# Measurable Sets form Algebra of Sets

## Theorem

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

Then the set of $\mu^*$-measurable sets is an algebra of sets.

## Proof

For a subset $S \subseteq X$, let $\relcomp X S$ denote the relative complement of $S$ in $X$.

We first prove the second property of an algebra of sets, as described on that page.

Let $S$ be $\mu^*$-measurable. For any subset $A \subseteq X$:

 $\ds \map {\mu^*} A$ $=$ $\ds \map {\mu^*} {A \cap S} + \map {\mu^*} {A \cap \relcomp X S}$ $\ds$ $=$ $\ds \map {\mu^*} {A \cap \relcomp X {\relcomp X S} } + \map {\mu^*} {A \cap \relcomp X S}$ Complement of Complement

as desired.

Now we prove the first property.

Suppose that $S_1$ and $S_2$ are $\mu^*$-measurable sets.

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

Since:

 $\ds A \cap \paren {S_1 \cup S_2}$ $=$ $\ds \paren {\paren {A \cap \paren {S_1 \cup S_2} } \cap S_1} \cup \paren {\paren {A \cap \paren {S_1 \cup S_2} } \setminus S_1}$ Set Difference Union Intersection and Union is Commutative $\ds$ $=$ $\ds \paren {A \cap \paren {\paren {S_1 \cup S_2} \cap S_1} } \cup \paren {\paren {\paren {S_1 \cup S_2} \cap A} \setminus S_1}$ Intersection is Associative and Intersection is Commutative $\ds$ $=$ $\ds \paren {A \cap \paren {\paren {S_1 \cup S_2} \cap S_1} } \cup \paren {\paren {\paren {S_1 \cup S_2} \setminus S_1} \cap A}$ Intersection with Set Difference is Set Difference with Intersection $\ds$ $=$ $\ds \paren {A \cap S_1} \cup \paren {\paren {S_2 \setminus S_1} \cap A}$ Intersection Absorbs Union and Set Difference with Union is Set Difference $\ds$ $=$ $\ds \paren {A \cap S_1} \cup \paren {\paren {S_2 \cap A} \setminus S_1}$ Intersection with Set Difference is Set Difference with Intersection $\ds$ $=$ $\ds \paren {A \cap S_1} \cup \paren {\paren {A \cap S_2} \setminus S_1}$ Intersection is Commutative $\ds$ $=$ $\ds \paren {A \cap S_1} \cup \paren {\paren {A \setminus S_1} \cap S_2}$ Intersection with Set Difference is Set Difference with Intersection

By subadditivity of an outer measure:

$\map {\mu^*} {A \cap \paren {S_1 \cup S_2} } \le \map {\mu^*} {A \cap S_1} + \map {\mu^*} {\paren {A \setminus S_1} \cap S_2}$

Thus:

 $\ds$ $=$ $\ds \map {\mu^*} {A \cap \paren {S_1 \cup S_2} } + \map {\mu^*} {A \setminus \paren {S_1 \cup S_2} }$ $\ds$ $=$ $\ds \map {\mu^*} {A \cap \paren {S_1 \cup S_2} } + \map {\mu^*} {\paren {A \setminus S_1} \setminus S_2}$ Set Difference with Union $\ds$ $\le$ $\ds \map {\mu^*} {A \cap S_1} + \map {\mu^*} {\paren {A \setminus S_1} \cap S_2} + \map {\mu^*} {\paren {A \setminus S_1} \setminus S_2}$ by the above argument $\ds$ $=$ $\ds \map {\mu^*} {A \cap S_1} + \map {\mu^*} {A \setminus S_1}$ Definition of Measurability of $S_2$ $\ds$ $=$ $\ds \map {\mu^*} A$ Definition of Measurability of $S_1$

The result follows by the subadditivity of an outer measure.

Alternatively, one could use the equality

 $\ds \map {\mu^*} {A \cap \paren {S_1 \cup S_2} }$ $=$ $\ds \map {\mu^*} {A \cap \paren {S_1 \cup S_2} \cap S_1} + \map {\mu^*} {A \cap \paren {S_1 \cup S_2} \setminus S_1}$ $\ds$ $=$ $\ds \map {\mu^*} {A \cap S_1} + \map {\mu^*} {A \cap S_2 \setminus S_1}$

to prove the result directly without the use of subadditivity.

$\blacksquare$