# measure of a closed and bounded subset of $\mathbb{R}^d$

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 $$K\supset \bigcap E_n$$.