Finding a domain with $C^1$ boundary

Question

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and let $F$ be a compact subset of $\Omega$. Does there exist an open subset $U$ of $\Omega$ such that $U$ has a $C^1$ boundary and satisfies $F \subset U \subset \overline U \subset \Omega$? Intuitively, we should always be able to find such a subset $U$ that possesses an arbitrarily smooth boundary, however I have no idea how to show this in a somewhat rigorous fashion.

Answer 1

We first fatten $F$ by $F' = \{x: \text{dist}(x,F)\le \delta\}$ for some small $\delta$ such that $F'\subset \Omega$.

Let $\eta$ be a standard mollifier and $\eta_\varepsilon(x) = \frac{1}{\varepsilon^n}\eta(\frac{x}{\varepsilon})$. If $f_\varepsilon = \eta_\varepsilon * \chi_{{}_{F'}},$ then for all $\varepsilon>0$ the functions $f_\varepsilon$ are smooth. The support of $f_{\varepsilon}$ will be compact and for $\varepsilon$ small $$\text{supp }f_{\varepsilon}\subset \Omega.$$

By Sard's theorem, the critical values of $f_\varepsilon$ have Lebesgue measure $0$, and so for each fixed $\varepsilon$ small enough we have $$\{x: f_{\varepsilon}(x)>t\}\subset\text{supp }f_\varepsilon\subset\Omega $$ is open and smooth for a.e. $t\in (0,1)$.

So for some $t>0$ and $\varepsilon$ small, $U = f_{\varepsilon}^{-1}((t,\infty))\subset \Omega$ is an open set with $C^\infty$ boundary. Since $f_\varepsilon(x) = 1$ for all $x\in F\subset F'$, this set $U$ is such a desired open set.