# 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.

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:

 $\displaystyle \mu(E_1\cup E_2)$ $=$ $\displaystyle \mu^*(E_1\cup E_2)$ since $E_1\cup E_2\in\mathfrak M(\mu^*)$ $\displaystyle$ $=$ $\displaystyle \mu^*((E_1\cup E_2)\cap E_1) + \mu^*((E_1\cup E_2) - E_1)$ by definition of a measurable set $\displaystyle$ $=$ $\displaystyle \mu^*(E_1) + \mu^*(E_2)$ since $E_1$ and $E_2$ are disjoint $\displaystyle$ $=$ $\displaystyle \mu (E_1) + \mu (E_2)$ since $E_1, E_2\in\mathfrak M(\mu^*)$

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

Since Additive and Countably Subadditive Function is 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.

$\blacksquare$

## Consequences

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