21.7. VECTOR MEASURES 679

Also,

Ai = ∪∞j=1Ai∩E j, so F (A j) =

∞

∑j=1

F (Ai∩E j)

and so by the triangle inequality, ∥F(Ai)∥ ≤ ∑∞j=1∥∥F(Ai∩E j)

∥∥. Therefore, by the above,

|F |(E∞)− ε <n

∑i=1

≥∥F(Ai)∥︷ ︸︸ ︷∞

∑j=1

∥∥F(Ai∩E j)∥∥= ∞

∑j=1

n

∑i=1

∥∥F(Ai∩E j)∥∥≤ ∞

∑j=1|F |(E j)

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, 21.7.29 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 21.7.3 A Banach space is said to have the Radon Nikodym property if whenever

(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.