67.9. SOME REPRESENTATION THEOREMS 2301

Proof: First say X = XD and replace g(Y) with XY−1(B). Let

µ (B)≡∫

Ω

XDXB (Y)dP

Then ∫Ω

XDXY−1(B)dP = P(D∩Y−1 (B)

)∫Rp

XB (y)dµ (y) = µ (B)≡∫

Ω

XDXB (Y)dP

=∫

Ω

XDXY−1(B)dP = P(D∩Y−1 (B)

)Thus ∫

Ω

XDXY−1(B)dP =∫

Ω

XDXB (Y)dP =∫Rp

XB (y)dµ (y)

Now let sn (y) ↑ g(y) , and let sn (y) = ∑mk=1 ckXBk (y) where Bk is a Borel set. Then∫

Rpsn (y)dµ (y) =

∫Rp

m

∑k=1

ckXBk (y)dµ (y) =m

∑k=1

ck

∫Rp

XBk (y)dµ (y)

=m

∑k=1

ckP(D∩Y−1 (Bk)

)∫

Ω

sn (Y)XDdP =m

∑k=1

ck

∫Ω

XDXBk (Y)dP =m

∑k=1

ckP(D∩Y−1 (Bk)

)which is the same thing. Therefore,∫

Ω

sn (Y)XDdP =∫Rp

sn (y)dµ (y)

Now pass to a limit using the monotone convergence theorem to obtain∫Ω

g(Y)XDdP =∫Rp

g(y)dµ (y)

Next replace XD with ∑mk=1 dkXDk (ω) ≡ sn (ω) , a simple function. Then from what was

just shown, ∫Ω

g(Y)m

∑k=1

dkXDk dP =m

∑k=1

dk

∫Ω

g(Y)XDk dP

=m

∑k=1

dk

∫Rp

g(y)dµk

where µk (B)≡∫

ΩXDkXB (Y)dP. Now let

νn (B)≡∫

Ω

m

∑k=1

dkXDkXB (Y) =∫

Ω

snXB (Y)dP