The dual of $L^1(G)$ for a locally compact group $G$

Question

I might be missing something, but most literature on topological groups and harmonic analysis that I've encountered mention that $L^\infty(G)$ can be naturally identified with the dual of $L^1(G)$ by means of the isomorphism $$\begin{array}{ccc}L^\infty(G)&\longrightarrow&L^1(G)^*\\f&\mapsto&\int fg\, \mathrm d\mu\end{array}$$ where $\mu$ is a fixed Haar measure. However, when dealing with measure spaces $(X,\mu)$ in general, a lot of literature also states that the above map is an isomorphism of $L^\infty(X)$ onto $L^1(X)$ if the measure $\mu$ is $\sigma$-finite. Are Haar measures always $\sigma$-finite (I would think not), and if not, then why does so much literature insist that the above identification works?

Answer 1

In case of non-separable groups $G$, you have to slightly re-define $L^\infty(G)$ for this to work. Identify two functions if they agree "locally almost everywhere", that is, if they agree a.e. on every set of finite measure.

Note: "almost everywhere" and "locally almost everywhere" are the same for functions with $\sigma$-finite support, and that includes all $L^p$ with $0<p<\infty$.

Reference: Hewitt & Stromberg, Real and Abstract Analysis. pages 346--353. Haar measure on a locally compact Hausdorff group is "decomposable" in the sense used there.