On theorem 1.19 in Folland's and completion of measure

[psie]

TL;DR

I'm struggling with a remark on page 35 in Folland's text. He makes a connection to a theorem that comes later, theorem 1.19, and I don't see connection.

Folland remarks on page 35 that each increasing and right-continuous function gives rise to not only a Borel measure $\mu_F$, but also a complete measure $\bar\mu_F$ which includes the Borel $\sigma$-algebra. He then says that the complete measure is the extension of the measure and that this should follow from the below theorem, theorem 1.19, where $\mathcal M_\mu$ is the domain of the complete (Lebesgue-Stieltjes) measure $\mu$. How?

(A $G_\delta$ set is a countable intersection of open sets, and a $F_\sigma$ set is a countable union of closed sets.)

From what I recall, I need to show that if $$N_1\subset E,E\in\mathcal M_\mu\text{ and }\mu(E)=0 \implies N_1\in\mathcal M_\mu.$$ The theorem, on the other hand, says that if $E\in\mathcal M_\mu$, then there exists a ($G_\delta$ set) $V\in\mathcal B_\mathbb R\subset \mathcal M_\mu$ such that $E=V\setminus N_1$ and $\mu(N_1)=0$.

I struggle with putting the pieces together.

EDIT: Upon closer thought, maybe I need to use (c) and the fact that the completion $\overline{\mathcal M}$ of a $\sigma$-algebra $\mathcal M$, where $\mathcal N=\{N\in\mathcal M:\mu(N)=0\}$, is $$\overline{\mathcal M}=\{E\cup F: E\in\mathcal M\text{ and } F\subset N\text{ for some } N\in\mathcal N\}.\tag1$$ Yet, I still don't see how theorem 1.19c and $(1)$ say the same thing.

 

[psie]

After some more thinking, I think I now understand how to show this.

$\overline{\mu}_F$ is the completion of $\mu_F$.

Recall that if $(X,\mathcal M,\mu)$ is a measure space and $\mathcal N=\{N\in\mathcal M:\mu(N)=0\}$, then the complete $\sigma$-algebra is $$\overline{\mathcal M}=\{E\cup F: E\in\mathcal M\text{ and } F\subset N\text{ for some } N\in\mathcal N\}.$$ We prove $\mathcal{M}_\mu=\overline{\mathcal{B}}_\mathbb{R}$. If $E\in\mathcal{M}_\mu$, then $E=H\cup N$, where $H$ is an $F_\sigma\in\mathcal{B}_\mathbb{R}$ and $\mu(N)=0$, so $E\in\overline{\mathcal{B}}_\mathbb{R}$. For the other inclusion, suppose $E\in\overline{\mathcal B}_\mathbb{R}$, then $E=F\cup N_1$ where $F$ is a Borel set and $N_1$ is a null set. But note, $F=H\cup N_2$, where $H$ is an $F_\sigma$ set (since $\mathcal B_\mathbb{R}\subset\mathcal M_\mu$). And since $N_1\cup N_2$ is a null set, $H\cup (N_1\cup N_2)=E\in\mathcal M_\mu$.