I'm having trouble understanding the logic of the following statement (taken from pg 53 of Lebesgue Integration on Euclidean Space by Frank Jones):
"Though unions of countably many closed sets are not necessarily closed, in this case it is easy to check that $F$ is closed. For any limit of $F$ must already be a limit point of some finite union $\bigcup_{k=1}^N K_k$, and the latter is closed."
Definitions: $A$ is a Lebesgue measurable subset of $\mathbb{R}^n$, $E_k$ is defined as $$ E_k = \{x \in \mathbb{R}^n: k-1\leq |x| \leq k\} $$ $K_k$ are compact sets and $G_k$ open sets satisfying \begin{align*} K_k \subset A \cap E_k \subset G_k \\ \lambda(G_k \setminus K_k) < \epsilon 2^{-k} \end{align*} $\lambda$ is the Lebesgue measure, and $F = \bigcup_{k=1}^\infty K_k$.