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 $$X$$:


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

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 $$\mu^{*}\left({A}\right) = \mu^{*}(B_{1}) + \mu^{*}(A \cap B_{1}^{C})$$.

Now assume for induction that $$\mu^{*}\left({A}\right) = \sum_{i=1}^{n} \left({\mu^{*} \left({B_{i}}\right)}\right) + \mu^{*}\left({A \cap \left({\bigcup_{i=1}^{n}S_{i}}\right)^{C}}\right)$$ for a given $$n$$.

Since $$S_{n+1}$$ is measurable, $$\mu^{*} \left({A \cap \left({\bigcup_{i=1}^{n} S_{n}}\right)^{C}}\right) = \mu^{*} \left({A \cap \left({\bigcup_{i=1}^{n}S_{n}}\right)^{C}\cap S_{n}}\right) + \mu^{*} \left({A \cap \left({\bigcup_{i=1}^{n+1}S_{i}}\right)^{C}}\right)$$.

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

So the hypothesis holds for all $$n$$.

By monotonicity of the outer measure, $$\mu^{*} \left({A \cap \left({\bigcup_{i=1}^{n} S_{i}}\right)^{C}}\right) \geq \mu^{*} \left({A \cap \left({\bigcup_{i=1}^{\infty}S_{i}}\right)^{C}}\right)$$.

Hence $$\mu^{*} \left({A}\right) \geq \sum_{i=1}^{n} \left({\mu^{*} \left({B_{i}}\right)}\right) + \mu^{*}\left({A \cap \left({\bigcup_{i=1}^{\infty}S_{i}}\right)^{C}}\right)$$.

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

But note that by countable subadditivity, $$\sum_{i=1}^{\infty} \left({\mu^{*} \left({B_{i}}\right)}\right) \geq \mu^{*} \left({\bigcup_{i=1}^{\infty} B_{i}}\right)$$.

Also note that $$\bigcup_{i=1}^{\infty}B_{i} = A \cap \left({\bigcup_{i=1}^{\infty} S_{i}}\right)$$.

Thus, we are done.