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 $\relcomp X S$ 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:

By subadditivity of an outer measure:
 * $\map {\mu^*} {A \cap \paren {S_1 \cup S_2} } \le \map {\mu^*} {A \cap S_1} + \map {\mu^*} {\paren {A \setminus S_1} \cap S_2}$

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.