# Growth and shrinking rate of measurable sets along the boundary

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

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

• What is $M$ - maybe $E$? May 21, 2021 at 13:00
• Yeah, sorry my bad. May 21, 2021 at 13:01

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.
• 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 $\mu(E_\epsilon^+) \geq 1 - c \delta^n$. May 21, 2021 at 13:50
• Hmm, how can the $q_i$ be chosen such that the balls don’t intersect though? May 21, 2021 at 13:51
• 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. May 21, 2021 at 13:59
• @LeoMoos I had it that way first as well, but then I noticed that per definition $E \subset E^+_\epsilon$.