# 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 |

Work In ProgressIn particular: I'm pretty sure the above, or at least significant parts of the above, have/has already been posted up as proofs in their own right. If I remember, or rather, when I'm in the mood, I may go through and find them.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{WIP}}` from the code. |

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.

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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.

$\blacksquare$