682 CHAPTER 24. THE BOCHNER INTEGRAL

Let θ (s) be given by θ (s)(x) = hx (s) if s /∈ N and let θ (s) = 0 if s ∈ N. By 24.33 itfollows that θ (s) ∈ X ′ for each s. Also

θ (s)(x) = hx (s) ∈ L1 (Ω)

so θ (·) is weak ∗ measurable. Since X ′ is separable, Theorem 24.1.15 implies that θ isstrongly measurable. Furthermore, by 24.33,

∥θ (s)∥ ≡ sup∥x∥≤1

|θ (s)(x)| ≤ sup∥x∥≤1

|hx (s)| ≤ 1.

Therefore,∫

Ω∥θ (s)∥d |F |< ∞ so θ ∈ L1 (Ω;X ′). Thus, if E ∈S ,∫

Ehx (s)d |F |=

∫E

θ (s)(x)d |F |=(∫

Eθ (s)d |F |

)(x). (24.34)

From 24.32 and 24.34, (∫

E θ (s)d |F |)(x) = F (E)(x) for all x ∈ X and therefore,∫E

θ (s)d |F |= F (E).

Finally, since F ≪ µ, |F | ≪ µ also and so there exists k ∈ L1 (Ω) such that

|F |(E) =∫

Ek (s)dµ

for all E ∈S , by the scalar Radon Nikodym Theorem. It follows

F (E) =∫

Eθ (s)d |F |=

∫E

θ (s)k (s)dµ.

Letting g(s) = θ (s)k (s), this has proved the theorem. ■Since each reflexive Banach spaces is a dual space, the following corollary holds.

Corollary 24.7.6 Any separable reflexive Banach space has the Radon Nikodym prop-erty.

It is not necessary to assume separability in the above corollary. For the proof of a moregeneral result, consult Vector Measures by Diestal and Uhl, [13].

24.8 The Riesz Representation TheoremThe Riesz representation theorem for the spaces Lp (Ω;X) holds under certain conditions.The proof follows the proofs given earlier for scalar valued functions.

Definition 24.8.1 If X and Y are two Banach spaces, X is isometric to Y if thereexists θ ∈L (X ,Y ) such that

∥θx∥Y = ∥x∥X .

This will be written as X ∼= Y . The map θ is called an isometry.

The next theorem says that Lp′ (Ω;X ′) is always isometric to a subspace of (Lp (Ω;X))′

for any Banach space, X .

682 CHAPTER 24. THE BOCHNER INTEGRALLet 6(s) be given by @(s) (x) =h,(s) if s ¢ N and let 0(s) = 0 if s EN. By 24.33 itfollows that 6 (s) € X’ for each s. Also8 (s) (x) = hy (s) € L' (Q)so @(-) is weak * measurable. Since X’ is separable, Theorem 24.1.15 implies that @ isstrongly measurable. Furthermore, by 24.33,|| (s)|| = sup |@(s)(x)| < sup |hx(s)| <1.I|x|<1 \|x||<1Therefore, fo ||0 (s)||d|F| <9 so 6 € L' (Q;X’). Thus, if E € SY,[os (s)d|F| = [ 8 (s) (x)d|F| = (/ @(s)d Fl) (x). (24.34)From 24.32 and 24.34, (J, 0 (s)d|F|) (x) = F (E) (x) for all x € X and therefore,[oar —F(E).Finally, since F < u,|F| <p also and so there exists k € L' (Q) such thatFI(E)= [ k(s)dufor all E € .Y, by the scalar Radon Nikodym Theorem. It followsF(E)= | o()alF|= [ 6(s)k(s)du.Letting g(s) = 0 (s)k(s), this has proved the theorem.Since each reflexive Banach spaces is a dual space, the following corollary holds.Corollary 24.7.6 Any separable reflexive Banach space has the Radon Nikodym prop-erty.It is not necessary to assume separability in the above corollary. For the proof of a moregeneral result, consult Vector Measures by Diestal and Uhl, [13].24.8 The Riesz Representation TheoremThe Riesz representation theorem for the spaces L? (Q;X) holds under certain conditions.The proof follows the proofs given earlier for scalar valued functions.Definition 24.8.1 If X and Y are two Banach spaces, X is isometric to Y if thereexists 08 € L (X,Y) such that|| Ox\ly = |llly -This will be written as X =Y . The map @ is called an isometry.The next theorem says that L” (Q;X’) is always isometric to a subspace of (L? (Q;X))!for any Banach space, X.