Let $\Omega$ be a open subset of $\mathbb{R}^{d}$ and $C_c^{m}(\Omega)$ the space of $m$-times continuously differentiable functions with compact support with $0 \leq m \leq \infty$. Denote by $\partial X$ the boundary of the set $X$ and by $\operatorname{meas}(X)$ the measure of the set $X$.
My question: Is it true that $\operatorname{meas}(\partial(\operatorname{supp}(f)))=0$ for all $f \in C_c^{m}(\Omega)$?
Intuitively it seems true. For example, if $n=1$ then $\partial(\operatorname{supp}(f))$ has two points. If $n=2$, then $\partial(\operatorname{supp}(f))$ looks like a deformed circumference. But I do not know how to give an analytical proof of this fact.
Observations: The support of $f$ is the set $\operatorname{supp}(f)=\overline{\{x \in \Omega:f(x) \neq 0\}}^{\Omega}$. Here $\operatorname{meas} (X)=0$ can be the Lebesgue measure or prove that for all $\varepsilon>0$ there is a sequence of $d$-dimensional cubes $C_1, C_2, \dots, C_i, \dots$ such that $\operatorname{supp}(f) \subset \bigcup_{i=1}^{\infty} C_i$ and $\sum_{i=1}^{\infty} \operatorname{vol} C_i<\varepsilon.$
This is false even for $C^\infty_c$ functions on the real line. Let $A$ be a "fat" Cantor set; it's obtained similarly to the traditional Cantor set, by starting with $[0,1]$ and repeatedly removing middle intervals, but taking those removed pieces to be not middle-thirds but middle-tiny-intervals, getting tinier from stage to stage, so rapidly that the left-over $A$ has positive measure (in contrast to the traditional Cantor set's zero measure).
Now build the function $f$ by putting a $C^\infty$ bump function on each of the removed middle intervals, making the heights of the bumps shrink rapidly so that the whole $f$ is $C^\infty$. It's non-zero exactly on those middle intervals, so the boundary of its support is all of $A$, which has positive measure.