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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.


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I haven't given it much thought, but for these kind of questions it's usually helpful to think of $\mathbb{R}\times \mathbb{R}_d$ (where $\mathbb{R}_d$ is $$\mathbb{R}$ equipped with the *discrete topology) and its natural measure (which is...?). – Mark Sep 21 '11 at 21:06
I think the title of this question could be improved a bit...maybe something like "Open $\sigma$-compact sets with finite measure" or some such thing. – George Lowther Sep 21 '11 at 22:38
up vote 7 down vote accepted

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.

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