Measure of a compact subset of $\mathbb{R}^d$

Question

Let $K \subset \mathbb{R}^d$ be closed and bounded.

Define $E_n = \{x \in \mathbb{R}^d \vert \exists y \in K$, $ \vert x - y \vert < \frac{1}{n} \}$

wish to show lim$_{n \rightarrow \infty}$ $m(E_n)$ = $m(K)$.

ATTEMPT:

So I wanted to use double sided inequality to show the equality holds.

NOTE: $E_1 \supset E_2 \supset E_2 ..$

I have a lemma stating: if additionally $m(E_1) < + \infty$ then

lim$_{n \rightarrow \infty}$ $m(E_n)$ = $m(\bigcap_{n = 1}^{\infty} E_n$) **

(is ** lemma applicable her?)

using this, If I can make a claim stating $K = \bigcap_{n=1}^{\infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety. Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!

Also, I have an observation, I would like to know is true:

Do all compact subsets of $\mathbb{R}^d$ have finite (Lebesgue) measure.

Answer 1

For each $x\in\mathbb{R}^{d}$ and $r>0$, let $B(x,r)=\{y\mid d(x,y)<r\}$ be the open ball centered at $x$ with radius $r$. Note that $$ E_{n}=\bigcup_{x\in K}B(x,\frac{1}{n}) $$ which is a bounded (because $K$ is bounded) open set. In particular $m(E_{n})<\infty$.

To show that $K=\cap_{n}E_{n}$. Clearly $E_{1}\supseteq E_{2}\supseteq\cdots\supseteq K$. We only need to show that $\cap_{n}E_{n}\subseteq K$. Let $x\in\cap_{n}E_{n}$. For each $n$, $x\in E_{n}\Rightarrow\exists y_{n}\in K$ such that $x\in B(y_{n},\frac{1}{n})$. Since $K$ is compact, for the sequence $(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and $y\in K$ such that $y_{n_{k}}\rightarrow y$ as $k\rightarrow\infty$. Now \begin{eqnarray*} d(x,y) & \leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\\ & < & \frac{1}{n_{k}}+d(y_{n_{k}},y)\\ & \rightarrow & 0, \end{eqnarray*} as $k\rightarrow\infty$. Therefore, $d(x,y)=0$ and hence $x=y\in K$. This shows that $\cap_{n}E_{n}\subseteq K$.

Answer 2

I doubt this is true

It is, and your method is fine, once you've also checked that $m(E_1) < +\infty$.

Do all compact subsets of $\mathbb{R}^d$ have finite (Lebesgue) measure.

Yes. This follows from the Heine-Borel Theorem: all compact subsets of $\mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $\mathbb{R}^d$ are measurable, but indeed they are.