Gradient is zero on the preimage of a negligible set

Question

Let $f\in W^{1,1}(\Omega)$, where $\Omega\subseteq\mathbb{R^n}$ is open, and $E\subset\mathbb{R}$ a negligible Borel subset, i.e. $\mathcal{L}^1(E)=0$.

Is it always the case that $Df\equiv 0$ a.e. on $f^{-1}(E)$?

The same result when $E$ is a singleton is well-known.

I need this as a lemma to prove that if $h:\mathbb{R}\to \mathbb{R}$ is Lipschitz then $h\circ f\in W^{1,1}(\Omega)$ (assuming $\Omega$ bounded or $h(0)=0$) and $\nabla(h\circ f)=(h' \circ f)\nabla f$, which I am quite confident to be true.

Answer 1

For example, let $\Omega \subset \mathbb R^2$ be a bounded open set comfortably distant from the origin and define $f : \Omega \to \mathbb R^2$ by $f(x) = \dfrac{x}{|x|}$. The image of $f$ lies on the unit sphere, hence $\mathcal L^2(f(\Omega)) = 0$.

The differential $Df$ is rather easily computed and seen to be nonzero on $\Omega$.