Outer Measure of Limit of Increasing Sequence of Sets
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Theorem
Let $\mu^*$ be an outer measure on a set $X$.
Let $\sequence {S_n}$ be an increasing sequence of $\mu^*$-measurable sets, and let $S_n \uparrow S$ (as $n \to \infty$).
Then for any subset $A \subseteq X$:
- $\ds \map {\mu^*} {A \cap S} = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$
Proof
By the monotonicity of $\mu^*$, it suffices to prove that:
- $\ds \map {\mu^*} {A \cap S} \le \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$
Assume that $\map {\mu^*} {A \cap S_n}$ is finite for all $n \in \N$, otherwise the statement is trivial by the monotonicity of $\mu^*$.
Let $S_0 = \O$.
Then $x \in S$ if and only if there exists an integer $n \ge 0$ such that $x \in S_{n + 1}$.
Taking the least possible $n$, it follows that $x \notin S_n$, and so:
- $x \in S_{n + 1} \setminus S_n$
Therefore:
- $\ds S = \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n}$
From Intersection Distributes over Union:
- $\ds A \cap S = A \cap \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n} = \bigcup_{n \mathop = 0}^\infty \paren {A \cap \paren {S_{n + 1} \setminus S_n} }$
Therefore:
\(\ds \map {\mu^*} {A \cap S}\) | \(\le\) | \(\ds \sum_{n \mathop = 0}^\infty \map {\mu^*} {A \cap \paren {S_{n + 1} \setminus S_n} }\) | Definition of Countably Subadditive Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {\map {\mu^*} {A \cap S_{n + 1} } - \map {\mu^*} {A \cap S_{n + 1} \cap S_n} }\) | Definition of Measurable Set of Arbitrary Outer Measure | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {\map {\mu^*} {A \cap S_{n + 1} } - \map {\mu^*} {A \cap S_n} }\) | Intersection with Subset is Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n} - \map {\mu^*} {A \cap \O}\) | Telescoping Series | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}\) | Intersection with Empty Set and Definition of Outer Measure |
$\blacksquare$