Is that correct $R^2\cong R$ as measurable spaces? If we consider $R$ and $R^2$ with Borel $\sigma$-algebras, is there measurable map from $R$ to $R^2$ with measurable inverse?
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4$\begingroup$ See here $\endgroup$– Mateusz KwaśnickiCommented Aug 9, 2018 at 9:27
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1$\begingroup$ @Mateusz Kwaśnicki: that one is about the Borel sigma-algabras, not Lebesgue. $\endgroup$– QfwfqCommented Aug 9, 2018 at 9:33
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3$\begingroup$ @Qfwfq: Technically: yes. However, if measure spaces are isomorphic, so are their completions. $\endgroup$– Mateusz KwaśnickiCommented Aug 9, 2018 at 9:40
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2$\begingroup$ @MateuszKwaśnicki: what do you mean? the completion depends on the measure. Maybe what you say is true for divisible measures? $\endgroup$– Pietro MajerCommented Aug 9, 2018 at 9:45
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6$\begingroup$ @PietroMajer: The linked Math.SE post asserts that the measure spaces $(\mathbb{R}^n, \operatorname{Bor}(\mathbb{R}^n), \operatorname{Leb}^n)$ are all isomorphic. Isomorphic measures have isomorphic completions. (My second comment mentioned measure spaces, not just measurable spaces). $\endgroup$– Mateusz KwaśnickiCommented Aug 9, 2018 at 9:57
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