Let $E\subset\mathbb{R}^{n}$ be a bounded set and let $$ B(E,\varepsilon)=\{x\in\mathbb{R}^{n}:\text{dist}(x,E)\leq \varepsilon\}. $$ Is it true that $$ |B(E,\varepsilon)\setminus E)|\leq C\varepsilon $$ for some constant $C=C(E)$ and $\varepsilon\leq 1$? I know that when $E$ is a convex body, then the result follows from Minkowski - Steiner formula, but what about the general case? Here $|\cdot|$ is the Lebesgue measure.
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Just to be clear, you intend here that C depends on E I believe. – user24142 Aug 04 '20 at 07:27
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@user24142: I think that's what OP intended by "$C = C(E)$" perhaps. – Brian Tung Aug 04 '20 at 07:29
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My eyes must have passed straight over that... apologies. – user24142 Aug 04 '20 at 07:31
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In general this is not true. Consider, for example, in dimension $n=1$ the set $E = \mathbb{Q}\cap [0,1]$. If you prefer an open set, you can take the set of fat rationals in $[0,1]$.
Rigel
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Do You know what we should assume on the set E to make this statement true? – Isak Aug 04 '20 at 07:53
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I don't know what are the most general assumptions (other than smoothness or convexity). – Rigel Aug 04 '20 at 08:07
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@Isak: maybe this paper can be of your interest: http://cvgmt.sns.it/paper/1391/ – Rigel Aug 04 '20 at 09:08
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@Isak: Besides the paper Rigel suggested, see also: I. Ya. Oleksiv and N. I. Pesin, Finiteness of the Hausdorff measure of level sets of bounded subsets of a Euclidean space (Russian), Matematicheskie Zametki 37 #3 (March 1985), 422-431, 462 [English translation in Mathematical Notes 37 #3 (March 1985), 237-242]. MR 86k:28005; Zbl 573.28010 – Dave L. Renfro Aug 04 '20 at 10:40