# Outer Measure of Limit of Increasing Sequence of Sets

## Theorem

Let $\mu^*$ be an outer measure on a set $X$.

Let $\left\langle{S_n}\right\rangle$ be an increasing sequence of $\mu^*$-measurable sets, and let $S_n \uparrow S$ (as $n \to \infty$).

Then for any subset $A \subseteq X$:

$\displaystyle \mu^* \left({A \cap S}\right) = \lim_{n \to \infty} \mu^* \left({A \cap S_n}\right)$

## 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:

$\ds S = \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n}$
$\ds A \cap S = A \cap \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n} = \bigcup_{n \mathop = 0}^\infty \paren {A \cap \paren {S_{n + 1} \setminus S_n} }$

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

$\blacksquare$