# If $E$ is compact, then $m(E) = \lim\limits_{n \to \infty} m(\mathcal{O}_n)$

Suppose $E\subset \mathbb{R}^d$ is a given set, $m$ is the Lebesgue measure, and $\mathcal{O}_n$ is the open set:

$$\mathcal{O}_n = \{x : d(x, E) < 1/n\}.$$

The goal is for me to show that if $E$ is compact, then $m(E) = \lim\limits_{n \to \infty} m(\mathcal{O}_n)$.

I am having trouble not only visualizing these sets, but also intuitively realizing what this means. In other words, I have no idea how to begin this proof.

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Is this about sets in the real line? Is $m$ the Lebesgue measure? Also, I assume that you want $m(\mathcal O_n)$ in the limit. –  Martin Argerami Oct 1 '12 at 16:15
Sorry for the lack of specification. Here E is a subset of $\mathbb{R}^d$, and m is the Lebesgue measure. Your assumption of m $(\mathcal{O}_n)$ is correct as well. –  Domonic Mei Oct 2 '12 at 8:23

The role of compactness is to guarantee that $m(\mathcal O_n)<\infty$. Then, as $\mathcal O_1\supset\mathcal O_2\supset\cdots$ and $E$ is closed, we get that $E=\bigcap_n\mathcal O_n$, and the result follows by continuity of the measure.
The fact that $m(O_n)$ is finite for some $n$ is not true in full generality. The OP should give more precisions about what properties are satisfied by the measure $m$, and what is the underlying space. –  Ahriman Oct 1 '12 at 16:31
Sorry about the lack of clarity. I made the edit above: $E\subset\mathbb{R}^d$ and $m$ is the Lebesgue measure. –  Domonic Mei Oct 2 '12 at 8:31