# Measurable Sets form Sigma-Algebra

## Theorem

Let $\mu^*$ be an outer measure on a set $X$.

Then the set $\map {\mathfrak M} {\mu^*}$ of all $\mu^*$-measurable subsets of $X$ is a $\sigma$-algebra.

## Proof

First, note that $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets).

It remains to be shown that $\map {\mathfrak M} {\mu^*}$ is closed under countable union.

Because $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets), the union of any two $\mu^*$-measurable sets is $\mu^*$-measurable.

Using mathematical induction, it directly follows that the finite union of $\mu^*$-measurable sets is $\mu^*$-measurable.

Let $\sequence {S_n}$ be a sequence of $\mu^*$-measurable subsets of $X$.

Define $\ds S = \bigcup_{n \mathop = 1}^\infty S_n$.

We wish to prove that $S$ is $\mu^*$-measurable.

For all $n \in \N$, the set $T_n = S_1 \cup S_2 \cup \cdots \cup S_n$ is $\mu^*$-measurable.

By Subset of Union, the sequence $\sequence {T_n}$ is increasing.

Also, $T_n \uparrow S$ (as $n \to \infty$) where $\uparrow$ denotes the limit of an increasing sequence of sets.

Let $A$ be any subset of $X$.

$A = \paren {A \cap S} \cup \paren {A \setminus S}$

So by the subadditivity of $\mu^*$, it suffices to prove that $\map {\mu^*} A \ge \map {\mu^*} {A \cap S} + \map {\mu^*} {A \setminus S}$ for any subset $A \subseteq X$.

Then:

 $\ds \map {\mu^*} A$ $=$ $\ds \map {\mu^*} {A \cap T_n} + \map {\mu^*} {A \setminus T_n}$ Definition of Measurability of $T_n$ $\ds$ $\ge$ $\ds \map {\mu^*} {A \cap T_n} + \map {\mu^*} {A \setminus S}$ Set Complement inverts Subsets and Monotonicity of $\mu^*$

Letting $n \to \infty$, the result follows by Outer Measure of Limit of Increasing Sequence of Sets.

$\blacksquare$