Proof, [Theorem 6.15, Fol99]. By Theorem 16.3.3, the given map is isometric. Thus it is sufficient to show that it is surjective. Let $\phi \in L^{p}(X; H)^{*}$.

(Finite): First suppose that $\mu$ is finite, then $\Sigma(X, \cm; H) \subset L^{p}(X; H)$, and $\phi$ induces an $H$-valued measure on $(X, \cm)$, absolutely continuous with respect to $\mu$. By the Radon-Nikodym Theorem, there exists $g \in L^{1}(X; H)$ such that for each $f \in \Sigma(X, \cm; H)$,

By Theorem 16.3.3, $g \in L^{q}(X; H)$.

(Arbitrary): In the case of (a), by Lemma 16.3.2, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^{p}(X; H)$, $\dpn{f, \phi}{L^p(X; H)}= \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a localisable set satisfying the same restriction condition.

Let $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. By the finite case, there exists $g_{F} \in L^{q}(F; H)$ such that for every $f \in L^{p}(X; H)$,

In the case of (a), there exists a countable exhaustion $\seq{F_n}\subset \cm$ of $A$ with sets of finite measure. For each $n \in \natp$, a representative of $g_{F_n}$ may be taken to have separable range. In the case of (b), such a representative may be chosen for every $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. Thus by the gluing lemma for measurable functions, there exists a measurable function $g: X \to H$ such that $g|_{F}= g_{F}$ almost everywhere for all $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$.

If $q < \infty$, then $g \in L^{q}(X; H)$ by the Monotone Convergence Theorem. Otherwise,

Hence $g \in L^{q}(X; H)$ with $\norm{g}_{L^q(X; H)}\le \norm{\phi}_{L^1(X; H)^*}$.

Finally, let $f \in L^{p}(X; H)$, then there exists $\seq{F_n}\subset \cm$ such that $F_{n} \upto \bracsn{f \ne 0}\cap A$ and $\mu(F_{n}) < \infty$ for all $n \in \natp$. In which case, by the Dominated Convergence Theorem,

Therefore the mapping is surjective, and hence an isomorphism.$\square$