12.10. ALTERNATIVE TREATMENT OF PRODUCT MEASURE 323

Theorem 12.10.11 If (X ,S ,µ) and (Y,F,λ ) are both σ finite measure spaces, then forevery E ∈S ×F,

a.) x→ λ (Ex) is µ measurable, y→ µ(Ey) is λ measurable.b.)∫

X λ (Ex)dµ =∫

Y µ(Ey)dλ .

Proof: Let X = ∪∞n=1Xn,Y = ∪∞

n=1Yn where,

Xn ⊆ Xn+1,Yn ⊆ Yn+1,µ (Xn)< ∞,λ (Yn)< ∞.

LetSn = {A∩Xn : A ∈S }, Fn = {B∩Yn : B ∈F}.

Thus (Xn,Sn,µ) and (Yn,Fn,λ ) are both finite measure spaces.Claim: If E ∈S ×F, then E ∩ (Xn×Yn) ∈Sn×Fn.Proof: Let

Mn = {E ∈S ×F : E ∩ (Xn×Yn) ∈Sn×Fn} .Clearly Mn contains the algebra of elementary sets. It is also clear that Mn is a monotoneclass. Thus Mn = S ×F.

Now let E ∈S ×F. By Theorem 12.10.10,∫Xn

λ ((E ∩ (Xn×Yn))x)dµ =∫

Yn

µ((E ∩ (Xn×Yn))y)dλ (12.10.49)

where the integrands are measurable. Also

(E ∩ (Xn×Yn))x = /0

if x /∈ Xn and a similar observation holds for the second integrand in 12.10.49 if y /∈ Yn.Therefore, ∫

Xλ ((E ∩ (Xn×Yn))x)dµ =

∫Xn

λ ((E ∩ (Xn×Yn))x)dµ

=∫

Yn

µ((E ∩ (Xn×Yn))y)dλ

=∫

Yµ((E ∩ (Xn×Yn))

y)dλ .

Then letting n→ ∞, the monotone convergence theorem implies b.) and the measurabilityassertions of a.) are valid because

λ (Ex) = limn→∞

λ ((E ∩ (Xn×Yn))x)

µ (Ey) = limn→∞

µ((E ∩ (Xn×Yn))y).

This proves the theorem.This theorem makes it possible to define product measure.

Definition 12.10.12 For E ∈S ×F and (X ,S ,µ),(Y,F,λ ) σ finite,

(µ×λ )(E)≡∫

Xλ (Ex)dµ =

∫Y

µ(Ey)dλ