Is the intersection of the following closed and open set closed? Generally?

Question

Ok, I have been informed that the below lemma is incorrect. I needed it to prove the following statement. Could someone else provide a proof?

Statement: If m(E) is finite, there exists a compact set K with $K\subset E$ and $m(E-K)\le \epsilon$.

m(E) is the Lebesgue measure of set E.

Thanks!

Lemma: Let F be a closed subset of $\mathbb{R^d}$. Let $B_n(0)$ be a ball of radius n centered at the origin. Let $K_n=F\cap B_n$. Show $K_n$ is compact.

Answer 1

This is definitely false, for $F = \mathbb{R}^n$ you get $F \cap B_n = B_n$ which is not compact.

However, the closed ball $\overline{B}_n(0)$ is compact and the intersection between a compact and a closed set is always compact. Therefore $\overline{B}_n(0) \cap F$ is compact.

Another option would be to consider $\overline{F \cap B_n(0)}$. This set is closed and bounded since the closure of a bounded set is bounded aswell.

Answer 2

Take $K_n:=\overline{F\cap B_n}$, then $$\lim_{n\to\infty}\lambda(K_n)=\lambda(F),$$ where $\lambda$ denotes the Lebesgue measure. It should be easy to conclude from here.

Answer 3

(The OP drastically changed the question after this answer was posted.)

I'm not sure if you intended the $n$ in the superscript of $\mathbb{R}^n$ to be the same as the $n$ in $B_{n}(0)$.

Nevertheless: Your question asks about the intersection of a closed and open set, so I assume that by "ball" you mean an open ball.

The statement is false.

Let $n = 1$ so that we are dealing with $\mathbb{R}$.

Then $B_{1}(0)$ is the open interval $(-1, 1) \subset \mathbb{R}$.

Let $F$ be the closed subset $[-1, 1] \subset \mathbb{R}$.

Then $K_{1} = [-1, 1] \cap (-1,1) = (-1, 1)$ is the same open set (i.e., open interval) as $B_{1}(0)$.

In particular: This provides a counterexample to the statement in question.