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Given $F \subset \mathbb R^n$ non empty and $\epsilon > 0$. Let $F_\epsilon$ be $\epsilon$-parallel set of $F$, $$F_\epsilon := \{x \in \mathbb R^n:d(x,F)\le\epsilon\},$$ with $d(x,F):= \inf_{y\in F}d(x,y)$. Using monotone property we have $$\mathcal L^n(F_{\epsilon_1})\le \mathcal L^n(F_{\epsilon_2}),$$ for $\epsilon_1 < \epsilon_2$. Since $\mathcal L^n(F)\le\mathcal L^n(F_\epsilon)$ for all $\epsilon > 0$, the limit $\lim_{\epsilon\to0}\mathcal L^n(F_\epsilon)$ always exists. In general the limit is not $\mathcal L^n(F)$. For example, in $\mathbb R$ if we take $F=\mathbb Q\cap[0,1]$ we have $\lim_{\epsilon\to0}\mathcal L(F_\epsilon)=1$. The question is if we take $F$ as a compact set, can we assert $\lim_{\epsilon\to0}\mathcal L^n(F_\epsilon)=\mathcal L^n(F)?$ Thank you.

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  • $\begingroup$ Is $\mathcal L^n$ the Lebesgue measure on $\Bbb R^n$? $\endgroup$ – Davide Giraudo Sep 23 '12 at 8:59
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It's true when $F$ is compact. To see that, we just have to take the limit for the sequence $\{k^{—1}\}$ by monotonicity.

  • As $F$ is bounded, $F_1$ is of finite measure.
  • As $F$ is closed, $\bigcap_{k\geq 1} F_{k^{-1}}=F$.
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