# Measure Space from Outer Measure

## Theorem

Suppose $\mu^*$ is an outer measure on a set $X$.

Let $\map {\mathfrak M} {\mu^*}$ be the collection of $\mu^*$-measurable sets.

Let $\mu$ be the restriction of $\mu^*$ to $\map {\mathfrak M} {\mu^*}$.

Then $\struct {X, \map {\mathfrak M} {\mu^*}, \mu}$ is a measure space.

## Proof

First, note that $\map {\mathfrak M} {\mu^*}$ is a $\sigma$-algebra over $X$.

Next, choose $E_1, E_2 \in \map {\mathfrak M} {\mu^*}$ such that $E_1 \cap E_2 = \O$.

Thus:

 $\ds \map \mu {E_1 \cup E_2}$ $=$ $\ds \map {\mu^*} {E_1 \cup E_2}$ as $E_1 \cup E_2 \in \map {\mathfrak M} {\mu^*}$ $\ds$ $=$ $\ds \map {\mu^*} {\paren {E_1 \cup E_2} \cap E_1} + \map {\mu^*} {\paren {E_1 \cup E_2} - E_1}$ Definition of Measurable Set $\ds$ $=$ $\ds \map {\mu^*} {E_1} + \map {\mu^*} {E_2}$ since $E_1$ and $E_2$ are disjoint $\ds$ $=$ $\ds \map \mu {E_1} + \map \mu {E_2}$ since $E_1, E_2 \in \map {\mathfrak M} {\mu^*}$

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 $\struct {X, \map {\mathfrak M} {\mu^*} \mu}$ 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.