304 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRAL

10.14 Iterated IntegralsThis is about what can be said for the σ algebra of product measurable sets. First it isnecessary to define what this means.

Definition 10.14.1 A measure space (Ω,F ,µ) is called σ finite if there are mea-surable subsets Ωn such that µ (Ωn)< ∞ and Ω = ∪∞

n=1Ωn.

Next is a σ algebra which comes from two σ algebras.

Definition 10.14.2 Let (X ,E ) ,(Y,F ) be measurable spaces. That is, a set witha σ algebra of subsets. Then E ×F will be the smallest σ algebra which contains themeasurable rectangles, sets of the form E×F where E ∈ E , F ∈F . The sets in this newσ algebra are called product measurable sets.

Definition 10.14.3 Given two finite measure spaces, (X ,E ,µ) and (Y,F ,ν) ,onecan define a new measure µ×ν defined on E ×F by specifying what it does to measurablerectangles as follows:

(µ×ν)(A×B) = µ (A)ν (B)

whenever A ∈ E and B ∈F .

We also have the following important proposition which holds in every context inde-pendent of any measure.

Proposition 10.14.4 Let E ⊆ E ×F be product measurable E ×F where E is a σ

algebra of sets of X and F is a σ algebra of sets of Y . then if Ex ≡ {y ∈ Y : (x,y) ∈ E} andEy ≡ {x ∈ X : (x,y) ∈ E} , then Ex ∈ E and Ey ∈F .

Proof: It is obvious that if K is the measurable rectangles, then the conclusion of theproposition holds. If G consists of the sets of E ×F for which the proposition holds,then it is clearly closed with respect to countable disjoint unions and complements. This isobvious in the case of a countable disjoint union since

(∪iE i

)x = ∪iE i

x, similar for y. Asto complement, if E ∈ G , then Ex ∈F and so

(EC)

x = (Ex)C ∈F . It is similar for y. By

Dynkin’s lemma, G ⊇ E ×F . However G was defined as a subset of E ×F so these areequal. ■

Let (X ,E ,µ) and (Y,F ,ν) be two finite measure spaces. Define K to be the set ofmeasurable rectangles, A×B, A ∈ E and B ∈F . Let

G ≡{

E ⊆ X×Y :∫

Y

∫X

XEdµdν =∫

X

∫Y

XEdνdµ

}(10.25)

where in the above, part of the requirement is for all integrals to make sense.Then K ⊆ G . This is obvious.Next I want to show that if E ∈ G then EC ∈ G . Observe XEC = 1−XE and so∫

Y

∫X

XEC dµdν =∫

Y

∫X(1−XE)dµdν =

∫X

∫Y(1−XE)dνdµ

=∫

X

∫Y

XEC dνdµ

which shows that if E ∈ G , then EC ∈ G .