Let $\Omega\subseteq\mathbb{R}^N$ be an open and bounded set with uniform Lipschitz boundary. Consider the signed distance function:
$d:\mathbb{R}^N\to\mathbb{R},\ d(x)=\begin{cases} \mathrm{dist}(x,\partial\Omega),\ x\notin\Omega \\ 0,\ x\in\partial\Omega \\ -\mathrm{dist}(x,\partial\Omega),\ x\in\Omega\end{cases}$
It is well-known that $d$ is almost everywhere differentiable (being 1-Lipschitz) and $|\nabla d(x)|\leq 1$. Also it is known that $|\nabla d(x)|=1$ in a neighbourhood of $\partial\Omega$ if $\Omega$ has a $C^2$ boundary (see Gilbard&Trudinger, page 355). Of course we know that $\partial\Omega$ is a compact set with finite perimeter and with null Lebesgue measure.
My question is the following: Is it true that the set $\{x\in \mathbb{R}^N\ |\ \exists y_1\neq y_2\in\partial\Omega\ \text{with}\ d(x)=|x-y_1|=|x-y_2|, x\notin\Omega \ \text{and} \ d(x)=-|x-y_1|=-|x-y_2|, x\in\Omega\}$ has a null Lebesgue measure?
This fact would guarantee that $|\nabla d(x)|=1$ almost everywhere for $\Omega$ a Lipschitz domain. Is this true?
