Timeline for are there measure preserving mapping in this case?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 25, 2018 at 16:53 | comment | added | Gerald Edgar | For the case where $h$ need not be measure preserving, the hypothesis that $f, g$ have the same distribution is irrelevant: it is enough that $f$ and $g$ have the same range. But, as Nik notes, the selection $h$ may fail to be Borel measurable: but it may be chosen universally measurable. | |
| Jul 25, 2018 at 15:56 | comment | added | Nik Weaver | Ah, I didn't notice that. Yes, you certainly have a counterexample for $h$ measure preserving. The problem in the text seems more subtle --- here I feel the existence of $h$ should follow from some measurable selection theorem which could only fail for really pathological $g$, if it fails at all. | |
| Jul 25, 2018 at 14:46 | comment | added | Gerald Edgar | Perhaps we have to distinguish between the problem in the title (measure preserving) and the problem in the text that does not say $h$ is measure preserving. | |
| Jul 25, 2018 at 14:04 | comment | added | Nik Weaver | I thought of this too, but then I tried to show no $h$ could exist and discovered that it's easy to find an $h$ that works. | |
| Jul 25, 2018 at 11:10 | history | answered | Gerald Edgar | CC BY-SA 4.0 |