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My reference is L. Simon's Lectures on Geometric Measure Theory. He defines a measure on a set $X$ as a countably subadditive function $\mu:2^X\to[0,\infty]$ with $\mu(\emptyset)=0.$

When $X$ is a locally compact and separable topological space, he defines a measure $\mu$ on $X$ to be Radon if it is Borel regular (i.e. every set is contained in a Borel set with the same measure, and all Borel sets are measurable) and finite on compact sets.

Why make the assumption that $X$ is locally compact and separable? They seem extraneous to me.

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First of all, I think $\mu$ should only be defined on a subset of $2^X$, namely the Borel sets.

Second, I think you want a definition so that the Banach space of Borel measures is the dual of $C_0(X)$, the continuous functions that converge to $0$ at $\infty$ when $X$ is embedded in its one point compactification.

As I recall, some of the conditions can be relaxed if you use Baire measures. The Baire sets are the minimal sigma algebra for which continuous functions are measurable. I don't remember precisely what this does, but this might allow one to remove the assumption of separable.

Without local compactness, the assumption that the measure is finite on compact subsets probably becomes a bit meaningless. (Or the topology on the one point compactification becomes meaningless.) ???

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  • $\begingroup$ I don't understand the first line of your answer. If $\mu$ is defined only on the Borel sets, its Borel regularity is vacuous. Also, it is quite reasonable (and common) to define measures on $2^X$, without insisting on additivity (i.e., they are what other people call outer measures). $\endgroup$ – Post No Bulls Dec 8 '13 at 6:02
  • $\begingroup$ Oh yes, sorry. Anyway, I think the rest of what I said was mostly correct. $\endgroup$ – Stephen Montgomery-Smith Dec 8 '13 at 15:44

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