10.14. ITERATED INTEGRALS 305

Next I want to show G is closed under countable unions of disjoint sets of G . Let {Ai}be a sequence of disjoint sets from G . Then, using the monotone convergence theorem asneeded, ∫

Y

∫X

X∪∞i=1Aidµdν =

∫Y

∫X

∑i=1

XAidµdν =∫

Y

∑i=1

∫X

XAidµdν

=∞

∑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, (10.26)

Thus G is closed with respect to countable disjoint unions.From Lemma 9.3.2, G ⊇ σ (K ) , the smallest σ algebra containing K . Also the com-

putation in 10.26 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. (10.27)

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 10.14.5 Let f : X ×Y → [0,∞] be measurable with respect to the σ al-gebra, σ (K )≡ E ×F just defined and let µ×ν be the product measure of 10.27 whereµ and ν are finite measures on (X ,E ) and (Y,F ) respectively. Then∫

X×Yf 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 10.14.6 Let (X ,E ,µ) ,(Y,F ,ν) both be σ finite. Thus there exist dis-joint measurable 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 ofE with Xn and Fn similarly defined. Then letting K consist of all measurable rectanglesA×B for A ∈ E ,B ∈ F , and letting E ×F ≡ σ (K ) define the product measure of Econtained in this σ algebra as (µ×ν)(E)≡ ∑n ∑m (µn×νm)(E ∩ (Xn×Ym)) .

Lemma 10.14.7 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))

10.14. ITERATED INTEGRALS 305Next I want to show ¥ is closed under countable unions of disjoint sets of Y. Let {A;}be a sequence of disjoint sets from Y. Then, using the monotone convergence theorem asneeded,Re duav = | | Fr,dudv= | | Paduavhh VniAiGh y Oo AGH fp» x H=Y[ [ tauav=¥ | [ %avawi=1/¥ JX i=10X “Y=[¥ [ aavan= [Ya avan = ff Fp ,aavan. (10.26)JX jy JY X JY j=] JXSYThus ¥ is closed with respect to countable disjoint unions.From Lemma 9.3.2, Y D 0 (.#), the smallest o algebra containing .%. Also the com-putation in 10.26 implies that on o (.%) one can define a measure, denoted by wu x v andthat for every EC o(.%),(Ux Vv) (E)=[ | %eduav= | | %eavau. (10.27)with each iterated integral making sense.Next is product measure. First is the case of finite measures. Then this will extend to ofinite measures. The following theorem is Fubini’s theorem.Theorem 10.14.5 Let f : xX x Y — [0,00] be measurable with respect to the o al-gebra, 0(H) =& x F just defined and let U x v be the product measure of 10.27 whereLand Vv are finite measures on (X,&) and (Y, #) respectively. ThenNop fH XY) =| [ tauav= [| ravau.Proof: Let {s,} be an increasing sequence of o(.4) = & x ¥ measurable simplefunctions which converges pointwise to f. The above equation holds for s, in place of ffrom what was shown above. The final result follows from passing to the limit and usingthe monotone convergence theorem. MfOf course one can generalize right away to measures which are only o finite. This isalso called Fubini’s theorem.Definition 10.14.6 Let (x,&,),(Y,.F%,v) both be o finite. Thus there exist dis-joint measurable X, with Ur_,Xn and disjoint measurable Y,, with Ur_jYn = Y such thatL, Vv restricted to Xn,Yn respectively are finite measures. Let En be intersections of sets of& with X, and F;, similarly defined. Then letting consist of all measurable rectanglesAxB forA € &,B € F, and letting € x F = 0(#) define the product measure of Econtained in this o algebra as (Ux V)(E) = Yn Yim (Mn X Vn) (EM (Xn X Yin))-Lemma 10.14.7 The above definition yields a well defined measure on & X FProof: 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 X;, Y; on whichthe measures are finite, then(uxv)(E) = LEdLu n X Vm) (EN (Xn Xe X Yn Yi)nm kyyyy (My x W/) (EN (Xn Xe X Ym Yi)k lnm