Timeline for Measurable functions with non measurable image
Current License: CC BY-SA 3.0
12 events
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| Apr 18, 2018 at 22:44 | comment | added | Piotr Hajlasz | @user39115 Regarding the last question. It is possible to construct such a function. Just the construction from my answer needs to be slightly adjusted. | |
| Apr 18, 2018 at 22:36 | comment | added | user39115 | Many thanks. As you say such a function has to map a null set to a non measurable set, I guess: Any measurable and computable map in the unit interval sends Borel sets to Lebesgue sets? | |
| Apr 18, 2018 at 22:26 | vote | accept | user39115 | ||
| Apr 18, 2018 at 21:51 | comment | added | Nate Eldredge | A note: your function $f$ cannot be Borel. The image of $[0,1]$ (or any Borel set) under any Borel function is an analytic set, and analytic sets are always Lebesgue measurable (indeed, universally measurable), though they need not be Borel. In particular, modifying your function on a null set can make the image measurable. So such a function has to map a null set to a non-measurable set. | |
| Apr 18, 2018 at 21:36 | history | edited | user39115 | CC BY-SA 3.0 |
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| Apr 18, 2018 at 21:17 | comment | added | user39115 | If I understand correctly, basically the examples are: send a Lebesgue measurable set of measure zero of cardinality continuum onto a non-measurable set and the complement onto a Lebesgue measurable set of measure zero. | |
| Apr 18, 2018 at 20:32 | comment | added | Piotr Hajlasz | @user39115 If $V$ is not Lebesgue measurable, then it cannot be Borel since all Borel sets are Lebesgue measurable. | |
| Apr 18, 2018 at 20:30 | comment | added | user39115 | This is a good idea. Is the non Lebesgue measurable set a Borel set? If it is not, then I understand your example, if it is, then I do not understand it... | |
| Apr 18, 2018 at 20:23 | answer | added | Piotr Hajlasz | timeline score: 8 | |
| Apr 18, 2018 at 19:54 | comment | added | Michael Greinecker | If your function doesn't have to be measure preserving, you can just take a function that maps a measure zero set with the cardinality of the continuum (the Cantor set, say) onto some nonmeasurable set and maps everything else to $0$. That function will be Lebesgue measurable but have nonmeasurable range. | |
| Apr 18, 2018 at 19:37 | history | edited | user39115 | CC BY-SA 3.0 |
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| Apr 18, 2018 at 19:25 | history | asked | user39115 | CC BY-SA 3.0 |