Timeline for Why do we want maps to be measurable (in countably-additive setting)
Current License: CC BY-SA 3.0
21 events
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| S Aug 9, 2013 at 7:18 | history | bounty ended | SBF | ||
| S Aug 9, 2013 at 7:18 | history | notice removed | SBF | ||
| S Aug 8, 2013 at 7:18 | history | bounty started | SBF | ||
| S Aug 8, 2013 at 7:18 | history | notice added | SBF | Reward existing answer | |
| Jul 29, 2013 at 9:06 | comment | added | Benoît Kloeckner | @mbsq: I only wrote that when considering Lebesgue measure it was more relevant to ask which sets are Lebesgue-measurable than to ask whether there is an extension of Lebesgue measure to all parts of the reals (possibly not giving zero mass to Lebesgue-negligible sets). Now, I do think that Lebesgue measure on $[0,1]$ is important, if only because it is a particularly natural instance of the unique (up to isomorphism) atomless probability measure on a standard $\sigma$-algebra. Deciding whether all subsets of $[0,1]$ are Lebesgue-measurable has consequences for all such proba. | |
| Jul 29, 2013 at 8:17 | history | edited | SBF | CC BY-SA 3.0 |
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| Jul 26, 2013 at 16:09 | answer | added | Alexander Pruss | timeline score: 4 | |
| Jul 19, 2013 at 9:05 | vote | accept | SBF | ||
| Jul 18, 2013 at 0:54 | comment | added | Monroe Eskew | Benoit, When dealing with abstract probability theory, why is the Lebesgue measure so important? Probability spaces need not be geometrical. | |
| Jul 17, 2013 at 11:11 | comment | added | Benoît Kloeckner | @mbsq: thanks for the correction (but the word "probability" should probably not be there). That said, for the present purpose the Lebesgue measurability of all subsets of $\mathbb{R}$ is certainly more relevant than arbitrary extensions of Lebesgue measure. | |
| Jul 17, 2013 at 8:42 | history | edited | SBF | CC BY-SA 3.0 |
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| Jul 16, 2013 at 23:43 | comment | added | Monroe Eskew | It is equiconsistent with a two-valued measurable cardinal, a proposition that has withstood the test of time pretty well. | |
| Jul 16, 2013 at 21:58 | comment | added | Michael Greinecker | @mbsq How do you know that is consistent? | |
| Jul 16, 2013 at 20:08 | comment | added | Monroe Eskew | Pietro and Benoit, you are both incorrect. It is consistent with ZFC that there are sigma additive probability measures defined on the whole power set, even extending Lebesgue measure on R. See "real valued measurable cardinals." | |
| Jul 16, 2013 at 19:47 | comment | added | Benoît Kloeckner | @Pietro Majer: Well, one can have a sigma additive measure defined on the whole power set, only at the expense of replacing choice by dependent countable choice, which is already quite powerful. | |
| Jul 16, 2013 at 19:05 | answer | added | Yuri Bakhtin | timeline score: 11 | |
| Jul 16, 2013 at 15:06 | comment | added | SBF | @PietroMajer: so we can't define some/most interesting measures on power sets while having the countably additive $\implies$ $\sigma$-fields $\implies$ measurability? | |
| Jul 16, 2013 at 15:04 | comment | added | Pietro Majer | Well, already the case of the Lebesgue measure shows that in general one can't hope to have a sigma additive measure defined on the whole power set. Once one agrees to have measures defined on sigma algebras, I'd say measurability of maps is indeed a must. | |
| Jul 16, 2013 at 14:33 | comment | added | SBF | @PietroMajer: indeed, and his research is also cited in the section in Dubins and Savage that I've mentioned in OP. So do you mean that the requirement of measurability is mostly a technical drawback of a $\sigma$-additive setting? | |
| Jul 16, 2013 at 14:23 | comment | added | Pietro Majer | Even earlier, Bruno De Finetti (en.wikipedia.org/wiki/Bruno_de_Finetti) also tried to settle the Theory of Probability without assuming countable additivity, with deep philosophical arguments. But after all, mathematically, the study of additive probabilities may be brought back to sigma-additivity. | |
| Jul 16, 2013 at 13:25 | history | asked | SBF | CC BY-SA 3.0 |