Gradient zero a.e on the the zero set In Brezis Functional Analysis Page 314 Point 4 it is given that for u in $W^{1,p}(\Omega)$ where $\Omega$ is any open set then $\nabla u=0$ a.e on the set where $\{u(x)=k\}$, k is a constant.
How does one prove this fact.
 A: Let $A = \{u = k\}$.  Let $\varphi_n : \mathbb{R} \to [0,1]$ be a smooth function supported inside $[k-\frac{1}{n}, k+\frac{1}{n}]$, with $\varphi_n(k)=1$.  Then $\varphi_n(u) \to 1_A$, pointwise and boundedly, so for each $i$, since $\partial_i u \in L^p$, dominated convergence gives $\varphi_n(u) \partial_i u \to 1_A \partial_i u$ in $L^p$ norm.  
Let $\psi_n(t) = \int_{-\infty}^t \phi_n(s)\,ds$ so that $\psi_n' = \varphi_n$; note $\psi_n \to 0$ uniformly.  Let $v \in C^\infty_c(\Omega)$ be arbitrary.  By the chain rule and integration by parts we have
$$\int (\varphi_n(u) \partial_i u) v = \int \partial_i (\psi_n(u)) v = -\int \psi_n(u) \partial_i v.$$
As $n \to \infty$ the left side converges to $\int (1_A \partial_i u) v$ and the right side converges to $0$.  So $\int (1_A \partial_i u) v = 0$.  Since $v$ was arbitrary and $C^\infty_c(\Omega)$ is dense in $L^q(\Omega)$, we conclude that $1_A \partial_i u = 0$ almost everywhere, which is the desired statement.
(Self-plagiarism disclosure: this proof is the very same argument as Lemma 7.4 of these notes of mine; I think I learned the argument from Nualart's The Malliavin Calculus and Related Topics, but it seems to be pretty standard.)
