12.9. PRODUCT MEASURES 315

In all the above there would be no change in writing dνdµ instead of dµdν . The sameresult would be obtained. This proves the theorem.

Now let f : X×Y → [0,∞] be µ×ν measurable and∫f d (µ×ν)< ∞. (12.9.46)

Let s(x,y) ≡ ∑mi=1 ciXEi (x,y) be a nonnegative simple function with ci being the nonzero

values of s and suppose0≤ s≤ f .

Then from the above theorem, ∫sd (µ×ν) =

∫ ∫sdµdν

In which ∫sdµ =

∫XNC (y)sdµ

for N a set of ν measure zero such that y→∫

XNC (y)sdµ is ν measurable. This followsbecause 12.9.46 implies (µ×ν)(Ei)< ∞. Now let sn ↑ f where sn is a nonnegative simplefunction and ∫

snd (µ×ν) =∫ ∫

XNCn(y)sn (x,y)dµdν

wherey→

∫XNC

n(y)sn (x,y)dµ

is ν measurable. Then let N ≡ ∪∞n=1Nn. It follows N is a set of ν measure zero. Thus∫

snd (µ×ν) =∫ ∫

XNC (y)sn (x,y)dµdν

and letting n→ ∞, the monotone convergence theorem implies∫f d (µ×ν) =

∫ ∫XNC (y) f (x,y)dµdν

=∫ ∫

f (x,y)dµdν

because of completeness of the measures, µ and ν . This proves Fubini’s theorem.

Theorem 12.9.11 (Fubini) Let (X ,S ,µ) and (Y,T ,ν) be complete measure spaces andlet

(µ×ν)(E)≡ inf{∫ ∫

XR (x,y)dµdν : E ⊆ R ∈R

}2

2Recall this is the same as

inf

{∞

∑i=1

µ (Ai)ν (Bi) : E ⊆ ∪∞i=1Ai×Bi

}in which the Ai and Bi are measurable.