680 CHAPTER 24. THE BOCHNER INTEGRAL

because{

Ai∩E j}n

i=1 is a partition of E j.Since ε > 0 is arbitrary, this shows

|F |(∪∞j=1E j)≤

∞

∑j=1|F |(E j).

Also, 24.30 implies that whenever the Ei are disjoint, |F |(∪nj=1E j)≥ ∑

nj=1 |F |(E j). There-

fore,∞

∑j=1|F |(E j)≥ |F |(∪∞

j=1E j)≥ |F |(∪nj=1E j)≥

n

∑j=1|F |(E j).

Since n is arbitrary,

|F |(∪∞j=1E j) =

∞

∑j=1|F |(E j)

which shows that |F | is a measure as claimed. ■

Definition 24.7.3 A Banach space is said to have the Radon Nikodym property ifwhenever

(Ω,S ,µ) is a finite measure space

F : S → X is a vector measure with |F |(Ω)< ∞

F ≪ µ

then one may conclude there exists g ∈ L1 (Ω;X) such that

F (E) =∫

Eg(s)dµ

for all E ∈S .

Some Banach spaces have the Radon Nikodym property and some don’t. No attempt ismade to give a complete answer to the question of which Banach spaces have this property,but the next theorem gives examples of many spaces which do. This next lemma was usedearlier. I am presenting it again.

Lemma 24.7.4 Suppose ν is a complex measure defined on S a σ algebra where(Ω,S ) is a measurable space, and let µ be a measure on S with |ν (E)| ≤ rµ (E) andsuppose there is h ∈ L1 (Ω,µ) such that for all E ∈S ,

ν (E) =∫

Ehdµ, ,

Then |h| ≤ r a.e.

Proof: Let B(p,δ )⊆ C\B(0,r) and let E ≡ h−1 (B(p,δ )) . If µ (E)> 0. Then∣∣∣∣ 1µ (E)

∫E

hdµ− p∣∣∣∣≤ 1

µ (E)

∫E|h(ω)− p|dµ < δ

Thus,∣∣∣ ν(E)

µ(E) − p∣∣∣< δ and so |ν (E)− pµ (E)|< δ µ (E) which implies

|ν (E)| ≥ (|p|−δ )µ (E)> rµ (E)≥ |ν (E)|