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Notation: $|A|$ is the Lebesgue measure of $A \subset \mathbb{R}^d$, and $A_\delta = \{ x : \text{dist}(x,A) \leq \delta \} $ is the $\delta$-neighborhood of $A$.

I want to show that there is a constant $C$ such that $|E_{2\delta}| \leq C |E_{\delta}|$ for all compact sets $E \subseteq \mathbb{R}^d$ and all $\delta > 0$.

I have been told I need to apply the Hardy-Littlewood maximal theorem to $f = \mathbb{1}_{E_{\delta}}$. For $p > 1$, this gives the estimate $||Mf||_p^p \leq C ||f||_p^p = C|E_{\delta}|$. For $p = 1$, it is the estimate $\lambda \cdot |\{x : Mf(x) > \lambda \}| \leq C ||f||_1 = C |E_{\delta}|$. Here $$ Mf(x) = \sup_{r > 0} \frac{1}{B(x,r)} \int_{B(x,r)} |f(y)| dy. $$

I think I should try to show that $|E_{2 \delta}| \leq C ||Mf||_p^p$. I see that $$Mf(x) = \sup_{r > 0} \frac{|E_{\delta} \cap B(x,r)| }{|B(x,r)|}.$$ But I am stuck after that.

I know I need to use the compactness of $E$ somehow. My thought is that I should look at a finite covering by balls $B(x,r)$ or balls $B(x,\delta)$ or something like that.

Can someone please help?

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up vote 2 down vote accepted

You can use the Vitali covering lemma to show this. Let $E_\delta = \{x \in R^d: dist(x,A) < \delta\}$; it's a little easier to work with the $E_\delta$ and the result is entirely equivalent since $E_{\delta} \subset A_{\delta} \subset A_{2\delta} \subset E_{4\delta}$ and you can just apply the $E_{\delta}$ result twice.

By Vitali, there is a countable collection of disjoint balls $B(x_n,2\delta)$ with $x_n \in A$ for all $n$ such that $$E_{2\delta} = \cup_{x \in A}B(x,2\delta) \subset \cup_n B(x_n,10\delta)$$ So we have $$|E_{2\delta}| \leq \sum_n |B(x_n,10\delta)| $$ $$= 10^d \sum_n |B(x_n,\delta)|$$ Since these balls are disjoint subsets of $E_{\delta}$ this is at most $$ \leq 10^d |E_{\delta}|$$ This gives the desired result.

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Can Vitali's covering Lemma be deduced – admchrch Feb 11 '11 at 15:41
It can indeed, see – Glen Wheeler Feb 11 '11 at 17:57
Sorry, that comment was half-finished. I meant "from the maximal theorem." But, upon thinking about it more, I see that Zarrax's argument is likely best. Indeed, the maximal theorem is usually deduced using Vitali's lemma. – admchrch Feb 11 '11 at 21:13

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