# Does any compact subset belong to Borel Sigma Algebra?

Does any compact subset belong to Borel Sigma Algebra? Why?

Here Borel Sigma algebra is defined to be the sigma algebra generated by all open subsets in the topological space.

Thanks and regards!

UPDATE:

I found that there are two kinds of definitions for a Borel sigma algebra: one generated by all open subsets, and the other by all compact subsets. These two definitions are equivalent for most typical spaces, including any locally compact separable metrizable space (or more generally any σ-compact space),

Also in the definition for inner regular measure, is it correct that all the compact subsets are required to be Borel measurable, where the Borel sigma algebra is defined to be generated by all open sets instead of compact sets?

I saw in Wikipedia, inner regular measure is defined to be on Hausdorff space. Is a Haursdorff space a locally compact separable metrizable space or a σ-compact space? Or in a Hausdorff space, the sigma algebra generated by all compact subsets is a proper subset of the sigma algebra generated by all open subsets?

Thanks!

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It's best if you post new questions as new questions. In answer to your new questions, please read the definition of "Hausdorff": en.wikipedia.org/wiki/Hausdorff_space. Hausdorff does not imply locally compact, separable, metrizable, or $\sigma$-compact. However, in a Hausdorff space, compact sets are closed, so the $\sigma$-field generated by the compacts is contained in the Borel $\sigma$-field. This containment may or may not be strict. –  Nate Eldredge Nov 17 '10 at 6:19