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 $\partial \{ x \in \Omega : \text{dist}(x , \partial \Omega) > \alpha\} = \{ x \in \Omega : \text{dist}(x , \partial \Omega) = \alpha\}$?

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    $\begingroup$ 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 $\mathrm{dist}(0, \partial \Omega) = \alpha$. $\endgroup$ – Nate Eldredge Jan 20 at 17:25
  • $\begingroup$ @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 $\alpha = 1$. $\endgroup$ – Nate Eldredge Jan 20 at 19:31
  • $\begingroup$ @NateEldredge Yes, I misread the question. $\endgroup$ – user126920 Jan 20 at 19:56
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    $\begingroup$ 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 $\Omega$. $\endgroup$ – Nate Eldredge Jan 20 at 20:16
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    $\begingroup$ So is that really what you meant to ask? $\endgroup$ – Nate Eldredge Jan 20 at 20:16

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