I'm trying to understand a statement in the Brezis book, that $L^1(\Omega, \mu)$ is not reflexive. He divides the problem into two parts: one in which $\mu$ is a non atomic measure and another in which $\mu$ is purely atomic, i.e, $\Omega$ consists of a infinite countable number of distinct atoms.
My question: In the second case in which $\Omega$ consists of a infinite countable number of distinct atoms, how to show that $L^1(\Omega)$ isomorphic to $l¹$?
Suppose that $\mu$ is purely atomic and there are countably many atoms, call them $x_1, x_2, \ldots$. Let $f\in L_1(\Omega)$. Then
$$\int\limits_{\Omega} f(x)\,\mu({\rm d}x) = \sum_{n=1}^\infty \int\limits_{\{x_n\}} f(x)\,\mu({\rm d}x) = \sum_{n=1}^\infty f(x_n).$$
This shows that the map $f\mapsto (f(x_n))_{n=1}^\infty\in \ell_1$ is an isomorphism. As point out by Daniel Fisher, it need not be isometric. To make it such, just define
$$f\mapsto \left(\frac{f(x_n)}{\mu(\{x_n\})}\right)_{n=1}^\infty.$$