Regular compact domains of a Riemannian manifolds

Question

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.

Answer 1

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.