24.2. THE BOCHNER INTEGRAL 655

Now if E ̸= /0, consider {XEC fn}∞

n=1 . Then XEC fn is measurable and the sequenceconverges pointwise to XE f everywhere. Therefore, from the first part, there exists aset of measure less than ε,F such that on FC,{XEC fn} converges uniformly to XEC f .Therefore, on (E ∪F)C , { fn} converges uniformly to f . ■

24.2 The Bochner Integral24.2.1 Definition and Basic Properties

Definition 24.2.1 Let ak ∈ X , a Banach space and let a simple function ω→ x(ω)be

x(ω) =n

∑k=1

akXEk (ω) (24.3)

where for each k, Ek is measurable and µ (Ek)< ∞. Thus this is a measurable finite valuedfunction zero off a set of finite measure. Then define∫

Ω

x(ω)dµ ≡n

∑k=1

akµ (Ek).

Proposition 24.2.2 Definition 24.2.1 is well defined, the integral is linear on simplefunctions and ∥∥∥∥∫

Ω

x(ω)dµ

∥∥∥∥≤ ∫Ω

∥x(ω)∥dµ

whenever x is a simple function.

Proof: It suffices to verify that if ∑nk=1 akXEk (ω) = 0,then ∑

nk=1 akµ (Ek) = 0. Let

f ∈ X ′. Then

f

(n

∑k=1

akXEk (ω)

)=

n

∑k=1

f (ak)XEk (ω) = 0

and, therefore,

0 =∫

Ω

(n

∑k=1

f (ak)XEk (ω)

)dµ =

n

∑k=1

f (ak)µ (Ek) = f

(n

∑k=1

akµ (Ek)

).

Since f ∈ X ′ is arbitrary, and X ′ separates the points of X , ∑nk=1 akµ (Ek) = 0 as hoped. It

is now obvious that the integral is linear on simple functions.As to the triangle inequality, say x(ω) = ∑

nk=1 akXEk (ω) where the Ek are disjoint.

Then from the triangle inequality,∥∥∥∥∫Ω

x(ω)dµ

∥∥∥∥=∥∥∥∥∥ n

∑k=1

akµ (Ek)

∥∥∥∥∥≤ n

∑k=1∥ak∥µ (Ek) =

∫Ω

∥x(ω)∥dµ ■

Definition 24.2.3 A strongly measurable function x is Bochner integrable if thereexists a sequence of simple functions xn converging to x pointwise and satisfying∫

Ω

∥xn (ω)− xm (ω)∥dµ → 0 as m,n→ ∞. (24.4)