What can we say about the measure of the boundary of a level set of an infinitely differentiable function?

Question

Let $u$ be function in $\mathbb{R}^n$ such that $u: U \rightarrow \mathbb{R}$ (here $U \subset \mathbb{R}^n$), $u \in C^{\infty}(\overline{U})$. Suppose $m(\partial U) = 0$ and $U$ has a $C^1$ boundary. Is it true that $m(\partial \{z: u(z) = 0 \}) = 0$. Here $m$ is the usual lebesgue measure in $\mathbb{R}^n$ i.e $m(A) = \int_{A}1_{A}(x) dx$ and $\partial A$ denotes the boundary of $A$.

Answer 1

For any closed subset $F\subseteq \mathbb{R}^n$, there exists a smooth function $u : \mathbb{R}^n \to \mathbb{R}$ such that $F$ is the zero-set of $u$.

Now pick a compact set $F \subset U$ so that $\partial F$ has positive measure and let $u$ be any smooth function having $F$ as zero-set of $u$.

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Proof of the first claim: We know that the bump function

$$ \varphi(x) = \begin{cases} \exp\left(-1/(1-\|x\|^2)\right), & \|x\| < 1 \\ 0, & \|x\| \geq 1 \end{cases} $$

is smooth. Now fix a closed set $F \subseteq \mathbb{R}^n$ and let $V = \mathbb{R}^n \setminus F$ be its complement. Then there exists a countable family of open balls $\{ B(x_k, r_k) : k = 1, 2, \cdots \}$ such that $V = \bigcup_{k=1}^{\infty} B(x_k, r_k)$. Using this, for each multi-index $\alpha \in \mathbb{N}_0^{n}$ with $|\alpha| = \alpha_1 + \cdots + \alpha_n$, we define

$$ u_{\alpha}(x) = \sum_{k=1}^{\infty} \frac{(r_k \wedge 1)^k}{k!} \cdot \frac{1}{r_k^{|\alpha|}} (\partial^{\alpha}\varphi)\left(\frac{x - x_k}{r_k} \right), $$

where $a \wedge b := \min\{a, b\}$. Then we make the following observations.

• The sum defining $u_{\alpha}$ converges uniformly for each $\alpha \in \mathbb{N}_0^{n}$. So it follows that $u = u_0$ is infinitely differentiable with $\partial^{\alpha} u = u_{\alpha}$.

• For each $x \in B(x_k, r_k)$, we have $u(x) \geq \frac{(r_k\wedge 1)^k}{k!} \phi\left(\frac{x-x_k}{r_k}\right) > 0$.

• If $x \notin U$, then $u(x) = 0$ since each summand vanishes.

Combining altogether, $u$ is a non-negative smooth function such that $u(x) > 0$ if and only if $x \in U$. Therefore $F$ is exactly the zero-set of $u$.