**Definitions:**


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 $E \setminus \partial E\_\varepsilon$.


**Question:**


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


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



Growth and shrinking rate of measurable sets along the boundary
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.


