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
The second property of an algebra of sets, as described on that page, follows directly from the definition of a $\mu^*$-measurable set.

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

Now we prove the first property.

Suppose $E_1$ and $E_2$ are $\mu^*$-measurable sets. Let $A$ be any subset of $X$.

Since:
 * $A \cap \left({E_1 \cup E_2}\right) = \left({A \cap E_1}\right) \cup \left({A \cap E_2 \setminus E_1}\right) $

we have, by the subadditivity of an outer measure:
 * $\mu^* \left({A \cap \left({E_1 \cup E_2}\right)}\right) \leq \mu^* \left({A \cap E_1}\right) + \mu^* \left({A \cap E_2 \setminus E_1}\right)$

Thus:

{[explain|How is $\mu^* \left({A \setminus \left({E_1 \cup E_2}\right)}\right) = \mu^* \left({A \setminus E_1 \setminus E_2}\right)$ De Morgan's Laws? There is a result which proves this, it's just a matter of finding it.}}

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.