680 CHAPTER 21. THE BOCHNER INTEGRAL

Lemma 21.7.4 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) and supposethere 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)|

which contradicts the assumption. Hence h−1 (B(p,δ )) is a set of µ measure zero forall such balls contained in C \B(0,r) and so, since countably many of these balls coverC\B(0,r), it follows that µ

(h−1

(C\B(0,r)

))= 0 and so |h(ω)| ≤ r for a.e. ω .

Theorem 21.7.5 Suppose X ′ is a separable dual space. Then X ′ has the Radon Nikodymproperty.

Proof: By Theorem 21.1.16, X is separable. Let D be a countable dense subset of X .Let F≪ µ,µ a finite measure and F a vector measure and let |F |(Ω)< ∞. Pick x ∈ X andconsider the map

E→ F (E)(x)

for E ∈S . This defines a complex measure which is absolutely continuous with respect to|F |. Therefore, by the earlier Radon Nikodym theorem, there exists fx ∈ L1 (Ω, |F |) suchthat

F (E)(x) =∫

Efx (s)d |F |. (21.7.30)

Also, by definition ∥F (E)∥ ≤ |F |(E) so |F (E)(x)| ≤ |F |(E)∥x∥ . By Lemma 21.7.4,| fx (s)| ≤ ∥x∥ for |F | a.e. s. Let D̃ consist of all finite linear combinations of the form∑

mi=1 aixi where ai is a rational point of F and xi ∈ D. For each of these countably many

vectors, there is an exceptional set of measure zero off which | fx (s)| ≤ ∥x∥. Let N be theunion of all of them and define fx (s)≡ 0 if s /∈ N. Then since F (E) is in X ′, it is linear andso for ∑

mi=1 aixi ∈ D̃,∫

Ef∑

mi=1 aixi (s)d |F |= F (E)

(m

∑i=1

aixi

)=

m

∑i=1

aiF (E)(xi) =∫

E

m

∑i=1

ai fxi (s)d |F |

and so by uniqueness in the Radon Nikodym theorem,

f∑mi=1 aixi (s) =

m

∑i=1

ai fxi (s) |F | a.e.