12.9. PRODUCT MEASURES 311

Then by Lemma 12.9.4, 12.9.36, and the observation that ∪∞i=1Ri ∈R,

(µ×ν)(∪∞i=1Si) ≤ (µ×ν)(∪∞

i=1Ri)

= ρ (∪∞i=1Ri)≤

∞

∑i=1

ρ (Ri)

≤∞

∑i=1

((µ×ν)(Si)+

ε

2i

)=

(∞

∑i=1

(µ×ν)(Si)

)+ ε.

Since ε is arbitrary, this proves the lemma.By Caratheodory’s procedure, it follows there is a σ algebra of subsets of X ×Y, de-

noted here by S ×T such that (µ×ν) is a complete measure on this σ algebra. The firstthing to note is that every rectangle is in this σ algebra.

Lemma 12.9.8 Every rectangle is (µ×ν) measurable.

Proof: Let S⊆ X×Y. The following inequality must be established.

(µ×ν)(S)≥ (µ×ν)(S∩ (A×B))+(µ×ν)(S\ (A×B)) . (12.9.39)

The following claim will be used to establish this inequality.Claim: Let P,A×B ∈R. Then

ρ (P∩ (A×B))+ρ (P\ (A×B)) = ρ (P) .

Proof of the claim: From Lemma 12.9.3, P=∪∞i=1A′i×B′i where the A′i×B′i are disjoint.

Therefore,

P∩ (A×B) =∞⋃

i=1

(A∩A′i

)×(B∩B′i

)while

P\ (A×B) =∞⋃

i=1

(A′i \A

)×B′i∪

∞⋃i=1

(A∩A′i

)×(B′i \B

).

Since all of the sets in the above unions are disjoint,

ρ (P∩ (A×B))+ρ (P\ (A×B)) =

∫ ∫ ∞

∑i=1

X(A∩A′i)(x)XB∩B′i

(y)dµdν +∫ ∫ ∞

∑i=1

X(A′i\A)(x)XB′i

(y)dµdν

+∫ ∫ ∞

∑i=1

XA∩A′i(x)XB′i\B (y)dµdν