On the locality of the gradient of a Sobolev function

Question

Let $\Omega$ be an open subset of $\mathbb{R}^n$. Suppose that $u,v\in W^{1,1}(\Omega)$ and that $u = v$ on a Borel subset $E\subseteq \Omega$.

Question: is it true that $\nabla u= \nabla v$ a.e. in $E$?

Answer 1

A possible proof goes like follows. First, note that it is sufficient to prove that if $u=0$ a.e. on a Borel subset $E$ of $\Omega$ then $\nabla u=0$ a.e. in $E$. Now, it is possible to show [EG, Theorem 4.4] that if $u\in W^{1,1}(\Omega)$ then $\nabla u=0$ a.e. on the zero-level set $\{u=0\}$. But this concludes the proof because if $u=0$ a.e. in $E$ then $E\subseteq \{u=0\}$.

[EG] Evans Lawrence C. and Ronald F. Gariepy. Measure theory and fine properties of functions. Chapman and Hall/CRC, 2015.