOM VARIABLES 725

= P(∩n−1

k=1A jk ∩BC)+P(∩n−1

k=1A jk ∩B)= P

(∩n−1

k=1A jk ∩BC)+ n−1

∏k=1

P(A jk

)P(B)

and so

P(∩n−1

k=1A jk ∩BC)= n−1

∏k=1

P(A jk

)(1−P(B)) =

n−1

∏k=1

P(A jk

)P(BC)

showing if B ∈ G(A j1 ··· ,A jn−1

), then so is BC. It is clear that G(A j1 ··· ,A jn−1

) is closed with

respect to disjoint unions also. Here is why. If{

B j}∞

j=1 are disjoint sets in G(A j1 ···A jn−1

),

P(∪∞

i=1Bi∩∩n−1k=1A jk

)=

∑i=1

P(Bi∩∩n−1

k=1A jk

)=

∑i=1

P(Bi)n−1

∏k=1

P(A jk

)=

n−1

∏k=1

P(A jk

) ∞

∑i=1

P(Bi) =n−1

∏k=1

P(A jk

)P(∪∞

i=1Bi)

Therefore, by the π system lemma, Lemma 9.3.2 G(A j1 ···A jn−1

) = F jn . This proves the

induction step in going from r−1 to r. ■What is a useful π system for B (E) , the Borel sets of E where E is a Banach space?Recall the fundamental lemma used to prove the Pettis theorem. It was proved on Page

649.

Lemma 26.5.2 Let E be a separable real Banach space. Sets of the form

{x ∈ E : x∗i (x)≤ α i, i = 1,2, · · · ,m}

where x∗i ∈ D′, a dense subspace of the unit ball of E ′ and α i ∈ [−∞,∞) are a π system,and denoting this π system by K , it follows σ (K ) = B (E). The sets of K are examplesof “cylindrical” sets. The D′ is that set for the proof of the Pettis theorem.

Proof: The sets described are obviously a π system. I want to show σ (K ) containsthe closed balls because then σ (K ) contains the open balls and hence the open sets andthe result will follow. Let D′ be described in Lemma 24.1.7. As pointed out earlier it canbe any dense subset of B′. Then

{x ∈ E : ∥x−a∥ ≤ r}=

{x ∈ E : sup

f∈D′| f (x−a)| ≤ r

}

=

{x ∈ E : sup

f∈D′| f (x)− f (a)| ≤ r

}= ∩ f∈D′ {x ∈ E : f (a)− r ≤ f (x)≤ f (a)+ r}

= ∩ f∈D′ {x ∈ E : f (x)≤ f (a)+ r and (− f )(x)≤ r− f (a)}which equals a countable intersection of sets of the given π system. Therefore, every closedball is contained in σ (K ). It follows easily that every open ball is also contained in σ (K )because

B(a,r) = ∪∞n=1B

(a,r− 1

n

).

26.5. BANACH SPACE VALUED RANDOM VARIABLES 725n—-1=P (MS1Ag, OBS) +P (MLA je 1B) = P (MAR OB) + TTP (Ai) PB)k=1and son—\P (EAB) = TTP As) PB) = TTP Ai) PB)Ajy_| Ji Aj |respect to disjoint unions also. Here is why. If {B iti are disjoint sets in Va, A ). dy in=1aynshowing if BE Ws ) then so is BC. It is clear that Ws ) is closed withco °° n—-1P (U2 BAZAR) = YP (BAAR) = YP (Bi) [] P(A;,)i=l i=l k=1n—1= I] P (Aj, )k=1Therefore, by the 7 system lemma, Lemma 9.3.2 4oo n—1» P(Bi) = TTP (Ai) PUB)=1i== #;. This proves thei ~Aj,_1) Jninduction step in going from r—1 tor.What is a useful z system for & (E), the Borel sets of E where E is a Banach space?Recall the fundamental lemma used to prove the Pettis theorem. It was proved on Page649.Lemma 26.5.2 Let E bea separable real Banach space. Sets of the form{x € E: xi (x) < a@j,i=1,2,---,m}where x; € D', a dense subspace of the unit ball of E' and a; € [—-°,°°) are a 1 system,and denoting this 1 system by X, it follows 0 (#) = &(E). The sets of H are examplesof “cylindrical” sets. The D' is that set for the proof of the Pettis theorem.Proof: The sets described are obviously a 2 system. I want to show o (.%) containsthe closed balls because then o (.%) contains the open balls and hence the open sets andthe result will follow. Let D’ be described in Lemma 24.1.7. As pointed out earlier it canbe any dense subset of B’. Then(se Bilal sr) = |ve sup ial <r}fed!= {re Ess ir)~ sa) <r}fed’= Npeo XEE: fla)—rsfax)sfl@tr}=Nyep {x € E: f(x) < f(a) +r and (—f)(x) <r—f(a)}which equals a countable intersection of sets of the given 7 system. Therefore, every closedball is contained in o (.%). It follows easily that every open ball is also contained in o (.%)because. 1B(a,r) =U7_|B ar—= J.