Measure Space from Outer Measure

Theorem
Suppose $$\mu^*\ $$ is an outer measure on a set $$X\ $$. Let $$\mathfrak M(\mu^*)$$ be the collection of $\mu^*\ $-measurable sets, and let $$\mu\ $$ be the restriction of $$\mu^*\ $$ to $$\mathfrak M(\mu^*)$$.

Then $$(X, \mathfrak M(\mu^*), \mu)$$ is a measure space.

Proof
First, note that $\mathfrak M(\mu^*)$ is a $\sigma$-algebra over $X\ $.

Next, choose $$E_1, E_2\in\mathfrak M(\mu^*)$$ such that $$E_1\cap E_2 = \varnothing$$.

Thus:

$$ $$ $$ $$

Therefore $$\mu\ $$ is additive. But because it is constructed from an outer measure, it is also countably subadditive.

Since Additive and Countably Subadditive Functions are Countably Additive, it follows that $$\mu\ $$ is countably additive.

Finally, also because it is constructed from an outer measure, $$\mu\ $$ is nonnegative.

Hence $$(X, \mathfrak M(\mu^*), \mu)$$ meets all the criteria of a measure space.

Consequences
It immediately follows from the theorem that the Lebesgue measure forms a measure over the collection of Lebesgue measurable sets of reals.