Part of proof 11.10 in Rudin's Principles of Mathematical Analysis There is a part of proof 11.10 that I don't get in Rudin's Principles of Mathematical Analysis (3rd edition).
The whole theorem is the following two statements:


*

* $\mathcal{M}\left(\mu\right)$ is a $\sigma$-ring.

* $\mu^*$ is countably additive on $\mathcal{M}\left(\mu\right)$.


I'm having difficulties with the second statement. In particular, there is a lemma inside of the proof that claims:
$\left[\left(\mu^*\left(A\right)<\infty\right)\wedge\left(A\in\mathcal{M}\left(\mu\right)\right)\right]\Rightarrow A\in\mathcal{M}_F\left(\mu\right)$.
I don't understand the proof of that lemma.
In particular, on page 308 after equation (38), the setence that goes: "and since $B_n\in\mathcal{M}_F\left(\mu\right)$, it is easily seen that $A\in\mathcal{M}_F\left(\mu\right)$."
Well, not so easily by me. I tried proving that particular statement and couldn't. This should somehow follow from the fact that $\lim_{n\to\infty}d\left(A,B_n\right)=0$, but I don't see it. I tried using the fact that $B_n\in\mathcal{M}_F\left(\mu\right)$ to say that $\exists \left\{C_{nj}\right\}_{j\in\mathbb{N}}$ such that $C_{nj}\in\mathcal{E}\forall j\in\mathbb{N}$ and $\lim_{j\to\infty}d\left(C_{nj},B_n\right)=0$ and that we can pick $C_{nj}$ to be mutually disjoint, to define $D_n:=\bigcup_{i=1}^{n}\bigcup_{j=1}^{n}C_{ij}$. Thus $D_n\in\mathcal{E}$. Then somehow if we could show that $\lim_{n\to\infty}d(A,D_n)=0$ then $A\in\mathcal{M}_F\left(\mu\right)$. But I cannot prove that $\lim_{n\to\infty}d(A,D_n)=0$.
 A: It turns out the answer is quite simple.
The insight came from the last paragraph of page 306, where Rudin explains that $\mathcal{M}_{F} \left(\mu\right)$ is really just the closure of $\mathcal{E}$, in the metric space of $d$ and modulo sets which are $d\left(A,\,B\right)=0$ (since that would be the only way to satisfy the requirement for a metric space that iff $d\left(A,\,B\right)=0$ then $A=B$).
So, in particular, define $D_n$ as $C_{nn}$ (instead of what it is defined above) and the result follows from the fact that $D_n$ is a sequence of elementary sets and the triangle inequality for $d$: $D_n\to A$. This means that $A\in\mathcal{M}_{F} \left(\mu\right)$.
A: The key point is that For if $ A=\bigcup A^{'}_n $ with $A^{'}_n \in M_F(\mu)$, then A can be represented as the union of a countable collection of disjoint sets of $A_n \in M_F(\mu)$, that is $$ A=\bigcup ^{\infty}_{n=1} A_n$$ so $$ d(A,B_n)=\mu^{*}(\bigcup^{\infty}_{i=n+1} A_i)=\sum^{\infty}_{i=n+1}\mu^{*}(A_i)\rightarrow 0 $$ put $(B_n=A_1\bigcup...\bigcup A_n) \in M_F(\mu)$. Hence $A \in M_F$
