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?
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2 Answers
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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.
answered Jun 5, 2014 at 15:14
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The support of any $L_1$ function is $\sigma$-finite. A countable union of $\sigma$-finite sets is $\sigma$-finite. So a sequence of $L_1$ functions is supported by a $\sigma$-finite set. Now you are in the "easy extension" mentioned.
answered Jun 5, 2014 at 23:56