12.9. PRODUCT MEASURES 313
It follows from the dominated convergence theorem that
limn→∞
XNC (y)∫
XQn (x,y)dµ = XNC (y)∫
XP (x,y)dµ
and soy→XNC (y)
∫XP (x,y)dµ
is also measurable. By completeness of ν ,
y→∫
XP (x,y)dµ
must also be ν measurable and so it makes sense to write∫ ∫XP (x,y)dµdν
for every P ∈R1. Also, by the dominated convergence theorem,∫ ∫XP (x,y)dµdν =
∫XNC (y)
∫XP (x,y)dµdν
= limn→∞
∫XNC (y)
∫XQn (x,y)dµdν
= limn→∞
∫ ∫XQn (x,y)dµdν
= limn→∞
ρ (Qn) ∈ [(µ×ν)(S) ,(µ×ν)(S)+1/n]
for all n. Therefore,
ρ (P)≡∫ ∫
XP (x,y)dµdν = (µ×ν)(S) .
The sets of R1 are µ×ν measurable because these sets are countable intersections ofcountable unions of rectangles and Lemma 12.9.8 verifies the rectangles are µ×ν measur-able. This proves the Lemma.
The following theorem is the main result.
Theorem 12.9.10 Let E ⊆X×Y be µ×ν measurable and suppose (µ×ν)(E)<∞. Then
x→XE (x,y) is µ measurable a.e. y.
Modifying XE on a set of measure zero, it is possible to write∫XE (x,y)dµ.
The function,
y→∫
XE (x,y)dµ
is ν measurable and(µ×ν)(E) =
∫ ∫XE (x,y)dµdν .
Similarly,
(µ×ν)(E) =∫ ∫
XE (x,y)dνdµ.