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 $X$.
Theorem 2.8.2 from "Geometric Measure Theory" of Federer. Let $\mathcal{K}$ be a countable family of subsets of $X$ and let $\sigma\colon\mathcal{K}\to(0,1)$ a function such that for each $K\in\mathcal{K}$ it holds the following property: for each open subset $U\subseteq X$ there exists a countable and disjointed subfamily $\mathcal{G}\subseteq\mathcal{F}$ such that $\bigcup_{G\in\mathcal{G}}G\subseteq U$ and $\mu\left((U\cap K)\setminus\bigcup_{G\in\mathcal{G}}G\right)\le\sigma(K)\mu(U\cap K)$. Then $\mathcal{F}$ is $\mu$-adequate for $\bigcup_{K\in\mathcal{K}}K$.
The Proof starts by saying that "since $X$ is the union of countably many bounded sets, we may assume that each member of $\mathcal{K}$ is bounded". I don't undersant why I can assume this. Can anyone explain me why, please?
In other words how can I deduce the general from the particular case in which each member of $\mathcal{K}$ is bounded?
Attempt. Let $\{B_n\}_{n\in\mathbb{N}}$ a sequence of open bounded subsets of $X$ that covers $X$. Let us consider the family $\mathcal{K}'=\{K\cap B_n\mid K\in\mathcal{K}, n\in\mathbb{N}\}$. I define a function $\sigma'\colon\mathcal{K}'\to(0,1)$ as $\sigma'(K\cap B_n):=\sigma(K)$ for all $K\in\mathcal{K},n\in\mathbb{N}$. Using hypothesis we have that for each $K\in\mathcal{K}$ and for each $n\in\mathbb{N}$, it holds that for each open subset $U\subseteq X$ there exists a countable and disjointed subfamily $\mathcal{G}\subseteq\mathcal{F}$ such that $\bigcup_{G\in\mathcal{G}}G\subseteq U\cap B_n$ and $\mu\left((U\cap B_n\cap K)\setminus\bigcup_{G\in\mathcal{G}}G\right)\le\sigma(K)\mu(U\cap B_n\cap K)=\sigma'(K\cap B_n)\cdot\mu(U\cap B_n\cap K)$. Then the family $\mathcal{K}'$ satisfies the hypothesis of the Theorem and each member of $\mathcal{K}'$ is bounded, then $\mathcal{F}$ is $\mu$-adequate for $\bigcup\mathcal{K}'=\bigcup\mathcal{K}$ and this concludes.
My question: am I sure that function $\sigma'$ is well defined?