24.7. VECTOR MEASURES 681

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 24.7.5 Suppose X ′ is a separable dual space. Then X ′ has the RadonNikodym property.

Proof: By Theorem 24.1.16, 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 |. (24.31)

Also, by definition ∥F (E)∥ ≤ |F |(E) so |F (E)(x)| ≤ |F |(E)∥x∥ . By Lemma 24.7.4,| 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.

and so, we can regard this as holding for all s /∈ N. Also, if x ∈ D̃, | fx (s)| ≤ ∥x∥. Now forx,y ∈ D̃, ∣∣ fx (s)− fy (s)

∣∣= ∣∣ fx−y (s)∣∣≤ ∥x− y∥

and so, by density of D̃, we can define

hx (s)≡ limn→∞

fxn (s) where xn→ x,xn ∈ D̃

For s ∈ N, all functions equal 0. Thus for all x, |hx (s)| ≤ ∥x∥. The dominated convergencetheorem and continuity of F (E) implies that for xn→ x, with xn ∈ D̃,∫

Ehx (s)d |F |= lim

n→∞

∫E

fxn (s)d |F |= limn→∞

F (E)(xn) = F (E)(x). (24.32)

It follows from the density of D̃ that for all x,y ∈ X ,s /∈ N, and a,b ∈ F, let xn→ x,yn→y,an→ a,bn→ b, with xn,yn ∈ D̃ and an,bn ∈Q or Q+ iQ in case F= C. Then

hax+by (s) = limn→∞

fanxn+bnyn (s) = limn→∞

an fxn (s)+bn fyn (s)≡ ahx (s)+bhy (s), (24.33)