real analysis - Show that $lim_{n rightarrow infty}m(O_n)=m(E)$ when $E$ is compact.

Below is an attempt at a proof of the following problem. Any feedback would be greatly appreciated. Thx!

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Let $E$ be a set and $O_n = \{x: d(x, E) < \frac{1}{n}\}$.

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.

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Suppose $E$ is compact. Then $m(E) < \infty$. Also, since each $O_n$ is an open set of $\mathbb{R}^d$, $m(O_n) < \infty$. 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, $x \in \bigcap_{n=1}^{\infty}O_n \iff \exists n> \frac{1}{\epsilon},\ n \in \mathbb{N} \iff x\in \{x \mid d(x,E) =0\}$.

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 $O_n = \{(x,y): d(x,E) < \frac{1}{n}\}$.

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 $O_n = \{x \mid d(x,E) < \frac{1}{n}\}$.

$\mathbb{R}-E$ 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 $lim_{n \rightarrow \infty} m(O_n) \neq m(E)$.

Answer

Your example for $E$ open and bounded is not correct. I really can't tell what you are doing, but it should be clear that $m((0,1)) = \lim_{n\to \infty}m(O_n)$ in this case.

Hint for an example: Let $E$ be an open dense subset of $(0,1)$ such that $m(E)<1/2.$,