322 CHAPTER 12. THE CONSTRUCTION OF MEASURES

smallest σ algebra containing these elementary sets. But M ⊆S ×Fby definition and soM = S ×F . This proves the theorem.

It follows from Lemma 14.1.2 that the elementary sets form an algebra because clearlythe intersection of two measurable rectangles is a measurable rectangle and

(A×B)\ (A0×B0) = (A\A0)×B∪ (A∩A0)× (B\B0),

an elementary set.

Theorem 12.10.10 If (X ,S ,µ) and (Y,F,λ ) are both finite measure spaces

(µ(X), λ (Y )< ∞)

then for every E ∈S ×F,a.) x→ λ (Ex) is µ measurable, y→ µ(Ey) is λ measurableb.)∫

X λ (Ex)dµ =∫

Y µ(Ey)dλ .

Proof: Let

M = {E ∈S ×F such that both a.) and b.) hold} .

Since µ and λ are both finite, the monotone convergence and dominated convergence the-orems imply M is a monotone class.

Next I will argue M contains the elementary sets. Let

E = ∪ni=1Ai×Bi

where the measurable rectangles, Ai×Bi are disjoint. Then

λ (Ex) =∫

YXE (x,y)dλ =

∫Y

n

∑i=1

XAi×Bi (x,y)dλ

=n

∑i=1

∫Y

XAi×Bi (x,y)dλ =n

∑i=1

XAi (x)λ (Bi)

which is clearly µ measurable. Furthermore,∫X

λ (Ex)dµ =∫

X

n

∑i=1

XAi (x)λ (Bi)dµ =n

∑i=1

µ (Ai)λ (Bi) .

Similarly, ∫Y

µ (Ey)dλ =n

∑i=1

µ (Ai)λ (Bi)

and y→ µ (Ey) is λ measurable and this shows M contains the algebra of elementary sets.By the monotone class theorem, M = S ×F . This proves the theorem.

One can easily extend this theorem to the case where the measure spaces are σ finite.