Suppose $E$ is a given set, and $O_n = \{x : d(x,E)&lt;\frac1n\}$.  Show that if $E$ is compact, then $$m(E) = \lim_{n \rightarrow \infty} m(O_n)$$  Also, show that this conclusion may be false for $E$ closed an unbounded, or $E$ open and bounded.  I'm gonna need some help on this one... this measure theory stuff is so tricky! Thanks everyone.
Below is an attempt at a proof of the following problem. Any feedback would be greatly appreciated. Thx!  Let $E$ be a set and $O_n = \{x: d(x, E) &lt; \frac{1}{n}\}$. Show  If $E$ is compact then $m(E) = lim_{n \rightarrow \infty} m(O_n)$.  This is false for $E$ closed and unbounded or $E$ open and bounded.   Note that all sets are real and measurable refers to Lebesgue measurable.  Suppose $E$ is compact. Then $m(E) &lt; \infty$. Also, since each $O_n$ is an open set of $\mathbb{R}^d$, $m(O_n) &lt; \infty$. If either of the following is true, we have that $lim_{n \rightarrow \infty}m(O_n)=m(E)$: $O_n \nearrow E$ $O_n \searrow E$ Now  $\bigcap_{n=1}^{\infty} O_n = \{ x:d(x,E) = 0\}$. That is, $x \in \bigcap_{n=1}^{\infty}O_n \iff \exists n&gt; \frac{1}{\epsilon},\ n \in \mathbb{N} \iff x\in \{x \mid d(x,E) =0\}$. Then we have that $\bigcap_{n=1}^{\infty}O_n = E$ and $O_n \searrow E$. Thus $ \ m(E) = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is closed and unbounded. Let $E = \{(x,y): y = 2 \}$ and $O_n = \{(x,y): d(x,E) &lt; \frac{1}{n}\}$. Since $E$ is a line in $\mathbb{R}^2$, $m(E) = 0$. Also, since the measure of a rectangle in $\mathbb{R}^d$ is its volume, $m(O_n) = \left| O_n \right| = \infty$, since the rectangle $O_n$ is unbounded. It is therefore apparent that $m(E) \not = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is open and bounded. Let $E=(0,1)$ and $O_n = \{x \mid d(x,E) &lt; \frac{1}{n}\}$. $\mathbb{R}-E$ is closed and $O_n \in \mathbb{R}-E$. Since $\mathbb{R}-E$ contains all its limit points $lim_{n \rightarrow \infty}O_n = O \in \mathbb{R}-E$. Thus $lim_{n \rightarrow \infty} m(O_n) \neq m(E)$.
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