It is a theorem of Steinhaus that for any finite measure $\mu$, the Banach space $L\_1(\mu)$ is weakly sequentially complete. Using the Radon-Nikodym theorem one can extend this easily to $\sigma$-finite measures. What about arbitrary measures. Is $L\_1(\mu)$ always weakly sequentially complete?



How does one prove that $L_1(\mu)$ is weakly sequentially complete for any measure?
Yes. You can reduce the general case to the separable case, and every separable $L\_1$ is clearly isometrically isomorphic to $L\_1(\mu)$ with $\mu$ a finite measure. 


For the reduction of the general case, let $X$ be the closed sublattice generated by your weakly null sequence. This is an abstract $L\_1$ space and so, by Kakutani's theorem, is isometrically isomorphic to an $L\_1$ space. 


