Measurable Sets form Sigma-Algebra

Theorem
Let $\mu^*$ be an outer measure on a set $X$, and let $M$ be the collection of measurable sets on $X$.

Then $M$ is a sigma-algebra.

Proof
First, note that M is an algebra.

It remains to be shown that for any sequence $S_n$ of measurable sets and any subset $A$ of $X$:


 * $\displaystyle \mu^* \left({A}\right) = \mu^* \left({A \cap \left({\bigcup_{i=1}^\infty S_i}\right)}\right) + \mu^* \left({A - \left({\bigcup_{i=1}^\infty S_i}\right)}\right)$

And by subadditivity of $\mu^*$, we only have to show that the left-hand side is greater than or equal to the right-hand side.

To do so, begin by defining:
 * $B_1 = \varnothing, B_{n+1} = A \cap S_1^C \cap S_2^C \cap \cdots \cap S_n^C \cap S_{n+1}$

where $S_i^C$ denotes the relative complement of $S_i$ in $X$.

Note that:
 * $\mu^*\left({A}\right) = \mu^* \left({B_1}\right) + \mu^* \left({A \cap B_1^C}\right)$

Now assume for induction that:
 * $\displaystyle \mu^*\left({A}\right) = \sum_{i=1}^n \left({\mu^* \left({B_i}\right)}\right) + \mu^* \left({A \cap \left({\bigcup_{i=1}^n S_i}\right)^C}\right)$

for a given $n$.

Since $S_{n+1}$ is measurable, we have:
 * $\displaystyle \mu^* \left({A \cap \left({\bigcup_{i=1}^n S_n}\right)^C}\right) = \mu^* \left({A \cap \left({\bigcup_{i=1}^n S_n}\right)^C \cap S_n}\right) + \mu^* \left({A \cap \left({\bigcup_{i=1}^{n+1}S_i}\right)^C}\right)$

Substituting the right-hand side into our inductive assumption, we see that:
 * $\displaystyle \mu^*\left({A}\right) = \sum_{i=1}^{n+1} \left({\mu^* \left({B_i}\right)}\right) + \mu^* \left({A \cap \left({ \bigcup_{i=1}^{n+1} S_i}\right)^C}\right)$

So the hypothesis holds for all $n$.

By monotonicity of the outer measure:
 * $\displaystyle \mu^* \left({A \cap \left({\bigcup_{i=1}^n S_i }\right)^C }\right) \ge \mu^* \left({A \cap \left({\bigcup_{i=1}^\infty S_i}\right)^C}\right)$

Hence:
 * $\displaystyle \mu^* \left({A}\right) \ge \sum_{i=1}^n \left({\mu^* \left({B_i}\right)}\right) + \mu^*\left({A \cap \left({\bigcup_{i=1}^\infty S_i}\right)^C}\right)$

Finally, since this inequality is true for any number of summands, it is true in the limit:
 * $\displaystyle \mu^*\left({A}\right) \ge \sum_{i=1}^\infty \left({\mu^* \left({B_i}\right)}\right) + \mu^*\left({A \cap \left({\bigcup_{i=1}^\infty S_i}\right)^C}\right)$

But note that by countable subadditivity:
 * $\displaystyle \sum_{i=1}^\infty \left({\mu^* \left({B_i}\right)}\right) \ge \mu^* \left({\bigcup_{i=1}^\infty B_i}\right)$

Also note that:
 * $\displaystyle \bigcup_{i=1}^\infty B_i = A \cap \left({\bigcup_{i=1}^\infty S_i}\right)$

Thus, we are done.