If $u_n \geq 0$ quasieverywhere, is also $u \geq 0$ quasieverywhere?

Question

Let $u_n \to u$ in $W^{1,p}(\Omega)$ on abounded domain, where eg. $p=2$.

If $u_n \geq 0$ quasieverywhere, is also $u \geq 0$ quasieverywhere?

Would weak convergence be sufficient for the limit to be non-negative too?

Quasieverywhere means it holds on all $x \in \Omega$ except on a set of capacity zero.

Answer 1

Yes, this is possible, since $u_n \to u$ in $W^{1,p}(\Omega)$ implies that there is a subsequence $u_{n_k}$ such that $u_{n_k}(x) \to u(x)$ for quasi-every $x \in \Omega$. Now, you first assertion follows easily.

Edit: It is also possible to consider the following: you can easily show $u(x) \ge 0$ for almost all $x \in \Omega$. Since $u$ is quasi-continuous, this implies $u(x) \ge 0$ for quasi-all $x \in \Omega$.

In the case of weak convergence, you can apply Mazur's lemma and get that a sequence of convex combinations of $u_n$ converges strongly to $u$. Together with the first assertion, this yields the claim.