If $u=v$ on $A \subset \Omega$, then $\nabla u = \nabla v$ on $A$ too Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and let $A \subset \Omega$ be measure nonzero. For $u, v \in H^1(\Omega)$, if $u=v$ (a.e) on $A$, how to prove that $\nabla u = \nabla v$ on $A$?
This is problematic since this is the weak gradient.
Reference also appreciated.
 A: The following is more or less the same as Evans, Partial Differential Equations, Chapter 5, Exercise 18.
Let $u \in H^1(\Omega)$. For $\varepsilon > 0$ let
$$
F_\varepsilon (z) := \begin{cases}
\left(z^{2}+\varepsilon^{2}\right)^{1/2}-\varepsilon, & z\geq0,\\
0, & z<0.
\end{cases}
$$
Show $F_\varepsilon\circ u \in H^1(\Omega)$, $(F_{1/n} \circ u)_n$ is Cauchy in $H^1(\Omega)$ and $F_1/n \circ u \rightarrow u^+$ a.e., where $u^+$ is the positive part of $u$.
Furthermore, show that
$$
\partial_j (F_\varepsilon \circ u)(x) \xrightarrow[\varepsilon \downarrow 0]{} \begin{cases}
\left(\partial_{j}u\right)\left(x\right), & u\left(x\right)>0,\\
0, & u\left(x\right)\leq0
\end{cases}
$$
almost everywhere.
This will yield $u^+ \in H^1(\Omega)$ with weak derivative
$$
\partial_j u^+(x) = \begin{cases}
\left(\partial_{j}u\right)\left(x\right), & u\left(x\right)>0,\\
0, & u\left(x\right)\leq0
\end{cases}
$$
Do the same for $u^-$ instead of $u^+$.
This will finally yield
$$
\partial_j u = \partial_j u^+ - \partial_j u^-,
$$
a.e., where we used that $u = u^+ - u^-$.
But the above identity implies $\partial_j u = 0$ a.e. on $\{x \mid u(x) = 0\}$.
Finally, apply the above to $u-v$.
A: Obviously, it suffices to show that $\,u=0\,$ a.e. on $A$ implies $\,\nabla u=0\,$ a.e. on $A$.  It is clear that the approximate derivative $ap\,\nabla u=0\,$ a.e. on $\{x\in\Omega\colon\, u(x)=0\}$ — for details see theorem 3 in section 6.1.3 of "Measure theory and fine properties of functions" by L.C. Evans and R.F. Gariepy. Hence by remark $(ii)$ in theorem 4 (Ibid.), the weak derivative $\nabla u=ap\,\nabla u=0\,$ a.e. on a subset $A\subset \{x\in\Omega\colon\,u(x)=0\}$.   Q.E.D.
