Let $f\in C^1(\mathbb R^d)$ for some $d\in\mathbb N$. Are we able to show that (under mild additional assumptions) $\partial\{f=0\}$ is a null set wrt the Lebesgue measure $\lambda^d$ on $\mathcal B(\mathbb R^d)$?
We may note that if $a\in\mathbb R^d$ is a regular point of $f$ (i.e. $\nabla f(a)\ne 0$), then $\{f=f(a)\}$ is a $(d-1)$-dimensional submanifold of $\mathbb R^d$ and hence is a $\lambda^d$-null set. But beyond this observation, I wasn't able to tackle the problem. Maybe we can show in general that the (topological) boundary of a $k$-dimensional ($k<d$) submanifold of $\mathbb R^d$ is a $\lambda^d$-null set?
Given any closed set $E$ in $\mathbb R^{d}$ there is a $C^{\infty}$ function such that $f(x)=0$ for all $x \in E$ and $f(x) >0$ for all $x \in E^{c}$. [See Exercise 3) in the chapter on Test Functions and Distributions in Rudin's FA]. In particular if we take $E$ to be a fat Cantor set in $\mathbb R$ then $\partial f^{-1}(\{0\})$ has positive measure.