308 CHAPTER 12. THE CONSTRUCTION OF MEASURES

B′i ∈ T . Also, the intersection of finitely many sets of R is a set of R. For ρ defined in12.9.31, it follows that 12.9.32 and 12.9.33 hold for any element of R. Furthermore,

ρ (Q) = ∑i

µ(A′i)

ν(B′i)= ∑

iρ(A′i×B′i

).

Proof: Let Q be given as above. Let A′1×B′1 =A1×B1. By Lemma 12.9.2, it is possibleto replace A2×B2 with finitely many disjoint rectangles, {A′i×B′i}

m2i=2 such that none of

these rectangles intersect A′1×B′1, each is a subset of A2×B2, and

∪∞i=1Ai×Bi =

(∪m2

i=1A′i×B′i)∪ (∪∞

k=3Ak×Bk)

Now suppose disjoint rectangles, {A′i×B′i}mpi=1 have been obtained such that each rectangle

is a subset of Ak×Bk for some k ≤ p and

∪∞i=1Ai×Bi =

(∪mp

i=1A′i×B′i)∪(∪∞

k=p+1Ak×Bk).

By Lemma 12.9.2 again, there exist disjoint rectangles {A′i×B′i}mp+1i=mp+1 such that each is

contained in Ap+1×Bp+1, none have intersection with any of {A′i×B′i}mpi=1 and

∪∞i=1Ai×Bi =

(∪mp+1

i=1 A′i×B′i)∪(∪∞

k=p+2Ak×Bk).

Note that no change is made in {A′i×B′i}mpi=1 . Continuing this way proves the existence of

the desired sequence of disjoint rectangles, each of which is a subset of at least one of theoriginal rectangles and such that

Q = ∪∞i=1A′i×B′i.

It remains to verify x→XQ (x,y) is µ measurable for all y and

y→∫

XQ (x,y)dµ

is ν measurable whenever Q ∈ R. Let Q ≡ ∪∞i=1Ai×Bi ∈ R. Then by the first part of

this lemma, there exists {A′i×B′i}∞

i=1 such that the sets are disjoint and ∪∞i=1A′i×B′i = Q.

Therefore, since the sets are disjoint,

XQ (x,y) =∞

∑i=1

XA′i×B′i(x,y) =

∞

∑i=1

XA′i(x)XB′i

(y) .

It follows x→XQ (x,y) is measurable. Now by the monotone convergence theorem,∫XQ (x,y)dµ =

∫ ∞

∑i=1

XA′i(x)XB′i

(y)dµ

=∞

∑i=1

XB′i(y)∫

XA′i(x)dµ

=∞

∑i=1

XB′i(y)µ

(A′i).