# Is the $\sigma$-finiteness condition necessary to ensure that $L^p(\mu)$ is reflexive?

Suppose $(X,\mu)$ is a measure space, and $p,q>0$ such that $1/p+1/q=1$. We know that if $\mu$ is $\sigma$-finite, then $L^p(\mu)$, $L^q(\mu)$ are reflexive and dual to each other. The proof could be found in Rudin's Real and Complex Analysis or Stein's Real Analysis, based on Radon-Nikodym's theorem. However, I see another proof from Brezis's book, although with assumption that $X=\Omega\subseteq\mathbb R^n$ is open and $\mu$ is the Lebesgue measure, it seems that the proof could be generalized to the case that $(X,\mu)$ is any measure space, no matter whether $\mu$ is $\sigma$-finite. For simplicity, we assume that $p\ge2$ and we only show that $L^p(\mu)$ is reflexive. We know from Rudin's text that $L^p(\mu)$ is a Banach space, no matter whether $\mu$ is $\sigma$-finite. The proof that $L^p(\mu)$ is reflexive is outlined as follows:

1. Clarkson's inequality: $$\left\lVert\frac{f+g}2\right\rVert_{L^p}^p+\left\lVert\frac{f-g}2\right\rVert_{L^p}^p\le\frac12\left(\lVert f\rVert_{L^p}^p+\lVert g\rVert_{L^p}^p\right)$$
2. $L^p(\mu)$ is uniformly convex, therefore reflexive.

I don't know where $\sigma$-finiteness is implicitly used here, or it's applicable to the case that $\mu$ is not $\sigma$-finite. I need some help.

Thanks!

$L^p (\mu)$ is always reflexive for $1<p<\infty$.
EDIT: For the cases $p=1$ or $p=\infty$, this is almost never true.
But what is still true in the case $p=1$ (if the measure is sigma finite) is that the dual space of $L^1$ is $L^\infty$. In the non sigma finite case, this can fail.
If I recall correctly, Rudin even proves it for non sigma-finite spaces for $1<p<\infty$. But maybe I am confusing his book with Folland's "Real Analysis".
• It should be Folland's book. In addition, $L^1$ isn't reflexive in general even if $\mu$ is $\sigma$-finite, say, $(L^1(\mathbb R))^*=L^\infty(\mathbb R)$. – Yai0Phah Nov 15 '14 at 2:32