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?