Let $E$ be a measurable, bounded subset of $\mathbb R^n$ with nonzero Lebesgue measure.

Denote by $\partial E$ the measure theoretic boundary of $E$, defined as the set of points in $\mathbb R^n$ where the measure theoretic density of $E$ is not $0$ or $1$.

For $\varepsilon > 0$, write $\partial E_\varepsilon$ for the set of points within distance at most $\varepsilon$ of $\partial E$.

Write $E^+_\varepsilon$ for the set $E \cup \partial E_\varepsilon$, and $E^-_\varepsilon$ for the set $E \setminus \partial E_\varepsilon$.


Is it true that for all bounded measurable sets $E$, we have

$\limsup_{\varepsilon \to 0} \frac{\mu(E^+_\varepsilon)\mu(E^-_\varepsilon)}{\mu(E)^2} \leq 1$?

  • $\begingroup$ What is $M$ - maybe $E$? $\endgroup$
    – Leo Moos
    May 21, 2021 at 13:00
  • $\begingroup$ Yeah, sorry my bad. $\endgroup$
    – Nate River
    May 21, 2021 at 13:01

1 Answer 1


I think one of the classic counterexamples works here, to show that this is false: Let $\{q_i\}_{i\in\mathbb{N}}$ dense in $[0,1]^n$, $\delta >0$ and construct $$E = \bigcup_{i\in\mathbb{N}} B_{\delta 2^{-i}}(q_i).$$

Then $\mu(E) \leq c\delta^n$, but $E$ is dense in $[0,1]^n$. If I am not completely mistaken (You might need to choose the $q_i$ so that the balls don't intersect), then for the measure theoretic boundary it is still true that $\overline{\partial E} \cap [0,1]^n = [0,1]^n \setminus \operatorname{int}(E)$. So in particular $[0,1]^n \subset E_\epsilon^+$ for any $\epsilon > 0$ and thus $\mu(E_\epsilon^+) \geq 1$. Finally, $E_\epsilon^-$ includes most volume of all balls such that $\delta 2^{-i} \gg \epsilon$, so you can show that $\mu(E_\epsilon^-) \to \mu(E)$. But then $$\limsup_{\epsilon \to 0} \frac{\mu(E_\epsilon^+) \mu(E_\epsilon^-)}{\mu(E)^2} \geq \frac{1 \cdot \mu(E)}{\mu(E)^2} > \frac{1}{c\delta^n}$$ which is unbounded.

  • $\begingroup$ This looks good. One small comment: I think the sentence starting 'In particular [...]' might be inaccurate - the interior of 'large' balls wouldn't be covered. However, the rest of the argument works when replacing it with the observation that $[0,1]^n \setminus \mathrm{int} E \subset E_{\epsilon}^+$ and $\mu(E_\epsilon^+) \geq 1 - c \delta^n$. $\endgroup$
    – Leo Moos
    May 21, 2021 at 13:50
  • $\begingroup$ Hmm, how can the $q_i$ be chosen such that the balls don’t intersect though? $\endgroup$
    – Nate River
    May 21, 2021 at 13:51
  • $\begingroup$ Although what Leo mentioned should still be true I suppose - the set of boundary points will be dense in the complement of $\text{int} E$ and so this will go through nicely. $\endgroup$
    – Nate River
    May 21, 2021 at 13:59
  • $\begingroup$ @LeoMoos I had it that way first as well, but then I noticed that per definition $E \subset E^+_\epsilon$. $\endgroup$
    – mlk
    May 21, 2021 at 14:07
  • $\begingroup$ @mlk Oops, little brain fart - you're completely right. $\endgroup$
    – Leo Moos
    May 21, 2021 at 14:10

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