Open $\sigma$-compact sets with finite measure Let $X$ be locally compact Hausdorff space and let $\mu$ be positive Borel measure, finite on compacts, outer regular with respect to open subsets, for each Borel set, and inner regular with respect to compact subsets, for each open set and for each Borel with finite measure. Is it true that for every compact $F$ there exists an open $\sigma$-compact $G$ such $F\subset G$ and $G$ has finite measure.
Thanks.

Yes. As $X$ is locally compact and $F$ is compact, there is an $f\in C_c(X)$ (i.e., $f\colon X\to\mathbb{R}$ with compact support) which is strictly positive on $F$. We can then take $G=\{x\in X\colon f(x) > 0\}$. This is contained in the support of $f$, which is compact, so has finite measure. Also,
$$
G=\bigcup_{n=1}^\infty\left\{x\in X\colon f(x)\ge1/n\right\}
$$
expresses $G$ as a countable union of compact sets, so it is $\sigma$-compact.
