# What is the boundary of the set $\{ x : dist (x ,\partial \Omega) > \alpha \}$ for a domain $\Omega$?

Let $$\Omega$$ is a bounded open domain in $$\mathbb R ^n$$, and $$\alpha \geq 0$$ a real number, and consider the set $$E_\alpha = \{ x \in \Omega : \text{dist}(x , \partial \Omega) > \alpha\}$$, which $$\text{dist}$$ is the usual distance between a point and a set. What is the boundary of $$E_\alpha$$?

In particular I want to know what conditions should be imposed on $$\Omega$$ to ensure that $$\partial E_\alpha = \{ x \in \Omega : \text{dist}(x , \partial \Omega) = \alpha\}$$?

What is situation with geometric measure theoretic boundary (the reduced boundary) $$\partial ^ \ast \{ x \in \Omega : \text{dist}(x , \partial \Omega) \}$$?

Revision from comments: If $$\Omega$$ isn't convex, there are simple counterexamples, thus let $$\Omega$$ be convex. Also, for larger values of $$\alpha$$, ($$\alpha > \text{diam} (\Omega)$$ for instance). Thus the revised version is: Let $$\Omega$$ is a bounded and convex open domain in $$\mathbb R ^n$$, and $$\alpha \geq 0$$ a real number, $$\alpha$$ is smaller than some $$\varepsilon > 0$$, what (regularity) assumptions should imposed on $$\Omega$$ and $$\partial \Omega$$ to ensure that $$\partial \{ x \in \Omega : \text{dist}(x , \partial \Omega) > \alpha\} = \{ x \in \Omega : \text{dist}(x , \partial \Omega) = \alpha\}$$?

• It can't just be regularity conditions. If $\Omega \subset \mathbb{R}^2$ is the annulus $1 < |x| < 2$ and $\alpha = 1$, then $0 \notin \partial E_\alpha$ even though $\mathrm{dist}(0, \partial \Omega) = \alpha$. – Nate Eldredge Jan 20 at 17:25
• @StanleySnelson: But it's the distance to $\partial \Omega$, not to $\bar{\Omega}$. So we would apparently need $\partial \Omega$ to be convex, which seems extremely restrictive. In fact, come to think of it, the desired statement is not satisfied by the unit ball when $\alpha = 1$. – Nate Eldredge Jan 20 at 19:31
• @NateEldredge Yes, I misread the question. – user126920 Jan 20 at 19:56
• In fact, as to the first question, there are no nontrivial sets $\Omega$ that satisfy this conclusion for all $\alpha$. Since $\bar{\Omega}$ is compact and distance to a closed set is continuous, let $\alpha = \max_{x \in \bar\Omega} \mathrm{dist}(x, \partial \Omega)$; since $\bar\Omega$ has nonempty interior, we have $\alpha > 0$. Then $E_\alpha$ is disjoint from $\Omega$ by construction of $\alpha$, and hence so is $\partial E_\alpha$ (since $\Omega$ is open). But the maximum above is attained at some point of $\Omega$. – Nate Eldredge Jan 20 at 20:16
• So is that really what you meant to ask? – Nate Eldredge Jan 20 at 20:16