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



> 
> **THEOREM 1.** Let $(\Omega, \Sigma, \mu)$ be a **finite** measure 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
$$
 \left [ \sum\_{m=1}^n \frac{|H\_m (f\_m)|}{\|f\_m\|\_{L\_{p}(\mu\_m, X)}} \right ]^q \le \left [ \frac{|H (f)|}{\|f\|\_{L\_{p}(\mu, X)}} \right ]^q.
$$



Does $L_{p}(\mu, X)^*=L_{q} (\mu, X^*)$ hold for $\sigma$-finite measure spaces?
Below is my formalization of @Nik's hints to finish the proof.




---


Let's prove that
$$
\sum\_{m=1}^M \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*} \le \|H\|^q\_{L\_{p}(\mu, X)^\*} \quad \forall M \in \mathbb N^\*.
$$


Let $\Omega' := \bigcup\_{m=1}^M \Omega\_m$. We define a measure $\mu'$ on $\Omega$ by
$$
\mu' (B) := \mu(B \cap \Omega') \quad \forall B \in \Sigma.
$$


Let $\varphi' : L\_{p}(\mu', X)^\* \to L\_{q} (\mu', X^\*)$ be the canonical isometric isomorphism, i.e.,
$$
K (f) = \int\_\Omega \langle \varphi'(K), f \rangle \mathrm d \mu' \quad \forall K \in L\_{p}(\mu', X)^\*, \forall f \in L\_{p}(\mu', X).
$$


Then we have



> 
> [Lemma](https://mathoverflow.net/questions/434787/if-h-in-l-q-mu-x-such-that-int-langle-h-f-rangle-mathrm-d-mu) Let $N \in \Sigma$ and $K \in L\_{p}(\mu', X)^\*$ such that $K(f1\_N)=0$ for all $f \in L\_{p}(\mu', X)$. Then $\varphi'(K)=0$ on $N$.
> 
> 
> 


We define $H' ,H'\_m \in L\_{p}(\mu', X)^\*$ by
$$
H' (f) := H (f 1\_{\Omega'}) \quad H'\_m (f) := H (f 1\_{\Omega\_m}) \quad \forall f \in L\_{p}(\mu', X).
$$


Then $H' = \sum\_{m=1}^M H'\_m$. Also,
$$
\| H'\|\_{L\_{p}(\mu', X)^\*} \le \|H\|\_{L\_{p}(\mu, X)^\*}
\quad \text{and} \quad 
\|H'\_m\|\_{L\_{p}(\mu', X)^\*} = \|H\_m\|\_{L\_{p}(\mu\_m, X)^\*}.
$$


By our **Lemma**, the supports of $\{\varphi'(H'\_m)\}\_{m=1}^M$ are pairwise disjoint and thus justifies $(\star)$ below. We have
$$
\begin{align}
\sum\_{m=1}^M \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*} &= \sum\_{m=1}^M \|H'\_m\|^q\_{L\_{p}(\mu', X)^\*} \\
 &= \sum\_{m=1}^M \|\varphi'(H'\_m) \|^q\_{L\_{q}(\mu', X^\*)} \\
&\overset{(\star)}{=} \bigg \| \sum\_{m=1}^M \varphi' (H'\_m) \bigg \|^q\_{L\_{q}(\mu', X^\*)} \\
&= \bigg \| \varphi'\bigg (\sum\_{m=1}^MH'\_m \bigg) \bigg \|^q\_{L\_{q}(\mu', X^\*)} \\
&= \bigg \|\sum\_{m=1}^MH'\_m \bigg\|^q\_{L\_{p}(\mu', X)^\*} \\
&= \| H'\|^q\_{L\_{p}(\mu', X)^\*} \\
&\le \|H\|^q\_{L\_{p}(\mu, X)^\*}.
\end{align}
$$


This completes the proof.


