Let $v \in H^1_0(\Omega)$ on a smooth bounded domain. Suppose we hknow that $v \geq 0$ quasieverywhere on $I \subset \Omega$, where $I$ is a closed subset, and also $v = 0$ a.e. in $I$.
Does it follow that $v = 0$ quasieverywhere on $I$?
I think it does since $v$ has a quasicontinuous representative and if a continuous function is zero a.e. it is zero everywhere.
No, this is not true. Let $I$ be a subset of a closed hyperplane. Then, the measure of $I$ is zero, hence $v = 0$ a.e. on $I$ is vacuously satisfied. Now, it is clear that $v \ge 0$ q.e. on $I$ does not imply $v = 0$ q.e. on $I$.