# Is a boundary which consists of boundaries of balls Lipschitz?

I am working through the proof of the following theorem in Sohr - The Navier Stokes Equations:

Let $\Omega \subseteq \mathbb{R}^n$ be an arbitrary domain with $n \geq 2$. Then there exists a sequence $(\Omega_j)_{j = 1}^\infty$ of bounded Lipschitz subdomains of $\Omega$ and a sequence $(\varepsilon_j)_{j = 1}^\infty$ of positive numbers with the following properties:

1. $\overline \Omega_j \subseteq \Omega_{j + 1}, j \in \mathbb{N}$,
2. $\operatorname{dist} (\partial \Omega_{j + 1}, \Omega_j) \geq \varepsilon_{j +1}, j \in \mathbb{N}$,
3. $\lim_{j \to \infty} \varepsilon_j = 0$,
4. $\Omega = \bigcup_{j = 1}^\infty \Omega_j$.

Now let $x_0 \in \Omega$ and $\tilde\Omega \subseteq \Omega \cap B_1(x_0)$ be the connected component of the intersection of $\Omega$ with the unit ball, which contains $x_0$. For given $\varepsilon > 0$ there exist balls $B_\varepsilon(x_1), \dots, B_\varepsilon(x_m)$ covering the compact boundary $\partial \tilde \Omega$. Lets assume, that $\varepsilon < 0$ has been chosen such that $$x_0 \in \hat \Omega := \tilde \Omega \setminus \bigcup_{j = 1}^m \overline B_\varepsilon(x_j).$$

The author states, that $\hat\Omega$ is a bounded Lipschitz domain, since its boundary consists of parts of the boundaries of balls. But is this really correct? I can imagine bounded domains consisting of parts of the boundaries, that aren't Lipschitz. Just consider a boundary with a cusp like the graph of $$\sqrt{(|x|)}, x \in[-1,1]$$

How can this argument be made rigorous?

If this is the only problem I am facing, I think one could eliminate the risk of having a boundary with cusps by inflating all balls $B_\varepsilon(x_1),\dots,B_\varepsilon(x_m)$ a little, since there are only finitely many. Are there other problems that can happen at boundaries constructed as in the presented way?

• If the boundary has to be covered by the open balls, then I don't think the situation you posit can occur - you have a gap at the origin. Still, I don't think this step of the argument is obvious. Oct 10, 2016 at 4:29