So I have the following result to prove:

Let $K \subseteq \mathbb{R}^d$ be closed and bounded and define

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

then show $\lim_{n \rightarrow \infty} m(E_n) = m(K)$.

So I noticed that the $E_n$ are a decreasing sequence of measurable sets thus we can write

$\lim_{n \rightarrow \infty} m(E_n) = m(\bigcap_{n=1}^\infty E_n)$ so I am left to show

$ m(\bigcap_{n=1}^\infty E_n)=m(K)$ and since $K$ is closed and bounded set of reals, $K$ needs to include all of its own limit points so these $x \in E_n$ need be in $K$? its like the $E_n$ form like an open bubble around $K$ and they shrink to being just $K$ just having trouble making that part rigorous, thanks in advance!


We will prove that $\bigcap E_n = K$, and the result will follow easily.

Since $K\subset E_n$ for all $n$, we have that $K\subset \bigcap E_n$. It remains to be proved that $K\supset \bigcap E_n$.

In fact, if there was $p\in\big(\bigcap E_n\big)\setminus K$, then $p\in E_n$ for all $n$, and it would exist a sequence $x_n\in K$ s.t. $\lvert x_n - p\rvert<1/n$.

From this we conclude that $x_n\to p$, but since $K$ is a compact set, it contains all of its own limit points, which implies that $p\in K$, and this is a contradiction! Therefore $K\supset \bigcap E_n$.


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