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 $\complement \left({S}\right)$ 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 $A \subseteq X$:

\(\displaystyle \mu^* \left({A}\right)\) \(=\) \(\displaystyle \mu^* \left({A \cap S}\right) + \mu^* \left({A \cap \complement \left({S}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \mu^* \left({A \cap \complement \left({\complement \left({S}\right)}\right)}\right) + \mu^* \left({A \cap \complement \left({S}\right)}\right)\) 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:

\(\displaystyle A \cap \left({S_1 \cup S_2}\right)\) \(=\) \(\displaystyle \left[{\left({A \cap \left({S_1 \cup S_2}\right)}\right) \cap S_1}\right] \cup \left[{\left({A \cap \left({S_1 \cup S_2}\right)}\right) \setminus S_1}\right]\) Set Difference Union Intersection and Union is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left[{A \cap \left({\left({S_1 \cup S_2}\right) \cap S_1}\right)}\right] \cup \left[{\left({\left({S_1 \cup S_2}\right) \cap A}\right) \setminus S_1}\right]\) Intersection is Associative and Intersection is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left[{A \cap \left({\left({S_1 \cup S_2}\right) \cap S_1}\right)}\right] \cup \left[{\left({\left({S_1 \cup S_2}\right) \setminus S_1}\right) \cap A}\right]\) Intersection with Set Difference is Set Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle \left({A \cap S_1}\right) \cup \left({\left({S_2 \setminus S_1}\right) \cap A}\right)\) Intersection Absorbs Union and Set Difference with Union is Set Difference
\(\displaystyle \) \(=\) \(\displaystyle \left({A \cap S_1}\right) \cup \left({\left({S_2 \cap A}\right) \setminus S_1}\right)\) Intersection with Set Difference is Set Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle \left({A \cap S_1}\right) \cup \left({\left({A \cap S_2}\right) \setminus S_1}\right)\) Intersection is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left({A \cap S_1}\right) \cup \left({\left({A \setminus S_1}\right) \cap S_2}\right)\) Intersection with Set Difference is Set Difference with Intersection

we have, by the subadditivity of an outer measure:

$\mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right) \leq \mu^* \left({A \cap S_1}\right) + \mu^* \left({\left({A \setminus S_1}\right) \cap S_2}\right)$

Thus:

\(\displaystyle \) \(=\) \(\displaystyle \mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right) + \mu^* \left({A \setminus \left({S_1 \cup S_2}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right) + \mu^* \left({\left({A \setminus S_1}\right) \setminus S_2}\right)\) Set Difference with Union
\(\displaystyle \) \(\le\) \(\displaystyle \mu^* \left({A \cap S_1}\right) + \mu^* \left({\left({A \setminus S_1}\right) \cap S_2}\right) + \mu^* \left({\left({A \setminus S_1}\right) \setminus S_2}\right)\) by the above argument
\(\displaystyle \) \(=\) \(\displaystyle \mu^* \left({A \cap S_1}\right) + \mu^* \left({A \setminus S_1}\right)\) Definition of Measurability of $S_2$
\(\displaystyle \) \(=\) \(\displaystyle \mu^* \left({A}\right)\) Definition of Measurability of $S_1$

The result follows by the subadditivity of an outer measure.

Alternatively, one could use the equality

\(\displaystyle \mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right)\) \(=\) \(\displaystyle \mu^* \left({A \cap \left({S_1 \cup S_2}\right) \cap S_1}\right) + \mu^* \left({A \cap \left({S_1 \cup S_2}\right) \setminus S_1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \mu^* \left({A \cap S_1}\right) + \mu^* \left({A \cap S_2 \setminus S_1}\right)\)

to prove the result directly without the use of subadditivity.

$\blacksquare$