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the Lesbegue measure is defined an the Borel Sigma-Algebra, so does the Borel measure. Can somebody point out the difference betweem them and give an example to observe that difference ? Thanks.

ivo
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    The $\sigma$-algebra on wich the Lebesguemeasure is defined has (much) more elements than the Borel $\sigma$-algebra. The measures agree on Borel $\sigma$-algebra, so the first can be looked at as an extension of the second (wich is more handsome). Here you find the main difference: "The Borel measure agrees with the Lebesgue measure on those sets for which it is defined; however, there are many more Lebesgue-measurable sets than there are Borel measurable sets. The Borel measure is translation-invariant, but not complete." – drhab Apr 29 '15 at 10:30
  • So the difference is then the completion of the Borel sigma algebra to a Lebesgue sigma algebra by defining all subsets of the null set to be measurable in the Lebesgue sigma algebra and thus extending the Borel measure to the ( completed ) Lebesgue measure ? Is this what you are saying ? Thanks for your comment. – ivo Apr 29 '15 at 11:09

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