Weak limit of an $L^1$ sequence We have functions $f_n\in L^1$ such that $\int f_ng$ has a limit for every $g\in L^\infty$. Does there exist a function $f\in L^1$ such that the limit equals $\int fg$? I think this is not true in general (really? - why?), then can this be true if we also know that $f_n$ belong to a certain subspace of $L^1$?

Perhaps surprisingly, the answer is yes.
More generally, given any Banach space $X$, a sequence $\{x_n\} \subset X$ is said to be weakly Cauchy if, for every $\ell \in X^*$, the sequence $\{\ell(f_n)\} \subset \mathbb{R}$ (or $\mathbb{C}$) is Cauchy.  If every weakly Cauchy sequence is weakly convergent, $X$ is said to be weakly sequentially complete.
Every reflexive Banach space is weakly sequentially complete (a nice exercise with the uniform boundedness principle).  $L^1$ is not reflexive, but it turns out to be weakly sequentially complete anyway.  This theorem can be found in P. Wojtaszczyk, Banach spaces for analysts, as Corollary 14 on page 140.  It works for $L^1$ over an arbitrary measure space.
