Let $\Omega \subset \mathbb{R}^n$ be open and $u \in W_{loc}^{1, 1}(\Omega)$. It is a well known fact that $\nabla u = 0$ a.e. on the set $$Z := \{x \in \Omega: u(x) = 0 \}.$$ One usually proves this by computing the weak gradients of the positive and negative parts of $u$ and showing that they are $0$ whenever $u$ is $0$. I saw a very nice proof of this fact for Lipschitz functions and I am wondering if there is a generalization of the proof. It goes as follows:
Suppose $u$ as defined above is in fact locally Lipschitz. By Rademacher's theorem it is differentiable a.e. in $\Omega$. Let $x \in Z$ be a Lebesgue point of $Z$, where $u$ is differentiable (this is true for a.e. point of $Z$). If $\nabla u(x) \neq 0$, then $\nabla u(x) \cdot \nu > \frac{1}{2}|\nabla u(x)|$ for all $\nu$ on some nonempty open subset $V$ of $\mathbb{S}^{n-1}$. It now follows from $$u(x + t\nu) = \nabla u(x) \cdot t\nu + o(t\nu)$$ that there exists a $t_0$ such that $u(x + t\nu) > 0$ for all $0 < t < t_0$ and $\nu \in V$. But then $x$ can not be a Lebesgue point of $Z$.
I am curious whether something similar can be done if $u$ is merely weakly differentiable. Hence:
Question: If $u \in W_{loc}^{1, 1}(\Omega)$ and $x \in Z$ is a Lebesgue point of $Z$ as well as of $\nabla u$, is it true that $\nabla u(x) = 0$?
More specifically, I would like to know if we can generalize the strategy of looking what happens locally around a Lebesgue point.
Okay, I think I figured it out. We need to use approximate differentiability of weakly differentiable functions. This means for a.e. $x \in \Omega$ $$\operatorname*{ap lim}_{y \to x} \frac{|u(y) - u(x) - \nabla u(x) \cdot (y - x)|}{|y - x|} = 0.$$
See Evans & Gariepy, Section 6.3.1 for a discussion. Given this, the proof is quite similar.