702 CHAPTER 24. THE BOCHNER INTEGRAL

Theorem 24.12.1 Let E be a separable Banach space and let X ∈ L1 (Ω;E,F )where X is measurable with respect to F and let G be a σ algebra which is contained inF . Then there exists a unique Z ∈ L1 (Ω;E,G ) such that for all A ∈ G ,∫

AXdP =

∫A

ZdP

Denoting this Z as E (X |G ) , it follows ∥E (X |G )∥ ≤ E (∥X∥ |G ) .

Proof: First consider uniqueness. Suppose Z′ is another in L1 (Ω;E,G ) which works.

Consider a dense subset of E {an}∞

n=1. Then the balls{

B(

an,∥an∥

4

)}∞

n=1must cover E \

{0}. Here is why. If y ̸= 0, pick an ∈ B(

y, ∥y∥5).

any

0Then ∥an∥≥ 4∥y∥/5 and so ∥an− y∥< ∥y∥/5. Thus y∈B(an,∥y∥/5)⊆B

(an,∥an∥

4

).

Now suppose Z is G measurable and∫

A ZdP = 0 for all A ∈ G . Then define the set A by

A≡ Z−1(

B(

an,∥an∥

4

))it follows 0 =

∫A Z−an +andP and so

∥an∥P(A) =

∥∥∥∥∫AandP

∥∥∥∥= ∥∥∥∥∫A(an−Z)dP

∥∥∥∥≤

∫Z−1

(B(

an,∥an∥

4

)) ∥an−Z∥dP≤ ∥an∥4

P(A)

which is a contradiction unless P(A) = 0. Therefore, letting

N ≡ ∪∞n=1Z−1

(B(

an,∥an∥

4

))= Z−1 (E \{0})

it follows N has measure zero and so Z = 0 a.e. This proves uniqueness because if Z,Z′

both hold, then from the above argument, Z−Z′ = 0 a.e.Next I will show Z exists. To do this recall Theorem 24.2.4 on Page 656 which is stated

below for convenience.

Theorem 24.12.2 An E valued function, X, is Bochner integrable if and only if Xis strongly measurable and ∫

Ω

∥X (ω)∥dP < ∞. (24.53)

In this case there exists a sequence of simple functions {Xn} satisfying∫Ω

∥Xn (ω)−Xm (ω)∥dP→ 0 as m,n→ ∞. (24.54)

Xn (ω) converging pointwise to X (ω),

∥Xn (ω)∥ ≤ 2∥X (ω)∥ (24.55)