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How is a Borel measure on Euclidean space defined?

I mean the Borel measure on the Borel sigma algebra generated by usual toplogy on an Euclidean space, which coincides with the normal definition of volume of subsets in the Euclidean space.

For Lebesgue measure on an Euclidean space, it is defined in terms of the outer measure on the power set of the Euclidean space.

If the Borel measure can be defined independent of the Lebesgue measure, then the Lebesgue measure can be defined in terms of the Borel measure by completion of measure. Right?

Thanks and regards!

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Riemann integration defines a positive linear functional on the space of continuous functions with compact support, which by Riesz's representation theorem comes from a positive Borel measure. The completion of this Borel measure is Lebesgue measure. –  Jonas Meyer Oct 27 '10 at 23:32

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up vote 4 down vote accepted

The Borel $\sigma$-algebra on $\mathbb{R}$ is generated by the ring of finite disjoint unions of half-open intervals. On this ring one can define a countably additive measure in the obvious way (although one should prove that it is countably additive). By Carathéodory's extension theorem this measure extends to the entire Borel $\sigma$-algebra, and by Dynkin's lemma this extension is unique. Then one can take products to get a Borel measure on $\mathbb{R}^n$ and complete to get Lebesgue measure on the Lebesgue $\sigma$-algebra.

(Of course, this isn't essentially different - the proof of Caratheodory that I've seen uses outer measure anyway.)

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