Timeline for Continuity on a measure one set versus measure one set of points of continuity
Current License: CC BY-SA 3.0
12 events
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| Oct 28, 2013 at 0:52 | comment | added | François G. Dorais | In addition, if the requirements are decreasd from Borel to measurable with respect to the completion of $\mu$, then the choices for values of $h$ on $U_0 \setminus D$ in the lemma can be done freely and the separability requirement on $X$ can be dropped entirely. | |
| Oct 27, 2013 at 23:06 | history | edited | François G. Dorais | CC BY-SA 3.0 |
Cleanup
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| Oct 27, 2013 at 21:40 | vote | accept | Nate Ackerman | ||
| S Oct 27, 2013 at 21:36 | history | suggested | Daniel Roy | CC BY-SA 3.0 |
Fixed typo "supose" => "suppose" and added parenthetical (remark) about the sufficiency of separability and completeness, since this raises the question as to whether there's a Borel mapping if X is merely metrizable, which is the setting of Kuratowski.
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| Oct 27, 2013 at 21:32 | review | Suggested edits | |||
| S Oct 27, 2013 at 21:36 | |||||
| Oct 27, 2013 at 20:31 | comment | added | François G. Dorais | @Gerald: Fixed. | |
| Oct 27, 2013 at 20:31 | history | edited | François G. Dorais | CC BY-SA 3.0 |
Fixed after Gerald's comment
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| Oct 27, 2013 at 19:52 | comment | added | Gerald Edgar | But there is still a small thing to do, to get $g$ defined on the original space (the one we have before the open set is deleted). | |
| Oct 27, 2013 at 19:49 | comment | added | François G. Dorais | @Gerald: Since $X$ is second-countable this is not a problem. I'm replacing $X$ by a closed subset of $X$ of $\mu$-measure one wherein all nonempty open sets are $\mu$-postive. | |
| Oct 27, 2013 at 19:27 | comment | added | Gerald Edgar | So ... you first discard all $\mu$-null open sets. To get the answer required, at the end we have to extend $g$ to those discarded sets as well, still retaining continuity at every point of $D$. | |
| Oct 27, 2013 at 17:35 | comment | added | François G. Dorais | Actually, Polish is too strong: this works if $Y$ is complete (but not necessarily separable) and $X$ is separable (but not necessarily complete). Gerald's first answer generalizes to show that completeness of $Y$ is necessary. I made much use of the separability of $X$ but I don't know to what extent it is necessary. | |
| Oct 27, 2013 at 17:05 | history | answered | François G. Dorais | CC BY-SA 3.0 |