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

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 $$A$$.

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 $$u$$.

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 $$u$$.

• I agree with your second part but I don't see why the first statement is true – acreativename Sep 15 at 1:10
• @acreativename, Although it is a well-known fact, I added a proof to show some of the basic ideas behind the claim. – Sangchul Lee Sep 15 at 1:33
• Sigh...I thought that this result would have been true and I would have used this to prove another question – acreativename Sep 15 at 1:39