222 CHAPTER 9. THE LEBESGUE INTEGRAL
=∞
∑i=1
∫Y
∫X
XAidµdν =∞
∑i=1
∫X
∫Y
XAidνdµ
=∫
X
∞
∑i=1
∫Y
XAidνdµ =∫
X
∫Y
∞
∑i=1
XAidνdµ =∫
X
∫Y
X∪∞i=1Aidνdµ, (9.16)
Thus G is closed with respect to countable disjoint unions.From Lemma 8.3.2, G ⊇ σ (K ) , the smallest σ algebra containing K . Also the com-
putation in 9.16 implies that on σ (K ) one can define a measure, denoted by µ × ν andthat for every E ∈ σ (K ) ,
(µ×ν)(E) =∫
Y
∫X
XEdµdν =∫
X
∫Y
XEdνdµ. (9.17)
with each iterated integral making sense.Next is product measure. First is the case of finite measures. Then this will extend to σ
finite measures. The following theorem is Fubini’s theorem.
Theorem 9.12.5 Let f : X ×Y → [0,∞] be measurable with respect to the σ alge-bra, σ (K ) ≡ E ×F just defined and let µ ×ν be the product measure of 9.17 where µ
and ν are finite measures on (X ,E ) and (Y,F ) respectively. Then∫X×Y
f d (µ×ν) =∫
Y
∫X
f dµdν =∫
X
∫Y
f dνdµ.
Proof: Let {sn} be an increasing sequence of σ (K ) ≡ E ×F measurable simplefunctions which converges pointwise to f . The above equation holds for sn in place of ffrom what was shown above. The final result follows from passing to the limit and usingthe monotone convergence theorem. ■
Of course one can generalize right away to measures which are only σ finite. This isalso called Fubini’s theorem.
Definition 9.12.6 Let (X ,E ,µ) ,(Y,F ,ν) both be σ finite. Thus there exist disjointmeasurable Xn with ∪∞
n=1Xn and disjoint measurable Yn with ∪∞n=1Yn = Y such that µ,ν
restricted to Xn,Yn respectively are finite measures. Let En be intersections of sets of E withXn and Fn similarly defined. Then letting K consist of all measurable rectangles A×Bfor A ∈ E ,B ∈F , and letting E ×F ≡ σ (K ) define the product measure of E containedin this σ algebra as (µ×ν)(E)≡ ∑n ∑m (µn×νm)(E ∩ (Xn×Ym)) .
Lemma 9.12.7 The above definition yields a well defined measure on E ×F .
Proof: This follows from the standard theorems about sums of nonnegative numbers.See Theorem 2.5.4. For example if you have two other disjoint sequences Xk,Yl on whichthe measures are finite, then
(µ×ν)(E) = ∑n
∑m
∑k
∑l(µn×νm)(E ∩ (Xn∩Xk×Ym∩Yl))
= ∑k
∑l
∑n
∑m(µk×ν l)(E ∩ (Xn∩Xk×Ym∩Yl))
and so the definition with respect to the two different increasing sequences gives the samething. Thus the definition is well defined. (µ×ν) is a measure because if the Ei are disjointE ×F measurable sets and E = ∪iEi,
(µ×ν)(E)≡∑n
∑m(µn×νm)(∪iEi∩ (Xn×Ym)) = ∑
n∑m
∑i(µn×νm)(Ei∩ (Xn×Ym))