724 CHAPTER 26. INDEPENDENCE

Now suppose Y is measurable with respect to σ (X1, · · · ,Xm) . Then there exist simple func-tions

Yn (ω) =mn

∑k=1

fkXBk (X (ω))≡ gn (X (ω))

where the Bk are Borel sets in ∏mi=1 Ei, such that Yn (ω)→ Y (ω) , each gn being Borel.

Thus gn converges on X (Ω) . Furthermore, the set on which gn does converge is a Borelset equal to

∩∞n=1∪∞

m=1∩p,q≥m

[∣∣∣∣gp−gq∣∣∣∣< 1

n

]which containsX (Ω) . Therefore, modifying gn by multiplying it by the indicator functionof this Borel set containing X (Ω), we can conclude that gn converges to a Borel functiong and, passing to a limit in the above, Y (ω) = g(X (ω))

Conversely, suppose Y (ω) = g(X (ω)) . Why is Y σ (X) measurable?

Y−1 (open) =X−1 (g−1 (open))=X−1 (Borel) ∈ σ (X) ■

26.5 Banach Space Valued Random VariablesRecall that for X a random variable, σ (X) is the smallest σ algebra containing all the setsof the form X−1 (F) where F is Borel. Since such sets, X−1 (F) for F Borel form a σ

algebra it follows σ (X) ={

X−1 (F) : F is Borel}.

Next consider the case where you have a set of σ algebras. The following lemma ishelpful when you try to verify such a set of σ algebras is independent. It says you onlyneed to check things on π systems contained in the σ algebras. This is really nice becauseit is much easier to consider the smaller π systems than the whole σ algebra.

Lemma 26.5.1 Suppose {Fi}i∈I is a set of σ algebras contained in F where F is aσ algebra of sets of Ω. Suppose that Ki ⊆Fi is a π system and Fi = σ (Ki). Supposealso that whenever J is a finite subset of I and A j ∈K j for j ∈ J, it follows P(∩ j∈JA j) =

∏ j∈J P(A j) . Then {Fi}i∈I is independent.

Proof: I need to verify that under the given conditions, if { j1, j2, · · · , jn} ⊆ I andA jk ⊆F jk , then P

(∩n

k=1A jk

)= ∏

nk=1 P

(A jk

). By hypothesis, this is true if each A jk ∈K jk .

Suppose it is true whenever there are at most r− 1 ≥ 0 of the A jk which are not in K jk .Consider ∩n

k=1A jk where there are r sets which are not in the corresponding K jk . Withoutloss of generality, say there are at most r− 1 sets in the first n− 1 which are not in thecorresponding K jk .

Pick(A j1 · · · ,A jn−1

)let

G(A j1 ···A jn−1

) ≡{

B ∈F jn : P(∩n−1

k=1A jk ∩B)=

n−1

∏k=1

P(A jk

)P(B)

}I am going to show G(

A j1 ···A jn−1

) is closed with respect to complements and countable

disjoint unions and then apply the Lemma on π systems. By the induction hypothesis,K jn ⊆ G(

A j1 ···A jn−1

). If B ∈ G(A j1 ···A jn−1

),n−1

∏k=1

P(A jk

)= P

(∩n−1

k=1A jk

)= P

((∩n−1

k=1A jk ∩BC)∪ (∩n−1k=1A jk ∩B

))