Does $u=v$ almost everywhere on $A \subset \Omega$ imply $u=v$ quasi-everywhere on $A$?

Question

Let $A \subset \Omega$ be a subset of the bounded domain $\Omega$. $A$ is not quasi-open.

Suppose that we have two functions $u, v \in H^1(\Omega)$ such that $$u=v \quad\text{a.e. on $A$}$$ Does it follow that $u=v$ quasi-everywhere on $A$?

Answer 1

No. Let $A$ be the intersection of a hypersurface with $\Omega$. Then, the measure of $A$ is zero, but $A$ has a positive capacity. Now, consider $u = 0$ and $v = 1$.