Let $\Omega$ be a bounded domain with sufficiently smooth boundary. Let $u \in W^{1, 2}_{0}(\Omega)$ and $F \in C^{\infty}(\mathbb{R} \rightarrow \mathbb{R})$ such that $F(u(x)) = 0$ for almost every $x \in \Omega$. Must $F(0) = 0$?
My thoughts are: I think the answer is yes. Suppose $u$ instead was smooth. The $F(u(x)) = 0$ for every $x \in \Omega$. By continuity, $F(u(x)) = 0$ for every $x \in \overline{\Omega}$. Since $u = 0$ on the boundary $\partial \Omega$, then there exists an $x_{0}$ such that $u(x_{0}) = 0$. Thus $F(u(x_{0})) = 0$ and hence $F(0) = 0$. However, I am not sure how to argue if $u$ was not smooth. My first thought was to approximate $u \in W^{1, 2}_{0}$ by smooth functions, but then I'll be working with a Sobolev norm.
Suppose $F(0)\ne 0$. Then there is $\epsilon>0$ such that $F(x)\ne 0$ when $|x|<\epsilon$. Since $F\circ u=0$ a.e., it follows that the set $\{x\in \Omega: |u(x)|<\epsilon\}$ has zero measure.
So, the question reduces to showing that a function $u$ with $|u|\ge \epsilon$ a.e. cannot be in $W^{1,2}_0(\Omega)$. One way to do this is to recall that the zero extension of a $W^{1,p}_0$ function is in $W^{1,p}_0(\mathbb R^n)$. But if $|u|\ge \epsilon$ a.e., on $\Omega$, the zero extension of $u$ has an essential (unfixable by redefining on a set of measure zero) jump discontinuity on almost every line crossing $\Omega$. Thus, it does not have a representative with the ACL property, meaning it's not in any Sobolev class.
Note that the smoothness of the boundary is not needed.