I am doing some exercises from my real analysis class and got stuck trying to prove the following equivalences:Let $ A \subseteq \mathbb{R} $. Prove that the following statements are equivalent:
a) $ A \in \mathcal{M} $.
b) There exists a sequence $ (F_n)_{n \in \mathbb{N}} $ of closed sets contained in $ A $ and a set $ Z $ of measure zero such that$A = \bigcup_{n \in \mathbb{N}} F_n \cup Z.$
c) There exists a sequence $ (G_n)_{n \in \mathbb{N}} $ of open sets containing $A $ and a set $ H $ of measure zero such that$A = \bigcap_{n \in \mathbb{N}} G_n \setminus H.$
I have already developed a proof for (a) $\implies$ (c), but got stuck trying to do (c) $\implies $(b). Here's what i got:
(a)$\implies$ (c):
$A$ is measurable. By $G$-regularity, for every $\epsilon >0$ there exists $G \supset A $ open s.t. $\mu (G\setminus A) < \epsilon$ .
Consider $G = \bigcap_n G_n $, then $G\subset G_n \implies G \setminus A \subset G_n \setminus A$.
Define $H = G\setminus A$ .
Therefore, $A =G\setminus (G\setminus A) = (\bigcap_n G_n) \setminus H$ and $\mu (H) = \mu (G\setminus A) = 0$
Here's my attempt at (c) $\implies$ (b):
Consider $G$ from previous case.
Define $F = G^{c}$, then $ F = (\bigcap_n Gn)^{c} = \bigcup_n (G_n)^c $ from De
As $G_n$ open for every $n \in \mathbb{N}$, $(G_n)^c = F$ closed.
Here I got stuck, I can't think of a way to build $A$ that satisfies what I'm being asked. I had done a proof using $F$-regularity, but I would be doing (a) $\implies$ (b). Any help would be appreciated. Thanks in advance.
-L