Let $u_n$ weakly converge to $u$ in $H^1(\Omega)$ where $\Omega$ is a bounded smooth domain. We have the following information:
$$u_n \geq 0 \text{ quasieverywhere on $\Omega_0$}$$ where $\Omega_0 \subset \Omega$ is a set that may be closed. Does it follow that $u \geq 0$ quasieverywhere on $\Omega_0$?
I think it should follow from the pointwise a.e. convergence of $u_n$ to $u$ and the fact that the countable union of capacity zero sets has capacity zero but I wonder if I am missing something.
I don't think that the pointwise a.e. convergence of $u_n$ helps. However, by Mazur's lemma, there is a sequence $\{v_n\}$ consisting of convex combinations of the $u_n$, such that $v_n \to u$ strongly in $H^1(\Omega)$. This sequence has a subsequence which converges pointwise q.e. Hence, the limit $u$ is non-negative up to a countable union of sets of capacity zero.