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. 

With the assumptions from my comment above:
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. 
