Measurable Sets form Algebra of Sets

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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


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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.

$\blacksquare$

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