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From wikipedia, Radon measure is defined as

Radon measure is a measure on the $\sigma$-algebra of Borel sets of a Hausdorff topological space $X$ that is finite on all compact sets, outer regular on all Borel sets, and inner regular on open sets.

I am reading a proof on the $\sigma$-compact, locally compact metric space, that a local finite Borel measure is a Radon measure. It seems to me the proof shows it is inner regular on all Borel sets, which is stronger than the definition of Radon measure. Am I correct? What is the correct definition of inner regularity? I think the correct statement is

If $X$ is a locally compact, $\sigma$-compact metric space and $\mu$ is a locally finite Borel measure on $X$, then it is both inner and outer regular (or simply regular).

Fellow InstituteOfMathophile
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  • Indeed, as you read in the answer you linked, in a locally compact $\sigma$-compact metric space, every locally finite Borel measure is automatically regular, which is, as you observed, stronger than just saying it’s a Radon measure. However, this doesn’t have to hold for all locally compact Hausdorff spaces. You can find a Radon measure which is not inner regular in the examples section of this Wikipedia article: https://en.wikipedia.org/wiki/Regular_measure – David Gao Apr 04 '24 at 06:14
  • (Another thing worth pointing out: while the standard definition nowadays requires a Radon measure to be outer regular on all Borel sets and inner regular on all open sets, there are alternative definitions which require inner regularity on all Borel sets and have no requirements on outer regularity. The two notions are not equivalent, as again seen in the same example in the Wikipedia article, but they do correspond to each other, in the sense that given a Radon measure in one sense, there is a unique Radon measure in the other sense that coincides on all open sets and all compact sets.) – David Gao Apr 04 '24 at 06:20

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