Is it possible for an open set in a locally compact Hausdorff space to not be the union of an increasing sequence of compact sets? If so, given a regular Borel measure on such a space, how is it that the limit of the measures of the compact sets is equal to measure of the open set? How far can this disconnect be pushed; that is, what conclusion can be made about the union of the increasing sequence of compact sets relative to the open set, other than the limit of the measures is the measure of the limit?

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    $\begingroup$ $\Bbb R$ with the discrete topology is locally compact Hausdorff but not $\sigma$-compact. $\endgroup$ – Stefan Hamcke Aug 11 '13 at 19:03
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    $\begingroup$ Note that for a locally compact Hausdorff space the following are equivalent: (1) There is a sequence $(U_i)_1^\infty$ of open sets such that $\overline{U_i}$ is compact and contained in $U_{i+1}$ and their union is $X$. (2) $X$ is $\sigma$-compact, i.e. a union of countably many compact sets. (3) The one-point-compactification of $X$ has a countable local base at $\infty$. $\endgroup$ – Stefan Hamcke Aug 11 '13 at 19:10
  • $\begingroup$ @StefanH.Thank you for the answer to the first question; I know property (1) is used in proving the existence of partitions of unity for a differentiable manifold. I still am stuck on the second and third questions. $\endgroup$ – Wayne Aug 11 '13 at 19:39
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    $\begingroup$ I no too little measure theory to be of any help regarding the other questions :-) $\endgroup$ – Stefan Hamcke Aug 11 '13 at 19:43

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