Measurable Sets form Algebra of Sets

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

Then the set of $\mu^*$-measurable sets is an algebra of sets.

Proof
For a subset $S \subseteq X$, let $\complement \left({S}\right)$ denote the relative complement of $S$ in $X$.

We first prove the second property of an algebra of sets, as described on that page.

Let $S$ be $\mu^*$-measurable. For any subset $A \subseteq X$:

as desired.

Now we prove the first property.

Suppose that $S_1$ and $S_2$ are $\mu^*$-measurable sets. Let $A$ be any subset of $X$.

Since:

we have, by the subadditivity of an outer measure:
 * $\mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right) \leq \mu^* \left({A \cap S_1}\right) + \mu^* \left({\left({A \setminus S_1}\right) \cap S_2}\right)$

Thus:

The result follows by the subadditivity of an outer measure.

Alternatively, one could use the equality

to prove the result directly without the use of subadditivity.