Equivalence of measurable sets I am reading L.F. Richardson's Measure and Integral. I am unable to see how a corollary of the following theorem follows:

Let $\mu$ be a finite, countable additive measure on a field
$\mathfrak A\subseteq \mathscr P (X)$. Then by Caratheodory Extension
Theorem, there is a countable additive extension $\mu^*$ of $\mu$
defined on $\sigma$-field $\mathfrak {A}^*$ containing $\mathfrak A$.
Then $S\in\mathscr P (X)$ is $\mu^*$-measurable iff there exists sets
$B_*$ and $B^*$ in the sigma field generated by $\mathfrak A$ such
that $B_*\subseteq S\subseteq B^*$ and $\mu^* (B_*)=\mu^*
 (S)=\mu^*(B^*)$.

In the proof of Caratheodory Extension Theorem, $\mu^*:\mathscr P(X)\to[0, \infty]$ is the function given by
$$\mu^*(B)=\inf \left\{\sum_{i=1}^{\infty} \mu(A_i) : B\subseteq \bigcup_{i\in \mathbb N}A_i, \, A_i\in \mathfrak A\right\}$$
and $\mathfrak A^*$ is the sigma algebra of $\mu^*$-measurable function.
This is the corollary:

Let $\mu$ be a $\sigma$-finite, countable additive measure on a field
$\mathfrak A\subseteq \mathscr P (X)$. Then by Caratheodory Extension
Theorem, there is a countable additive extension $\mu^*$ of $\mu$
defined on $\sigma$-field $\mathfrak {A}^*$ containing $\mathfrak A$.
Then $S\in\mathscr P (X)$ is $\mu^*$-measurable iff there exists sets
$B_*$ and $B^*$ in the sigma field generated by $\mathfrak A$ such
that $B_*\subseteq S\subseteq B^*$ and $\mu^* (B^*\setminus B_*)=0$.

To prove this, the author writes: "Apply $\sigma$-finiteness and take note that countable union of sets of measure zero is of measure zero". This does not help me however.
All I know is that since $\mu$ is $\sigma$-finite, there exist $\{X_i\}_{i\in \mathbb N}\subseteq \mathfrak A$ such that $X=\bigcup_{i\in \mathbb N} X_i$ and $\mu (X_i)<\infty$ for each $i\in \mathbb N$. Now I can consider $\mu$ on the algebra $\mathfrak A_i =\{ X_i \cap A: A\in \mathfrak A \}$. Since $\mathfrak A _i$ is a subset of $\mathfrak A$, $\mu$ is countably additive on $\mathfrak A_i$ and $\mu(X_i)<\infty$. I can use Theorem now but it does not help.
Hints on proving this will be appreciated!
 A: There are many questions at MSE related to Caratheodory's extension theorem with very good answers. I was not able to find postings that would for the specificity of your question (One may combines several postings to address the issues that you brought forth, but that may not be so useful for you). So here I post an answer that I hope addresses the particulars of your questions. Some details are left for you.

Starting with the outer measure $\mu^*$ you defined in your posting, you can show that $\mu^*$ extends $\mu$ as a measure to a a $\sigma$-algebra $\mathfrak{A}^*$ that contains $\mathfrak{A}$ and all $\mu$-null sets.
(a set $N\subset X$ is $\mu$-null if $\mu^*(N)=0$, which is equivalent to say that there is $B\in\mathfrak{A}$ such that $N\subset B$ and $\mu(B)=0$). Thus $\mathfrak{A}$ contains the $\sigma$-algebra generated by $\mathfrak{A}$ and the collection $\mathcal{N}_\mu$ of $\mu$-null sets, that is $$\sigma(\mathfrak{A}\cup\mathcal{N}_\mu)\subset\mathfrak{A}^*$$
This is regardless of whether $\mu$ us $\sigma$-finite or not.
It is not difficult to show that
$$\sigma(\mathfrak{A}\cup\mathcal{N}_\mu)=\big\{E\subset X: \exists E_1,\, E_2\in\mathfrak{A}\,\text{such that}\,E_1\subset E\subset E_2\,\text{and}\,\mu(E_2\setminus E_1)=0\big\} =\overline{\mathfrak{A}}$$
The details can be found here for example.

*

*First notice that $\mathfrak{A}\cup\mathcal{N}_\mu\subset\overline{\mathfrak{A}}$.

*If $E\in\overline{\mathfrak{A}}$, then there are $E_1,E_2\in \mathfrak{A}$ with $E_1\subset E\subset E_2$ such that $\mu(E_2\setminus E_1)=0$. Then, as $E\setminus E_1\subset E_2\setminus E_1$, we have that $\mu^*(E\setminus E_1)=0$. Hence $E=E_1\cup (E\setminus E_1)\in\sigma(\mathfrak{A}\cup\mathcal{N}_\mu)$.

In the case when $\mu$ is $\sigma$-finte, more can be said: $\mathfrak{A}^*=\overline{\mathfrak{A}}$. Suppose $X=\bigcup_nX_n$ where $X_n\in\mathfrak{A}$ and $0<\mu(X_n)<\infty$. It is easy to check that the extension of the outer measure $\mu^*$ restricted to subsets of $X_n$ coincides with $\mu^*_n(\cdot\cap X_n)$ on $\mathfrak{A}$ for each $n$.
Now, suppose $E\in\mathfrak{A}^*$ and define $E_n=E\cap X_n$. Then $E_n\in\mathfrak{A}^*$ and $\mu^*(E_n)<\infty$. From Caratheodory's construction, you can find $F_{n,m}\in\mathfrak{A}$ such that $E_n\subset F^n_{m+1}\subset F_{n,m}\subset X_n$ and
$$\mu(F_{n,m})<\mu^*(E_n)+2^{-m}$$
Then $F_n:=\bigcap_m F_{n,m}\in\mathfrak{A}$ and $\mu^*(E_n)=\mu(F_n)$. Similarly, for $X_n\setminus E_n$, you can find $G_n\in\mathfrak{A}$ such that $X_n\cap E^c_n=X_n\setminus E_n\subset G_n$ and $0=\mu^*(G_n\cap(X_n\cap E^c_n)^c)=\mu^*(G_n\cap E_n)$.
Then, $F'_n=X_n\setminus G_n\subset E_n\subset F_n$ and
$$\mu(F_n-F'_n)\leq \mu^*(F_n\setminus E_n)+\mu^*(E_n\setminus F'_n)=0+\mu(G_n\cap E_n)=0$$
To conclude, notice that

*

*$F'= \bigcup_nF'_n\subset E=\bigcup_nE_n\subset \bigcup_nF_n=F$.

*$F',F\in\mathfrak{A}$

*$F\setminus F'\subset \bigcup_n(F_n\setminus F'_n)$ and so, $\mu(F\setminus F')=0$.

Therefore $E\in\overline{\mathfrak{A}}$.
