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Is Lebesgue measure the only complete mesure extending Borel mesure on $\mathbb R$?

If yes, why? If no, what is an example of a complete measure extending Borel measure that is different from Lebesgue measure?

I raise this question following Completion of measure spaces - uniqueness.

mathcounterexamples.net
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1 Answers1

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And the answer is the same as in the other post. You need to add more conditions to have uniqueness. It is possible to define extensions of the Lebesgue measure, using Carathéodory's theorem, for example.

https://en.wikipedia.org/wiki/Carathéodory%27s_extension_theorem

Then you can take the completion of any of these extended measures, and there you go.

A. Pongrácz
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  • I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure. – mathcounterexamples.net Aug 17 '18 at 10:58
  • That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please. – A. Pongrácz Aug 17 '18 at 11:21
  • Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the σ-algebra generated by R, and this extension is unique if the measure is σ-finite. Here, Borel measure is already equipped with a $\sigma$-algebra. Hence Carathéodory extension may just be the Borel $\sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure? – mathcounterexamples.net Aug 17 '18 at 11:43
  • Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work. – A. Pongrácz Aug 17 '18 at 11:46