29.2. CONDITIONAL EXPECTATION AND INDEPENDENCE 785

Proof: First I claim that σ (Z,Y ) consists of (Z,Y )−1 (B) where B is Borel in W ×WN

where WN ≡∏∞i=1 W . To see this, let{

(Z,Y )−1 (B) : B is Borel in W 1+N}≡F .

Then F is a σ algebra and for VJ of the form ∏∞j=1 Vj where Vj = W for all j except for

those contained in a finite set J and for the other j,Vj is an open set, and U an open setin W, then (Z,Y )−1 (U×VJ) = Z−1 (U)∩Y −1 (VJ) ∈F so that, in particular, choosingVJ and U appropriately shows Z,Yk are all measurable with respect to F . Hence F ⊇σ (Z,Y ). Also, U ×VJ just described, where U is in a countable basis for W and eachVj is W or in a countable basis for W is a countable basis for the topology of W 1+N. Bydefinition, σ (Z,Y ) must contain (Z,Y )−1 (U×VJ) for U ×VJ in this countable basis.Therefore, σ (Z,Y ) must contain (Z,Y )−1 (O) for all O open and so also σ (Z,Y ) mustcontain (Z,Y )−1 (B) for B Borel, so σ (Z,X) ⊇ F . Also, this shows that for K thesesets in the countable basis for W 1+N,σ (K ) = σ (Z,Y ) =F . Now consider the followingcomputation. ∫

(Z,Y )−1(U×VJ)E (X |σ (Z,Y ))dP

≡∫(Z,Y )−1(U×VJ)

XdP

=∫

XY −1(VJ)XZ−1(U)XdP ∗

= P(Y −1 (VJ)

)∫XZ−1(U)XdP

∗∗= P

(Y −1 (VJ)

)∫E(XZ−1(U)X |Z

)dP =

∫XY −1(VJ)

E(XZ−1(U)X |Z

)dP

=∫

Z−1(U)XY −1(VJ)

E (X |Z)dP =∫(Z,Y )−1(U×VJ)

E (X |Z)dP

Then ∗ happens because X[Z∈U ]X is E measurable and XY −1(VJ)is F measurable and by

assumption, these are independent σ algebras. ∗∗ happens because Ω is σ (Z) measurableand the definition of conditional expectation. Next step happens because XY −1(VJ)

is F

measurable and E(XZ−1(U)X |Z

)is E measurable. Then the rest follows because XZ−1(U)

is σ (Z) measurable so it comes out of the conditional expectation.Now let G be those sets G of σ (Z,Y ) = σ (K ) such that∫

GXdP≡

∫G

E (X |σ (Z,Y ))dP =∫

GE (X |σ (Z))dP≡

∫G

E (X |Z)dP

As just shown, K ⊆ G . If G = ∪∞k=1Gk where the Gk are disjoint and the equation holds

for each Gk∫G

E (X |Z)dP =∫ ∞

∑k=1

XGk E (X |Z)dP =∞

∑k=1

∫Gk

E (X |Z)dP

=∞

∑k=1

∫Gk

E (X |σ (Z,Y ))dP =∫ ∞

∑k=1

XGk E (X |σ (Z,Y ))dP

=∫

GE (X |σ (Z,Y ))dP