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 $\complement \left({S}\right)$ 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 |
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Work In Progress: 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. (discuss) You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. When this has been done, the template should be removed from the code. |
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.
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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.
$\blacksquare$
