Simon (Geometric Measure Theory) says:
If $X$ is a locally compact, separable metric space and $\mu(K) < \infty$ for all $K$ compact, then $X=\cup_{i=1}^\infty U_i$ where $U_i$ are open and $\mu(U_i) < \infty$.
I tried showing this by using separability $X=\cup_{i=1}^\infty B(x_i,r)$ for some $r>0$. However, I am unable to prove that each of the open balls $B(x_i,r)$ is contained in a finite union of compact sets (using the local compactness property). This would ensure that the measures of the open balls are finite and so the assertion will be ture. In $\mathbb{R}^n$ this can be done.
I am not sure if this is true in general metric spaces. Any help on how to prove this assertion generally is much appreciated.
Hint: Since $X$ is locally compact and separable, there exists a countable basis $\{U_i\}_{i=1}^{\infty}$ for the topology such that $\overline{U}_i$ is compact for every $i$...