670 CHAPTER 24. THE BOCHNER INTEGRAL

Each iterated integral makes sense and

s1 →∫

Ω2

φ ( f (s1,s2)XNC (s1))dλ

= φ

(∫Ω2

f (s1,s2)XNC (s1)dλ

)(24.22)

is µ measurable because

(s1,s2) → φ ( f (s1,s2)XNC (s1))

= φ ( f (s1,s2))XNC (s1)

is product measurable. Now consider the function,

s1→∫

Ω2

f (s1,s2)XNC (s1)dλ . (24.23)

I want to show this is also Bochner integrable with respect to µ so I can factor out φ onceagain. It’s measurability follows from the Pettis theorem and the above observation 24.22.Also, ∫

Ω1

∥∥∥∥∫Ω2

f (s1,s2)XNC (s1)dλ

∥∥∥∥dµ

≤∫

Ω1

∫Ω2

∥ f (s1,s2)∥dλdµ

=∫

Ω1×Ω2

∥ f (s1,s2)∥d (µ×λ )< ∞.

Therefore, the function in 24.23 is indeed Bochner integrable and so in 24.21 the φ can betaken outside the last integral. Thus,

φ

(∫Ω1×Ω2

f (s1,s2)d (µ×λ )

)=

∫Ω1×Ω2

φ ◦ f (s1,s2)d (µ×λ )

=∫

Ω1

∫Ω2

φ ◦ f (s1,s2)dλdµ

=∫

Ω1

φ

(∫Ω2

f (s1,s2)XNC (s1)dλ

)dµ

= φ

(∫Ω1

∫Ω2

f (s1,s2)XNC (s1)dλdµ

).

Since X ′ separates the points,∫Ω1×Ω2

f (s1,s2)d (µ×λ ) =∫

Ω1

∫Ω2

f (s1,s2)XNC (s1)dλdµ.

The other formula follows from similar reasoning. ■