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.
Best Answer
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 $\cap_{n}E_{n}\subseteq K$.