# Regular compact domains of a Riemannian manifolds

In a Riemannian manifold $M$ a regular compact domain $D$ is a compact subset of $M$ with non empty interior and such that for every $p \in \partial D$ there exists $\left(U,\varphi\right)$ coordinate neighbourhood of $p$ such that $\varphi(U\cap \partial D)\subseteq \partial \mathbb{R}^{n}_{+}$ and $\varphi (U\cap D)\subseteq \mathbb{R}_{+}^{n}$ where $\mathbb{R}^{n}_{+}$ is the set such that $x_n\geq 0$. Now let $R<R'$. My question: is there a regular compact domain $D$ such that $B_R \subset D$ and $\partial D \subset B_{R'}$ (where $B_R$ is the metric ball of $M$ )? I'm thinking that if we can construct a smooth function $\varphi$ such that $\varphi = 1$ on $B_R$, $supp(\varphi)\subset B_{R'}$ and $|\nabla \varphi |\neq 0$ on $interior( supp(\varphi)) \diagdown \overline{B_R}$ then the subset $\Omega=\{p \in M:\varphi(p)\geq \frac{1}{2}\}$ is the regular compact domain looked for. We have to assume completeness of $M$ in my argument. But if we replace the metric balls with two compact subsets $K \subset interior K'$ it should work without completeness.

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With metric balls on non-complete manifolds you have an obvious obstruction: if $\overline{B_R}$ is not compact, $B_R$ cannot be contained in a compact domain. The second version of your question is more robust: if $K$ is compact, $\Omega$ is open with compact closure, and $K\subset \Omega$, then we can find a regular compact domain pinched between $K$ and $\Omega$. Indeed, let $(\varphi_\alpha,U_\alpha)$ be a smooth partition of unity on $\Omega$, where $\overline{U_\alpha}\subset \Omega$ for all $\alpha$. Let $\psi=\sum_{U_\alpha\cap K\ne\varnothing }\varphi_\alpha$. Then $\psi$ is smooth, compactly supported in $\Omega$, and is identically $1$ on $K$. By Sard's theorem there exists $\lambda\in (0,1)$ such that the set $D=\{x:\psi(x)\ge \lambda\}$ is a regular compact domain.