Timeline for Is arbitrary union of closed balls in $\mathbb{R}^n$ Lebesgue measurable?
Current License: CC BY-SA 3.0
24 events
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| May 22, 2016 at 11:28 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Oct 29, 2010 at 20:48 | vote | accept | CKD | ||
| Oct 29, 2010 at 1:44 | comment | added | David Roberts♦ | @CKD re your edit: did you originally mean that all the balls have the same radius, or varying radii? | |
| Oct 29, 2010 at 1:07 | history | edited | CKD | CC BY-SA 2.5 |
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| Oct 27, 2010 at 1:31 | comment | added | David Roberts♦ | @George, ah, I see. Thanks for clearing that up. | |
| Oct 27, 2010 at 1:27 | vote | accept | CKD | ||
| Oct 29, 2010 at 0:50 | |||||
| Oct 27, 2010 at 1:12 | history | edited | Andrey Rekalo | CC BY-SA 2.5 |
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| Oct 26, 2010 at 23:03 | comment | added | Joel David Hamkins | Yes, I was only talking about the counterexample as a counterexample to the Borel statement. | |
| Oct 26, 2010 at 22:52 | comment | added | BS. | @George&Joel: by definition, any subset of a measure zero set such as $R\times 1$ is Lebesgue measurable, so we don't have a counter-example here. I agree with the Borel counter-example though. | |
| Oct 26, 2010 at 22:51 | comment | added | George Lowther | @Faisal: indeed. I meant to say "Borel measurable" in that last comment. | |
| Oct 26, 2010 at 22:49 | answer | added | George Lowther | timeline score: 23 | |
| Oct 26, 2010 at 22:46 | comment | added | Faisal | @George: If S is a nonmeasurable subset of R, then Sx{0} is a nullset in R^2 and therefore Lebesgue (though not Borel) measurable. | |
| Oct 26, 2010 at 22:39 | comment | added | Joel David Hamkins | George, post your answer as an answer, so we can vote it up. You have answered the second part of the question. Also, it seems that your idea generalizes to all larger $n$ as well. | |
| Oct 26, 2010 at 22:35 | comment | added | George Lowther | ...and it is not Lebesgue measurable since its intersection with a Lebesgue measurable set is not measurable. | |
| Oct 26, 2010 at 22:31 | answer | added | Gerhard Paseman | timeline score: -6 | |
| Oct 26, 2010 at 22:30 | answer | added | Faisal | timeline score: 18 | |
| Oct 26, 2010 at 22:28 | comment | added | George Lowther | David - I think you misunderstood me. If B is the closed unit ball, $\cup\{B+(s,0):s\in S\}$ is a union of closed balls, and is not Borel measurable. | |
| Oct 26, 2010 at 22:21 | comment | added | David Roberts♦ | @George - that set isn't a union of closed balls in R^2, though... | |
| Oct 26, 2010 at 21:25 | comment | added | George Lowther | non-Borel measurability is easy enough, for N=2. Consider the union of unit balls centered at Sx{0} for a non-measurable subset S of the reals, and look at its intersection with Rx{1}. Lebesgue measurability looks a bit more interesting. | |
| Oct 26, 2010 at 21:18 | comment | added | CKD | We are considering non-trival closed balls. So a single point is not a closed ball. | |
| Oct 26, 2010 at 21:17 | history | edited | CKD | CC BY-SA 2.5 |
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| Oct 26, 2010 at 20:49 | comment | added | Homology | A point is a closed ball. Is every set measurable? | |
| Oct 26, 2010 at 20:48 | comment | added | Noah Stein | Is a point a closed ball? | |
| Oct 26, 2010 at 20:32 | history | asked | CKD | CC BY-SA 2.5 |