668 CHAPTER 21. THE BOCHNER INTEGRAL
and so from the usual Fubini theorem for complex valued functions,∫Ω1×Ω2
φ ◦ f (s1,s2)d (µ×λ ) =∫
Ω1
∫Ω2
φ ◦ f (s1,s2)dλdµ. (21.4.19)
Now also if you fix s2, it follows from the definition of strongly measurable and the prop-erties of product measure mentioned above that
s1→ f (s1,s2)
is strongly measurable. Also, by 21.4.18∫Ω2
∥ f (s1,s2)∥dλ < ∞
for s1 /∈ N. Therefore, by Theorem 21.2.4 s2 → f (s1,s2)XNC (s1) is Bochner integrable.By 21.4.19 and 21.2.5 ∫
Ω1×Ω2
φ ◦ f (s1,s2)d (µ×λ )
=∫
Ω1
∫Ω2
φ ◦ f (s1,s2)dλdµ
=∫
Ω1
∫Ω2
φ ( f (s1,s2)XNC (s1))dλdµ
=∫
Ω1
φ
(∫Ω2
f (s1,s2)XNC (s1)dλ
)dµ. (21.4.20)
Each iterated integral makes sense and
s1 →∫
Ω2
φ ( f (s1,s2)XNC (s1))dλ
= φ
(∫Ω2
f (s1,s2)XNC (s1)dλ
)(21.4.21)
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 . (21.4.22)
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 21.4.21.