We have a smooth bounded domain $\Omega$, and sets $A$, $B$ and $A^+$ such that $A=A^+ \cup B$ with $B$ a set of capacity zero. (See below for what these sets are precisely).
Let $$S=\{ v \in H^1_0(\Omega) : v \geq 0 \text{ q.e. on $A$ and } v=0 \text{ a.e. on } A^+\}$$
Apparently, it is possible to say that in fact $$S=\{ v \in H^1_0(\Omega) : v = 0 \text{ q.e. on $A$ }\}$$ but I don't see why. Can someone explain?
All I see is that $v =0 $ a.e. on $A$, because $B$ has capacity zero (and hence Lebesgue measure zero).
Here, q.e. means quasieverywhere (i.e., everywhere except on a set of capacity zero).
If it helps: In my case, $A= \{u = 0\}$, $$B=\{u = 0\} \cap \{-\Delta u -f =0 \}$$ and $$A^+ = \{u=0\}\setminus \{-\Delta u -f = 0\},$$ where $u \in H^1(\Omega)$ with $f, \Delta u \in L^2(\Omega)$ and $-\Delta u -f \geq 0$ a.e. on $\Omega$.