Given a compact set $E$ with non-empty interior in $R^d$ and some small positive number $r$, what kind of conditions should we put on the set $E$ so that for all $x\in E$, the volume of the intersection of $B(x,r)$ with $E$ is uniformly bounded away from $0$? Here $B(x,r)$ is the ball of radius $r$ centered at $x$.
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$\begingroup$ Some natural conditions are used to guarantee the existence of the Sobolev injections. See e.g. the book by Adams, Sobolev spaces. $\endgroup$– Romain GicquaudCommented Dec 11, 2020 at 7:01
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$\begingroup$ There is the "measure density" condition which requires that there is $\gamma > 0$ such that $|E \cap B(x,r)| \geq \gamma |B(x,r)|$ uniformly for $r$ sufficiently small, say, $r \leq 1$. It is sometimes also called "interior thickness condition" or just "thickness condition", or the set $E$ is called "$d$-set". As mentioned before, it is a quite important property in the theory of Sobolev spaces, see e.g. Sobolev embeddings, extensions and measure density condition. $\endgroup$– HannesCommented Dec 11, 2020 at 11:10
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$\begingroup$ Thanks Romain and Hannes for your very useful comment. $\endgroup$– Yonglong LiCommented Dec 13, 2020 at 6:14
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For any fixed $r$, $f(x):=vol(B(x,r)\cap E)$ is Lipschitz, hence continuous, with respect to $x$. Indeed, $\vert f(x)-f(y)\vert$ is at most the volume of the symmetrical difference of the balls. Hence, the minimum of $f$ is reached at some $x\in E$. So, it is enough that $f>0$. This will be the case if you suppose e.g. that $E$ is the closure of its interior, a hypothesis which looks natural in your case. For example, $\partial E$ being a $C^0$ ($d-1$)-dimensional submanifold of $R^d$ is enough.
