1894 CHAPTER 59. BASIC PROBABILITY

Proof: For the sake of brevity, denote by Y the vector (Y1, · · · ,Yk) and by y the vector(y1, · · · ,yk) and let ∏

kj=1Rp j ≡ RP. For E a Borel set of Rn,∫Y−1(E)

XdP =∫Rn×RP

XRn×E (x,y)xdλ (X,Y)

=∫

E

∫Rn

xdλ X|ydλ Y. (59.11.20)

Consider the functiony→

∫Rn

xdλ X|y.

Since dλ Y is a Radon measure having inner and outer regularity, it follows the abovefunction is equal to a Borel function for λ Y a.e. y. This function will be denoted by g.Then from 59.11.20∫

Y−1(E)XdP =

∫E

g(y)dλ Y =∫RP

XE (y)g(y)dλ Y

=∫

Ω

XE (Y(ω))g(Y(ω))dP

=∫

Y−1(E)g(Y(ω))dP

and since Y−1 (E) is an arbitrary element of σ (Y) , this shows that since X is σ (Y) mea-surable,

X = g(Y) P a.e.

What about the case where X is not necessarily measurable in σ (Y1, · · · ,Yk)?

Lemma 59.11.4 There exists a unique function Z(ω) which satisfies∫F

X(ω)dP =∫

FZ(ω)dP

for all F ∈ σ (Y1, · · · ,Yk) such that Z is σ (Y1, · · · ,Yk) measurable. It is denoted by

E (X|σ (Y1, · · · ,Yk))

Proof: It is like the above. Letting E be a Borel set in Rp,∫Y−1(E)

XdP =∫Rn×RP

XRn×E (x,y)xdλ (X,Y)

=∫

E

∫Rn

xdλ X|ydλ Y.

Now let g(y)≡ E (X|y1, · · · ,yk) be a Borel representative of∫Rn

xdλ X|y