This is an exercise on Evans PDE book, Ch5. It provides another way to prove $\nabla u =0$ a.e. on $u=0$ for $u\in H^1(\Omega)$ then the one in Evans & Griapy's book.
The statement is as following. Choose $\phi$ smooth on $R$ such that $\phi(z)=z$ if $|z|<1$ and $|\phi(z)|\leq |z|$ for all $z\in R$. Moreover, $\phi$ is constant if $|z|>1+\delta$ for some $\delta>0$. We also require that $\phi'$ is bounded.
Then given any $u\in H^1(\Omega)$ we define $u_\epsilon:=\epsilon \phi (u/\epsilon)$ and it is easy to verify that $u_\epsilon$ is uniformly bounded in $H^1$ and hence we have subsequence convergence weakly to some $u$.
It is also very quick to prove that $u_\epsilon\to 0$ weakly in $L^2$. However, I got stack at the point on how to prove $\nabla u_\epsilon\to 0$ weakly in $L^2$ as well. here is what I tried: for any $g\in L^2(\Omega)$, $$\lim_\epsilon \int_\Omega \partial_i u_\epsilon \,\,g dx = \lim_\epsilon \int_\Omega \partial_i \phi'(u/\epsilon)\partial_i u \,\,\,g dx= \int_\Omega \lim_\epsilon\partial_i \phi'(u/\epsilon)\partial_i u \,\,\,g dx = \int_{\{u=0\}}\partial_i u\,\,gdx$$ But I can not prove the limit is finally $0$... Any help would be very welcome!