21.6. MEASURABLE REPRESENTATIVES 677

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

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

Now∫

Ωx(s)dµ ∈ X = L1 (B,ν) so it makes sense to ask for (

∫Ω

x(s)dµ)(t), at least µ

a.e. t. To find what this is, note∥∥∥∥∫Ω

xn (s)dµ−∫

Ω

x(s)dµ

∥∥∥∥X≤∫

Ω

∥xn (s)− x(s)∥X dµ.

Therefore, since the right side converges to 0,

limn→∞

∥∥∥∥∫Ω

xn (s)dµ−∫

Ω

x(s)dµ

∥∥∥∥X=

limn→∞

∫B

∣∣∣∣(∫Ω

xn (s)dµ

)(t)−

(∫Ω

x(s)dµ

)(t)∣∣∣∣dν = 0.

But (∫Ω

xn (s)dµ

)(t) =

∫Ω

xn (s, t)dµ a.e. t.

Therefore

limn→∞

∫B

∣∣∣∣∫Ω

xn (s, t)dµ−(∫

Ω

x(s)dµ

)(t)∣∣∣∣dν = 0. (21.6.27)

Also, since xn→ y in L1 (Ω×B),

0 = limn→∞

∫B

∫Ω

|xn (s, t)− y(s, t)|dµdν ≥

limn→∞

∫B

∣∣∣∣∫Ω

xn (s, t)dµ−∫

Ω

y(s, t)dµ

∣∣∣∣dν . (21.6.28)

From 21.6.27 and 21.6.28∫Ω

y(s, t)dµ =

(∫Ω

x(s)dµ

)(t) a.e. t.

Theorem 21.6.1 Let X = L1 (B) where (B,F ,ν) is a σ finite measure space and let

x ∈ L1 (Ω;X)

Then there exists a measurable representative, y ∈ L1 (Ω×B), such that

x(s) = y(s, ·) a.e. s in Ω, the equation in L1 (B) ,

and ∫Ω

y(s, t)dµ =

(∫Ω

x(s)dµ

)(t) a.e. t.