Let $ (X,d)$ be a metric space and let $\mu $ be a Radon $\sigma$-finite measure on the Borel $\sigma$-algebra. I read that it's possible to find countable disjoint compact sets $\lbrace K_n\rbrace_{\mathbb{N}}$ and a $\mu$-null set $N$ such that $$ X=\bigcup_{\mathbb{N}}K_n\cup N. $$
I've tried to reach some results using inner regularity of $\mu$, but nothing. Is this statement true? How can i prove it?
The key assumption here is that $\mu$ is a Radon measure, meaning it is inner regular with respect to compact sets. Without this assumption, this is not true, not even if $\mu$ is finite (for instance, there are metric spaces supporting continuous measures in which all compact sets are finite).
Write $X=\bigcup_n X_n$, where each $X_n$ are disjoint Borel and of finite measure. Then recursively, choose a compact $K_{n,m}\subseteq X_n\setminus \bigcup_{m'<m} K_{n,m'}$ such that $\mu((X_n\setminus \bigcup_{m'<m} K_{n,m'})\setminus K_{n,m})<1/m$. Then $X_n\setminus \bigcup_{m} K_{n,m}$ is null, and so $X\setminus\bigcup_{n,m} K_{n,m}$ is null, and $K_{n,m}$ are clearly disjoint.
