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

nonontrivial 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