Measurable Sets form Sigma-Algebra

Theorem
Let $\mu^*$ be an outer measure on a set $X$.

Then the set $\mathfrak M \left({\mu^*}\right)$ of all $\mu^*$-measurable subsets of $X$ is a $\sigma$-algebra.

Proof
First, note that $\mathfrak M \left({\mu^*}\right)$ is an algebra (of sets).

It remains to be shown that $\mathfrak M \left({\mu^*}\right)$ is closed under countable union.

Because $\mathfrak M \left({\mu^*}\right)$ is an algebra (of sets), the union of any two $\mu^*$-measurable sets is $\mu^*$-measurable.

Using mathematical induction, it directly follows that the finite union of $\mu^*$-measurable sets is $\mu^*$-measurable.

Let $\left\langle{S_n}\right\rangle$ be a sequence of $\mu^*$-measurable subsets of $X$.

Define $\displaystyle S = \bigcup_{n \mathop = 1}^\infty S_n$.

We wish to prove that $S$ is $\mu^*$-measurable.

For all $n \in \N$, the set $T_n = S_1 \cup S_2 \cup \cdots \cup S_n$ is $\mu^*$-measurable.

By Subset of Union, the sequence $\left\langle{T_n}\right\rangle$ is increasing.

Also, $T_n \uparrow S$ (as $n \to \infty$) where $\uparrow$ denotes the limit of an increasing sequence of sets.

Let $A$ be any subset of $X$.

By Set Difference Union Intersection:
 * $A = \left({A \cap S}\right) \cup \left({A \setminus S}\right)$

So by the subadditivity of $\mu^*$, it suffices to prove that $\mu^* \left({A}\right) \ge \mu^* \left({A \cap S}\right) + \mu^* \left({A \setminus S}\right)$ for any subset $A \subseteq X$.

Then:

Letting $n \to \infty$, the result follows by Outer Measure of Limit of Increasing Sequence of Sets.