Given a right continuous and increasing function $F: \textbf{R} \to \textbf{R}$, there is an associated complete measure $\mu_{F}$ that is defined over a set $\mathcal{M}_{\mu}$ that is a strict superset of $\mathcal{B}_{\textbf{R}}$, the Borel $\sigma-$algebra on $\textbf{R}$. With reference to this scenario, Folland writes on page 37:
The following are equivalent:
1 $E \in \mathcal{M}_{\mu}$
2 $E = V \smallsetminus N_1$ where $V$ is a $G_{\delta}$ set and $\mu(N_1) = 0$
3 $E = H \cup N_2$ where $H$ is a $G_{\sigma}$ set and $\mu(N_2) = 0$
The following explanation is given for one direction of the proof.
(b) and (c) each imply (a) since $\mu$ is complete on $\mathcal{M}_\mu$.
This explanation seems predicated on what I label as a theorem:
For a complete measure $\mu$ defined on $\mathcal{M}_{\mu} \supset \mathcal{B}_{\textbf{R}}$ where $\mathcal{B}_{\textbf{R}}$ is the borel measure on $\textbf{R}$, if $$\mu(N) \underbrace{: =}_{\text{This is how $\mu$ is defined}} \inf \{ \sum_{j=1}^{\infty}F(b_j) - F(a_j) | E \subseteq \bigcup_{j=1}^{\infty}(a_j, b_j)\} = 0$$ then $N \in \mathcal{M}_{\mu}$.
Is this true and am I correct that it is necessary for Folland's assertion? Is my proof below adequate to prove it?
Let $\{\epsilon\}$ be a sequence such that $\epsilon \to 0$ but $\epsilon > 0$ for all $j$. For each $j$ there is a covering of $E$ by sets $\{ (a_j^{k}, b_{j}^{k})\}$ where $k \in \mathbf{N}$ such that
$$\mu(\bigcup_{k=1}^{\infty}(a_j^{k},b_j^{k})) \leq \sum_{k=1}^{\infty}\mu\left( (a_j^{k},b_j^{k})\right) = \sum_{k=1}^{\infty}F(b_j^{k}) - F(a_j^{k}) < \epsilon$$
Define
$B = \bigcap_{j=1}^{\infty}\bigcup_{k=1}^{\infty}(a_j^{k}, b_j^{k})$ and then we have that
for all $j$
$$\mu(B) \leq \mu(\bigcup_{k=1}^{\infty}(a_j^{k},b_j^{k})) \leq \sum_{k=1}^{\infty}\mu\left( (a_j^{k},b_j^{k})\right) = \sum_{k=1}^{\infty}F(b_j^{k}) - F(a_j^{k}) < \epsilon$$
As $j \to \infty$ then we have $\mu(B) \leq 0 \implies \mu(B) = 0$. Since $N_1 \subseteq B$, and $\mu$ is complete, it follows that $N_1 \in \mathcal{M}_{\mu}$.
