Does for $u\in L^1(\Omega)$ and every $t$ also hold $\nabla u \cdot 1_{\{u=t\}}=0$ a.e.?

Question

The result is known if $u$ is more regular e.g. $u \in W^{1,1}(\Omega)$. Is it also possible to extend such an result to mere integrable or even just measurable functions?

Unfortunately the result for $u\in W^{1,1}$ requires the strong regularity of $u$. Trying to apply the definition of distributional derivative is also bad since one can not differentiate the function $1_{\{u=t\}}$. Even approximating this function seems to fail

Answer 1

I don't think the question is correctly stated (what does it mean to multiply the distribution $\nabla u$ by the characteristic function of a measurable set?), but in any case the answer should be negative. Let $E$ be a fat (positive measure) Cantor set in $\mathbb R$. Its characteristic function $1_E$ is in $L^1$. The gradient $\nabla 1_E$ is a distribution with support $E$. This is the opposite of it being zero on $E$, whatever that could possibly mean.