Limit of an $L^2$ sequence We have functions $f_n\in L^2$ such that $f_n(x)$ tends to some $f(x)$ for almost all $x$, suppose that $||f_n||<M$. Does this mean that $f_n$ tend to $f$ weakly? In my previous question about $L^1$ I realized that the weak convergence fails; can one describe spaces for which this is true?
 A: Yes.
By Fatou's lemma we have $f \in L^2$, so by replacing $f_n$ by $f_n - f$ we can assume that $f=0$.
Fix $g \in L^2$ and $\epsilon > 0$.  By a standard dominated convergence argument, there is a set $A$ of finite measure with $\int_{A^c} |g|^2 < \epsilon$.  There is also a $K$ so large that $\int_{|g| > K} |g|^2 < \epsilon$.  Let $B = A \cap \{|g| \le K\}$; then $\lvert\int_{B^c} f_n g \rvert \le M \sqrt{2 \epsilon}$ by Cauchy--Schwarz, for every $n$.
On the other hand, for each $n$ we have $\int_B |f_n g|^2 \le K^2 M$.  So $\{f_n g\}$ is an $L^2$-bounded sequence of functions on the finite measure space $B$, hence it is uniformly integrable on $B$.  We also have $f_n g \to 0$ almost everywhere, and hence they converge in measure on $B$; by the Vitali convergence theorem $f_n g \to 0$ in $L^1(B)$, so $\int_B f_n g \to 0$.  Letting $\epsilon \to 0$ completes the proof.
Edit: With respect to your other question (when can you say that an $L^1$-bounded, a.e. convergent sequence must converge weakly in $L^1$), the answer is: only in trivial cases.  Essentially, if the measure space is anything but a finite union of atoms, you can construct a sequence of nonnegative $f_n \in L^1$ with $f_n \to 0$ a.e. but $||f_n||_{L^1} = 1$, so that taking $g=1 \in L^\infty$, you have $\int f_n g = 1$ for all $n$.  (Idea: let $f_n = \mu(A_n)^{-1} 1_{A_n}$ where either $\mu(A_n) \uparrow \infty$, or $\mu(A_n) \downarrow 0$ fast (in particular try $\sum \mu(A_n) < \infty$)).
