Let $E \subset \Bbb R^n$ and $O_i=\{x \in \Bbb R^n \mid d(x,E)< \frac1i\}$. Show that if $E$ is compact, then $m(E)=\lim_{i\to\infty} m(O_i)$. 
Let $E \subset \Bbb R^n$ and $O_i=\{x \in \Bbb R^n \mid d(x,E)< \frac1i\}, i \in \Bbb N$. Show that if $E$ is compact, then $m(E)=\lim_{i\to\infty} m(O_i)$. Does the statement hold if $E$ is closed, but not bounded and if $E$ is open, but bounded?

What I got was that since $\bigcap_{i} O_i = E$ and $O_1 \supset O_2 \supset \dots$, then we have that $$m(E)=m(\bigcap_{i} O_i)=\lim_{i \to \infty}m(O_i)$$ by the continuity of the Lebesgue measure. How is the fact needed that $E$ is compact here?
 A: To use the result that you mention, you must have that at least one of you $O_{i}$ has finite measure. Here you need compactness, for example take an open ball $B$ such that $E$ is compactly contained in $B$. Hence, for $i$ big enough $E \subset O_{i} \subset B$, which implies that $m(O_{i})<m(B)<\infty$. After this argument, you can invoke the result that you are using.
Edit: here is a counter-example that I always remember, consider $E=\mathbb{R} \times \{0\}$, $O_{i}=\mathbb{R} \times \{1/i\}$. Then $E= \cap O_{i}$, but (as subsets of $\mathbb{R}^{2}$) $m(E)=0$ and $m(O_{i})=\infty$ (and hence the limit is infinite!).
A: If $E$ is not compact then it is not closed or it is not bounded and for both cases there is a counterexample.

If $E$ is a hyperplane then it is closed but not bounded.
It is evident that $m(E)=0$ while $m(O_i)=\infty$ for every $i$.

If $E=\{x\in\mathbb Q^n\mid d(x,0)\leq1\}$ then it is bounded but not closed.
It is evident that $m(E)=0$ and: $$\bigcap_{i=1}^{\infty}O_i=\{x\in\mathbb R^n\mid d(x,0)\leq1\}$$and also:$$\lim_{i\to\infty}m(O_i)=m\left(\bigcap_{i=1}^{\infty}O_i\right)>0$$ (here we apply the continuity of the Lebesgue measure).
