Proof or counterexample: $L^p$-boundedness gives a.e. convergent subsequence? Let $\Omega\subset\mathbb{R}^{d}$ open and let $f_{n}\in L^{2}\left(\Omega\right)$ be bounded.
Then there is obviously a weakly convergent subsequence.
Is there also a subsequence converging almost everywhere?
Do you know of a counterexample, i.e., a weakly convergent (or bounded) sequence without any subsequence converging almost everywhere?
Thanks a lot in advance! :-)
 A: I wish to elaborate on the example provided by David Mitra.
Assume  that $\Omega=(0,2\pi)$ and $f_n(x)=\sin nx$, and assume that this sequence has a subsequence which converged almost everywhere, i.e., $f_{n_k}(x)\to f(x)$, almost everywhere in $[0,2\pi]$. 
Clearly $\lvert f(x)\rvert\le 1$, and $f$ measurable, as a limit of measurable functions. Note also that all the $f_n$'s and $f$ belong to all the $L^p[0,2\pi]$ spaces, since they are all bounded.
But now we are allowed to use Lebesgue Dominated Convergence Theorem, which implies that
$$
\int_E f_{n_k}\to \int_E f, \tag{1}
$$
for all measurable $E\subset [0,2\pi]$. But $\int_E \sin nx\,dx\to 0$, for all such $E$, due to Riemann Lebesgue Lemma, which implies that only $f\equiv 0$ can satisfy $(1)$. However, Lebesgue Dominated Convergence Theorem also implies that
$$
\int_0^{2\pi} \lvert f_{n_k}(x)-f(x)\rvert\,dx=\int_0^{2\pi} \lvert f_{n_k}(x)\rvert\,dx\to 0,
$$ 
which of course is not true as $\int_0^{2\pi} \lvert \sin nx\rvert\,dx=4$, for all $n$.
