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Let $u\in C^\infty(\mathbb{R}^d)$ and consider $$ E = \partial\{x:Du(x)=0\}. $$ I am interested in understanding whether $\mu(E)=0$ or not, $\mu$ being the $d$-dimensional Lebesgue measure. What I could prove so far is that there exists a point $x\in E$ such that, for all neighborhoods $U\in U(x)$ we have $$ \mu(E\cap \overline{U})>0. $$ Any idea on how to proceed with proving that $\mu(E)>0$, assumed that it is possible?

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The measure $\mu(E)$ may be positive or zero. It is easy to give examples of the case $\mu(E) = 0$, so below is an example in $C^\infty(\mathbb{R})$ with $\mu(E) > 0$.

In light of this post here, it suffices to exhibit a closed subset of $\mathbb{R}^n$ whose boundary has positive measure. In the spirit of what you pointed out in your question, the so-called Fat Cantor Set is closed with empty interior (hence it is its own boundary). Moreover, by computing a simple geometric sum one sees that it has positive measure.

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  • $\begingroup$ It is possible to construct functions whose levels sets are a given closed set. However, in my case the function needs to be of the form $f(x) = |Du(x)|^2$ or something else such that $\partial f^{-1}(0)=E$. So maybe this constraint cancel the construction you are talking about. $\endgroup$ – Tommaso Seneci Jun 12 '18 at 17:16
  • $\begingroup$ @TommasoSeneci Let $f \in C^{\infty}(\mathbb{R})$ such that $f^{-1}(0)$ is given by the fat cantor set. Such a function exists, as we agree. Then just let $F$ be the antiderivative of this function. $\endgroup$ – Pete Caradonna Jun 12 '18 at 18:15
  • $\begingroup$ you are right, thanks! $\endgroup$ – Tommaso Seneci Jun 12 '18 at 18:31

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