Let $X$ be a metric space and $\mu\colon\mathcal{P}(X)\to[0,+\infty]$ an outer measure over $X$ such that all open subsets of $X$ are $\mu$-measurable (then $\mu$ is a Borel measure) and $\mu$ is finite on the bounded subsets of $X$. Let $\mathcal{F}$ a family of closed subsets of $X$.
I'm trying to understand the proof of the following theorem.
Theorem 2.8.7 from Geometric Measure Theory of Federer. Let $\delta\colon\mathcal{F}\to[0,+\infty)$ be a bounded functions and let $\tau,\lambda\in(1,+\infty)$. Let $S$ be a subset of $X$ and suppose that $\mathcal{F}$ covers finely $S$. Suppose that $\mu(\widehat{F})<\lambda\mu(F)$ for all $F\in\mathcal{F}$. Then $\mathcal{F}$ is $\mu$-adequate for $S$.
Proof (by Federer).
We may assume that $S$ is bounded. Given any bounded open subset $U$ of $X$, we observe that the family $\mathcal{F}'$ of all $F\in\mathcal{F}$ such that $F\subseteq$ covers finely $U\cap S$. Applying theorems 2.8.4-2.8.6 of Federer we obtain that there exists a disjointed subfamily $\mathcal{G}\subseteq\mathcal{F}$ such that
$(U\cap S)\setminus\bigcup\mathcal{H}\subseteq\bigcup_{G\in\mathcal{G}\setminus\mathcal{H}}\widehat{G}$
for all finite subfamily $\mathcal{H}\subseteq\mathcal{G}$. "Since $\mu(U)<+\infty$ and $G\subseteq U$ and $\mu(G)>0$ for all $G\in\mathcal{G}$, the family $\mathcal{G}$ is countable."
Can anyone explain me why the assertion in bold holds, please? How can i deduce that $\mathcal{G}$ is countable?
The family $\mathcal{G}$ is composed of disjoints sets, so since $$ \bigsqcup_{G \in \mathcal{G} } G \subset U$$ we have $$ \mu \left(\bigsqcup_{G \in \mathcal{G} } G\right)= \sum_{G \in \mathcal{G}} \mu(G) \leq \mu(U) <\infty$$
If you read my comment, you know that the set $\{G \in \mathcal{G} \mid \mu(G) >0\}$ is countable, but this set is exactly $\mathcal{G}$, hence $\mathcal{G}$ is countable.