6
$\begingroup$

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?

$\endgroup$

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.