TITLE: When exactly is the dual of $L^1$ isomorphic to $L^\infty$ via the natural map?
QUESTION [35 upvotes]: The dual space to the Banach space $L^1(\mu)$ for a sigma-finite measure $\mu$ is $L^\infty(\mu)$, given by the correspondence 
$\phi \in L^\infty(\mu) \mapsto I_\phi$, where $I_\phi(f) = \int f \cdot \phi \,d\mu$ for $f \in L^1(\mu)$.
Now, we consider this for more general measures $\mu$. If $\mu$ is not semifinite, then there is a measurable subset $A$ with $\mu(A) = \infty$ but $\mu(B) = 0$ for every subset $B$ of $A$ with $\mu(B) < \infty$. Any $\phi$ supported on $A$ will then map to the zero functional, so the correspondence is not one-to-one.
If $\mu$ is semifinite, the correspondence is one-to-one, but is it onto? 
Using the Radon-Nikodym Thorem, any member of $L^1(B)^*$ for $\mu(B) < \infty$ can be represented by an $L^\infty(B)$ function class $[\phi_B]$. Unlike in the $\sigma$-finite measure case, however, I do not see an easy way to form a "patchwork" $\phi$ with $\phi|_B = \phi_B$ a.e.$[\mu]$. I suspect this may have an easy answer, but it eludes me. Or perhaps there is a well-known counterexample?

REPLY [30 votes]: To answer the question in the title, there is the following general theorem:

The natural map $I \colon L^\infty(X) \to (L^1(X))^\ast$ is

an isometric injection if and only if $(X,\Sigma,\mu)$ is semifinite.
an isometric isomorphism if and only if $(X,\Sigma,\mu)$ is localizable.


You already covered point 1 in your question (the isometry property being straightforward). The second point is more delicate.
A measure space $(X,\Sigma,\mu)$ is called localizable if it is semifinite and in addition the following condition holds:
For every family $\mathcal{F} \subseteq \Sigma$ there is $H \in \Sigma$ such that 

$F \setminus H$ is a null set for all $F \in \mathcal{F}$.
If $G \in \Sigma$ is such that $F \setminus G$ is a null set for all $F \in \mathcal{F}$ then $H \setminus G$ is a null set.

Loosely speaking, this property asserts that every family $\mathcal{F}$ of measurable sets has a smallest measurable envelope $H$ (up to null sets).
The definition of localizability implies via a slightly technical argument that one can glue together measurable functions. More precisely:

Let $\mathscr{F}$ be a family of functions such that each $f \in \mathscr{F}$ is defined and measurable on a measurable subset $D_f$ of the localizable measure space $(X,\Sigma,\mu)$. Suppose in addition that $f_1 = f_2$ a.e. on $D_{f_1} \cap D_{f_2}$ whenever $f_1,f_2 \in \mathscr{F}$. Then there is a measurable function $g \colon X \to \mathbb{R}$ whose restriction to $D_f$ satisfies $g|_{D_f} = f$ a.e. for all $f \in \mathscr{F}$.

Using this result one can patch together the Radon-Nikodym derivatives one obtains from restricting a continuous linear functional $\varphi \colon L^1(X) \to \mathbb{R}$ to the subspaces $L^1(F) \subseteq L^1(X)$ where $F$ runs through the subsets of finite measure of $X$.
The proof of the converse direction proceeds by a direct verification of the localizability property of $(X,\Sigma,\mu)$ the fact that $I$ is isometric implies semi-finiteness by point 1. of the theorem and the "envelope condition" uses surjectivity of $I$.
Details can be found in 243G on page 153 of this PDF. The gluing property is proved in 213N on page 28.