24.6. MEASURABLE REPRESENTATIVES 677

24.6 Measurable RepresentativesIn this section consider the special case where X = L1 (B,ν) where (B,F ,ν) is a measurespace and x ∈ L1 (Ω;X). Thus for each s ∈Ω, x(s) ∈ L1 (B,ν). In general, the map

(s, t)→ x(s)(t)

will not be product measurable, but one can obtain a measurable representative. This isimportant because it allows the use of Fubini’s theorem on the measurable representative.

By Theorem 24.2.4, there exists a sequence of simple functions, {xn}, of the form

xn (s) =m

∑k=1

akXEk (s) (24.26)

where ak ∈ L1 (B,ν) which satisfy the conditions of Definition 24.2.3 and

∥xn− xm∥L1(Ω,L1(B))→ 0 as m,n→ ∞ (24.27)

For such a simple function, you can assume the Ek are disjoint and then

∥xn∥L1(Ω,L1(B)) =m

∑k=1∥ak∥L1(B) µ (Ek) =

m

∑k=1

∫B|ak|dνµ (Ek)

=∫

Ω

∫B|ak (t)|dν (t)XEk (s)dµ (s)

=∫

Ω

∫B|xn|dνdµ

Also, each xn is product measurable. Thus from 24.27,

∥xn− xm∥L1(Ω,L1(B)) =∫

Ω

∫B|xn− xm|dνdµ

which shows that {xn} is a Cauchy sequence in L1 (Ω×B,µ×λ ) . Then there exists y ∈L1 (Ω×B,µ×λ ) and a subsequence still called {xn} such that

limn→∞

∫Ω

∫B|xn− y|dνdµ = lim

n→∞

∫Ω

∥xn− y∥L1(B) dµ

= ∥xn− y∥L1(Ω,L1(B)) = 0.

Now consider 24.27. Since limm→∞ xm (s) = x(s) in L1 (B) , it follows from Fatou’s lemmathat

∥xn− x∥L1(Ω,L1(B)) ≤ lim infm→∞∥xn− xm∥L1(Ω,L1(B)) < ε

for all n large enough. Hence

limn→∞∥xn− x∥L1(Ω,L1(B)) = 0

and sox(s) = y(s) in L1 (B) µ a.e. s

In particular, for a.e. s, it follows that

x(s)(t) = y(s, t) for a.e. t.