Baby Rudin 11.11 remarks (b) I am reading Baby Rudin's Chapter 11 (The Lebesgue Theory) and find some gaps hard to fills in the 11.11 (b) remark. (pp309)

If $A \in \mathscr{M}(\mu)$ and $\varepsilon >0$, there exists sets $F$ and $G$ such that 
  $$
F \subset A \subset G,
$$
$F$ is closed, $G$ is open, and 
  $$\mu(G-A)<\varepsilon, \mu(A-F) < \varepsilon.
$$
  The first inequality holds since $\mu*$ (outer measure) was defined by means of coverings by open elementary sets.

I find it hard to relate the hint "since $\mu*$ (outer measure) was defined by means of coverings by open elementary sets." to the proof of the first inequality above.
 A: $A \in \mathscr{M}(\mu) \Rightarrow A = \bigcup\limits_{k=1}^{\infty}A_i$ for some collection of $\{A_i\} \subset \mathscr{M}_F (\mu)$.
For each $n$, since $A_n \in \mathscr{M}_F (\mu) $, we can choose a countable covering $\{G_{ni}\}$ of elementary open sets with $\bigcup\limits_{i=1}^{\infty} G_{ni} \supset A_n$, $((\bigcup\limits_i G_{ni}) \in \mathscr{M}_F (\mu)$ (by 11.11(a)),  and $\mu((\bigcup\limits_i G_{ni}) -A_n)<\epsilon/2^n$.
Let $G=\bigcup\limits_{k}\bigcup\limits_{i}G_{ki}$. Then $G$ is open (countable union of open sets), and $$\mu(A -G)\leq \sum_k\mu(A_k-(\bigcup\limits_{i}G_{ki}))\leq \sum_k\epsilon/2^k=\epsilon$$
A: The hint is that since the outer measure is defined as the infimum over countable open coverings, we can choose an open covering that is arbitrarily close to the set $A$.

*

*Suppose $\mu(A)$ is finite. Let $\{A_n\} $ be a countable collection of open elementary sets covering $A$ such that
$$ \mu(A) \leq \sum_{n=1}^{\infty}\mu(A_n) \lt \mu(A) + \epsilon$$
Let $G$ be this covering ( $ G = \bigcup_{n=1}^{\infty} A_n$) which is an open set. By subadditivity of $\mu$, and $A \subset G$
$$ \mu(A) \leq \mu(G) \leq  \sum_{n=1}^{\infty}\mu(A_n)$$
Since $\mu(A)$ is finite, we have $\mu(G-A) = \mu(G) - \mu(A) < \epsilon$.


*For $A \in \mathscr{M}(\mu)$ in general where $A = \bigcup_{n=1}^{\infty} A_n$ with $\{A_n\}$ pairwise disjoint and $\mu(A_n) \lt +\infty$.
Let $G_n$ be an open set from (1) such that
$$\mu(G_n-A_n) \lt 2^{-n}\epsilon$$
Let $G = \bigcup_{n=1}^{\infty} G_n$ which is an open set containing $A$. Then $G - A = \bigcup_{n=1}^{\infty} (G_n - A)$ so that
$$\mu(G-A) \leq  \sum_{n=1}^{\infty}\mu(G_n -A)$$
From $(G_n-A) \subset (G_n-A_n)$, we have $\mu(G_n-A) \leq \mu(G_n-A_n)$ and so
$$\mu(G-A) \leq  \sum_{n=1}^{\infty}\mu(G_n -A_n) \lt \epsilon$$
For the closed case, simply consider $A^c$, with the observation that $F^c - A^c = A - F$.
