Let $\Omega$ be a domain in $\mathbb{R}^n$. Then I can find a bounded subdomain $\Omega_0$ in $\Omega$ with $r=\mathrm{dist}(\Omega_0,\partial\Omega)>0$. Here domain means an open and connected subset of $\mathbb{R}^n$.

I want to find a bounded Lipschitz domain $\Omega_1$ in $\mathbb{R}^n$ which contains $\Omega_0$ and contained in $\Omega$.

I have a rough idea about the proof, but I'm not satisfying this strategy because I cannot make a rigorous proof based on this idea.

Cover $\Omega_0 \cup \partial \Omega_0$ by a family of open cubes whose diameter is less than $r$. Then by compactness, there exists a finite family $\mathcal{F}$ of cubes which covers $\Omega_0 \cup \partial \Omega_0$. Enlarge each $Q$ in $\mathcal{F}$ so that for any $Q\in \mathcal{F}$, there exists $Q'\in\mathcal{F}$ with $Q\neq Q^\prime$ such that $Q\cap Q' \neq \varnothing$.

Define $\Omega_1$ be the union of the above cubes. Then $\Omega_1$ is clearly open and connected. So it is a domain in $\mathbb{R}^n$.

Is union of two Lipschitz domains (with nonemtpy intersection) is Lipschitz? If so, then since cubes are Lipschitz domain and $\Omega_1$ is a finite union of Lipschitz domains, so I can prove the claim.

It seems obvious in 2D by picture but I cannot make any progress. One of my trial was changing parameter when two domain meets.

I can find such a domain with $C^1$-boundary using Sard's lemma. So the existence is clear. But I want to find different method based on elementary arguments.

Thanks for help.


It is not true. Take two squares in $\mathbb{R^2}$ that touch only at one corner, say $Q_1=(0,1)\times (0,1)$ and $Q_2=(1,2)\times (1,2)$. Then each cube is Lipschitz but the union is not, since at the point $(1,1)$ you cannot write the boundary as the graph of a function.

  • $\begingroup$ Thanks you for your answer. But that case is not counterexample of my claim since $Q_1 \cap Q_2 =\varnothing$. I want to make two domain into one domain by overlapping. $\endgroup$ – Will Kwon Jul 4 '17 at 2:40
  • $\begingroup$ but it is easy to modify this example to make the two domains overlap. You just take an handle that connects $Q_1$ with $Q_2$ away from $(1,1)$. $\endgroup$ – Gio67 Jul 4 '17 at 2:46
  • $\begingroup$ Thanks. I didn't think that. $\endgroup$ – Will Kwon Jul 4 '17 at 2:49
  • 1
    $\begingroup$ but if the cubes meet along faces (instead of corners), then I think that their union would be Lipschitz, so your construction could work. It's tricky though. Maybe working with dyadic cubes might help. en.wikipedia.org/wiki/Dyadic_cubes $\endgroup$ – Gio67 Jul 4 '17 at 2:51

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