Outer Measure of Limit of Increasing Sequence of Sets/Proof 1

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

Assume that $\mu^* \left({A \cap S_n}\right)$ is finite for all $n \in \N$, otherwise the statement is trivial by the monotonicity of $\mu^*$.

Let $S_0 = \varnothing$. Then $x \in S$ iff there exists an integer $n \ge 0$ such that $x \in S_{n+1}$.

One can always find the least $n$, such that $x \in S_{n+1} \setminus S_n$.

Therefore:
 * $\displaystyle S = \bigcup_{n=0}^\infty \, \left({S_{n+1} \setminus S_n}\right)$

Because intersection distributes over union:
 * $\displaystyle A \cap S = A \cap \bigcup_{n=0}^\infty \, \left({S_{n+1} \setminus S_n}\right) = \bigcup_{n=0}^\infty \, \left({A \cap \left({S_{n+1} \setminus S_n}\right)}\right)$

Therefore: