Proposition 14.7.2. Let $X$ be a set, $\ce \subset 2^{X}$, and $\rho: \ce \to [0, \infty]$ such that $\emptyset \in \ce$ and $\rho(\emptyset) = 0$. Define
\[\mu^{*}: 2^{X} \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\rho(E_n) \bigg | \seq{E_n} \subset \ce, \bigcup_{n \in \natp}E_n \supset E}\]
then $\mu^{*}$ is an outer measure.
Proof. (OM1): $\emptyset \in \ce$ and $\rho(\emptyset) = 0$.
(OM2): Let $\seq{E_n}\subset 2^{X}$ and assume without loss of generality that $\sum_{n \in \natp}\mu^{*}(E_{n}) < \infty$. Let $\eps > 0$. For each $n \in \nat$, let $\bracs{F_{n, k}}_{k = 1}^{\infty} \subset \ce$ such that $\bigcup_{k \in \natp}F_{n, k}\supset E_{n}$ and $\sum_{n \in \natp}\rho(F_{n, k}) < \mu^{*}(E_{n}) + 2^{-n}\eps$, then
\begin{align*}\mu^{*}\paren{\bigcup_{n \in \natp}E_n}&\le \sum_{(n, k) \in \natp \times \natp}\rho(F_{n, k}) = \sum_{n \in \natp}\sum_{k \in \natp}\rho(F_{n, k}) \\&\le \sum_{n \in \natp}(\mu^{*}(E_{n}) + 2^{-n}\eps) \le \eps + \sum_{n \in \natp}\mu^{*}(E_{n})\end{align*}
Since $\eps$ is arbitrary, $\mu^{*}\paren{\bigcup_{n \in \natp}E_n}\le \sum_{n \in \natp}\mu^{*}(E_{n})$.$\square$