I'm reading *Theorem 1* at page 98 of *Vector Measures* by Joseph Diestel, John Jerry Uhl.

THEOREM 1.Let $(\Omega, \Sigma, \mu)$ be afinitemeasure space, $1 \leq p<\infty$, and $X$ be a Banach space. Then $L_{p}(\mu, X)^*=L_{q} (\mu, X^*)$ where $p^{-1}+q^{-1}=1$, if and only if $X^{*}$ has the Radon-Nikodým property with respect to $\mu$.

I would like to extend it to $\sigma$-finite measure space. However, I'm stuck at proving that $\varphi$ is an isometry. Could you elaborate on how to prove it?

Thank you so much!

**My attempt:** Let $(\Omega, \Sigma, \mu)$ is a $\sigma$-finite measure space. There is a countable measurable partition $(\Omega_n)$ of $\Omega$ such that $\mu(\Omega_n) < \infty$. Let $\mu_n(A) := \mu(A \cap \Omega_n)$ for all $n$. Then

- $\mu_n$ is concentrated on $\Omega_n$,
- $(\Omega, \Sigma, \mu_n)$ is a finite measure space, and
- $\mu = \sum \mu_n$.

By **Theorem 1**, for each $n$ there is an isometric isomorphism
$$
\varphi_n : L_{p}(\mu_n, X)^* \to L_{q} (\mu_n, X^*).
$$

For $H \in L_{p}(\mu, X)^*$, we define $H_n \in L_{p}(\mu_n, X)^*$ by $$ H_n (f) := H (f 1_{\Omega_n}) \quad \forall f \in L_{p}(\mu_n, X). $$

Notice that $\varphi_n (H_n)$ is just an equivalence class of $L_{q} (\mu_n, X^*)$. If $g$ is a representative of $\varphi_n (H_n)$, then $g$ can take **any** value of $X^*$ on $\Omega \setminus \Omega_n$ and thus $\|g\|_{L_{q} (\mu_m, X^*)}$ can be $+\infty$ for some $m \neq n$. To avoid this situation, we define
$$
\varphi : L_{p}(\mu, X)^* \to L_{q} (\mu, X^*), H \mapsto \sum_n \varphi_n (H_n) 1_{\Omega_n}.
$$

It's straightforward to verify $\varphi$ is an isomorphism. Let's prove that it is an isometry. We have $$ \begin{align} \| \varphi (H) \|_{ L_{q} (\mu, X^*)}^q &= \int \bigg \| \sum_n \varphi_n (H_n) 1_{\Omega_n} \bigg \|_{X^{*}}^q \mathrm d \mu \\ &= \sum_m \int \bigg \| \sum_n \varphi_n (H_n) 1_{\Omega_n} \bigg \|_{X^{*}}^q \mathrm d \mu_m \\ &= \sum_m \int \big \| \varphi_m (H_m) \big \|_{X^{*}}^q \mathrm d \mu_m \\ &= \sum_m \| \varphi_m (H_m) \|_{ L_{q} (\mu_m, X^*)}^q \\ &= \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*}. \end{align} $$

So it suffices to prove that $$ \|H\|^q_{L_{p}(\mu, X)^*} = \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*}. $$

By Hölder's inequality, $$ \begin{align} \left [ \frac{|H(f)|}{\|f\|_{L_{p}(\mu, X)}} \right ]^q = \frac{\big |\sum_m H_m(f) \big |^q}{\left [\sum_m \|f\|^p_{L_{p}(\mu_m, X)} \right]^{q/p}} \le \sum_m \left [ \frac{|H_m (f)|}{\|f\|_{L_{p}(\mu_m, X)}} \right ]^q \quad \forall f \in L_{p}(\mu, X). \end{align} $$

As such, $$ \|H\|^q_{L_{p}(\mu, X)^*} \le \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*}. $$

Fix $\varepsilon>0$. Pick $f_m \in L_{p}(\mu_m, X)$ such that $$ \left [ \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q > \|H_m\|^q_{L_{p}(\mu_m, X)^*} - \varepsilon 2^{-m}. $$

WLOG, we can assume $$ H_m (f_m) \ge 0 \quad \forall m \in \mathbb N^*. $$

Then $$ \sum_m \left [ \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q > \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*} - \varepsilon. $$

It remains to prove that $$ \sum_{m=1}^n \left [ \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q \le \|H\|^q_{L_{p}(\mu, X)^*}. $$

It suffices to prove that there is $f \in L_{p}(\mu, X)$ such that $$ \left [ \sum_{m=1}^n \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q \le \left [ \frac{|H (f)|}{\|f\|_{L_{p}(\mu, X)}} \right ]^q. $$

finitemeasure $\nu$ and a density $\varphi:\Omega \to (0,\infty)$? $\endgroup$6more comments