The outer measure on $X$ has a collection $M$ that is a $\sigma$-algebra This is part of Caratheodory's Theorem taken by Real Analysis, Folland
If $\mu^*$  is an outer measure on $X$, the collection $M$ of $\mu^*$-measurable sets is a $\sigma$-algebra.
We first need to prove that $M$ is closed under complements and closed under finite unions to show that that $M$ is an algebra. I have no problem proving this I just want to know how to prove that $M$ is a $\sigma$-algebra. I know that it will suffice to prove that $M$ is closed under countable disjoint unions. If anyone could give me some advice on how to prove this, please let me know.
 A: I don't how it is proved in Folland's book. Normally it is proved as follow. 
We only need to prove that for pairwise disjoint $A_n\in M$
$$
\mu^*(\bigcup_{n=1}^\infty A_n) \geqslant \sum_{n=1}^\infty \mu^*(A_n)\tag{1}
$$
as other half is implied in subadditivity of outer measure. 
First note that for outer measure, there is
$$
A_1\subset A_2\implies \mu^*(A_1)\leqslant \mu^*(A_2)\tag{2}
$$
Then for any disjoint $A_n$, since 
$$
\bigcup_{n\leqslant m}A_n \subset \bigcup_{n=1}^{\infty} A_n
$$
By additivity for finite unions and (2), there is
$$
\sum_{n\leqslant m}\mu^*(A_n)=\mu^*(\bigcup_{n\leqslant m} A_n)\leqslant \mu^*(\bigcup_{n=1}^{\infty}A_n)\tag{3}
$$
Let $m\to\infty$, (1) is proved.
EDIT: We need to prove that countable union of measurable sets is measurable in order to prove $M$ forms $\sigma$-algebra. 
Given $A_n\in M$. Let
$$
A=\bigcup_{n=1}^\infty A_n=\bigcup_{n=1}^\infty (A_n-\bigcup_{k=1}^{n-1}A_k)=\bigcup_{n=1}^\infty A_n'
$$
where $A_1'=A_1$. Clearly $A_n'$ are pairwise disjoint sets. Since $A_n'$ are formed from finite union and complement of measurable sets, they are all measurable. So by (3)
$$
\sum_{n =1}^{m}\mu^*(A_n')=\mu^*(\bigcup_{n =1}^{m} A_n')\leqslant \mu^*(\bigcup_{n=1}^{\infty}A_n')
$$
So $\sum_{n =1}^{\infty}\mu^*(A_n')$ converges. Thus for any $\epsilon>0$, there exists $N>0$ such that
$$
\sum_{n=N+1}^{\infty}\mu^*(A_n')<\epsilon
$$
Let
$$
A=\bigcup_{n=1}^{\infty}A_n'=\bigcup_{n=1}^{N}A_n'\cup \bigcup_{n=N+1}^{\infty}A_n'=C\cup D
$$
Then 
$$
\mu^*(D)=\mu^*(\bigcup_{n=N+1}^{\infty}A_n')\leqslant\sum_{n=N+1}^{\infty}\mu^*(A_n')<\epsilon
$$
Since $C$ is measurable, by theorem $12$ on p $41$ in Real Analysis by Royden 4th or by theorem $6$ on p $261$ in Introductory Real Analysis by Kolmogorov and Fomin, there is an elementary set $B$ such that
$$
\mu^*(B\bigtriangleup C)<\epsilon
$$
Also since
$$
A\bigtriangleup B\subset (B\bigtriangleup C)\cup D
$$
There is 
$$
\mu^*(A\bigtriangleup B)\leqslant\mu^*(B\bigtriangleup C)+\mu^*(D)<2\epsilon
$$
By the above theorems again, $A=\bigcup_{n=1}^{\infty}A_n$ is measurable. 
