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Let $(X,Z,\mu)$ be a standard borel measure, I will like to know if there is an isomorphism mod zero preserving the measures form $([0,1],B,\lambda)$ where $\lambda$ is absolutely continuous with respect to the Lebesgue measure in the case where $X$ is uncountable. Look that when $X$ is countable we can find such an isomorphism into the counting measure. If you know such result please give me a reference.

edited Jul 6 2011 at 0:15

asked Jul 5 2011 at 23:59

Carlos De la Mora
113●5

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This is proved e.g. in Kechris, Classical Descriptive Set Theory, Section 17.F (in the continuous case, the countable case is trivial). The result goes back to Hausdorff and was stated in this form by Rokhlin in his thesis, von Neumann should also be mentioned. If I'm not mistaken, Ch III of Arveson's An invitation to $C^{\ast}$-algebras also contains a proof (at least it can easily be extracted from the results in there). – Theo Buehler Jul 6 2011 at 0:23

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By the way: you want $\mu$ non-atomic in your first sentence. – Theo Buehler Jul 6 2011 at 2:27

Thank you very much, this is very useful and I am very excited about this being true. In fact I am looking for a statement where the measure could be atomic. But I guess that in this case will be the sum of the counting measure and the Lebesgue. – Carlos De la Mora Jul 6 2011 at 15:26

Yes, it's really surprising when one first hears about it, so it essentially says that there are not too many well-behaved measure spaces. Yes, you can just decompose your measure (which I assume finite or $\sigma$-finite) into atomic and non-atomic parts. Then you have a disjoint union of a continuous standard Borel space + at most countably many atoms. – Theo Buehler Jul 6 2011 at 16:27

Ok but the result in kechris needs the condition of $\mu$ be a probability measure and seems to be necessary condition. So can I replace $[0,1]$ by the reals and have a similar result? – Carlos De la Mora Jul 6 2011 at 21:35

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Standard Borel Space

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