Timeline for Can every set be measurable?
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| May 4, 2019 at 21:48 | comment | added | Gabe Goldberg | I think it is an interesting condition. Equivalently, you could ask whether it is consistent with ZF that (the powerset of) every set carries a (real-valued) measure that does not concentrate on a smaller set. You then no longer need to exclude countable sets. Note also that one cannot have a measure on a cardinal of cofinality $\omega$ such that every set of strictly smaller cardinality is null. (The AD dichotomy is due to Woodin and is proved in Caicedo-Ketchersid's "A trichotomy theorem in natural models of $\text{AD}^+$" as a consequence of a more general result due to Caicedo-Ketchersid.) | |
| May 4, 2019 at 6:13 | comment | added | arsmath | @GabeGoldberg I didn't know $AD^{L(\mathbb{R})}$ had such a pretty dichotomy, but you're right that I should tighten my nontriviality condition. I have trouble thinking about cardinality without choice. Do you think it's sufficient to rule out that the measure is not a push-forward from a set of strictly smaller cardinality? | |
| May 2, 2019 at 1:02 | comment | added | Andrés E. Caicedo | @arsmath As you see from Gabe's comment, you probably want to strengthen the nontriviality condition in a way that relates the size of $X$ and, say, the completeness of the measure. This requires some care when $X$ is nonwell-orderable. | |
| May 1, 2019 at 23:18 | comment | added | Gabe Goldberg | @RobertFurber I posted this as a comment because I think arsmath may want to modify the nontriviality condition in (1) to rule out the sort of example I have given. If this is not the case, I am happy to post my comment as an answer. | |
| May 1, 2019 at 21:40 | comment | added | Elliot Glazer | For (2), I assume you're only referring to completions of non-trivial measures (otherwise this fails for $X=2$). In which case, this can hold only vacuously, namely in a model where the closure of the singletons under countable union is the whole universe, e.g. Gitik's model. I suspect for both (1) and (2), what you are really interested in is measures that generalize Lebesgue measure, rather than arbitrary atomless probability measures. | |
| May 1, 2019 at 19:09 | comment | added | Robert Furber | @GabeGoldberg Could you post this as an answer? It seems to be what arsmath is looking for. | |
| S May 1, 2019 at 18:27 | history | suggested | Alex Kruckman | CC BY-SA 4.0 |
Fixed the first sentence, as confirmed in the comments
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| May 1, 2019 at 17:24 | review | Suggested edits | |||
| S May 1, 2019 at 18:27 | |||||
| May 1, 2019 at 15:59 | comment | added | Gabe Goldberg | The restriction of (1) to uncountable sets can hold. Assume every set of reals is Lebesgue measurable, $\omega_1$ is measurable, and every uncountable set admits an injection from either $\omega_1$ or $\mathbb R$. (These conditions hold in $L(\mathbb R)$ assuming $\text{AD}^{L(\mathbb R)}$.) Then there is a nontrivial measure on $P(X)$ for every uncountable set $X$, namely the pushforward of the measure on $P(\omega_1)$ or $P(\mathbb R)$ by an injection into $X$. | |
| May 1, 2019 at 15:20 | comment | added | arsmath | Yes. I apparently can't be trusted to write a simple sentence correctly. | |
| May 1, 2019 at 14:18 | comment | added | Alex Kruckman | "consistent with every Lebesgue measurable set being measurable" - you mean "consistent with every subset of the reals being Lebesgue measurable", right? | |
| May 1, 2019 at 13:37 | history | edited | arsmath | CC BY-SA 4.0 |
added 23 characters in body
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| May 1, 2019 at 13:36 | comment | added | arsmath | Right. I'll fix it. I assume that any consistency statement here will require even stronger assumptions. | |
| May 1, 2019 at 13:29 | comment | added | YCor | "Solovay model shows that ZF is consistent": I think it shows this assuming the consistency of ZF(C?)+ $\exists$ inaccessible cardinal. | |
| May 1, 2019 at 13:04 | comment | added | arsmath | Though I have to admit that I don't know if this is sensitive to the definition of "countable" without choice. | |
| May 1, 2019 at 12:56 | comment | added | arsmath | Good point. I need to rule out countable sets as an example. | |
| May 1, 2019 at 12:55 | history | edited | arsmath | CC BY-SA 4.0 |
Rule out trivial loophole.
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| May 1, 2019 at 12:39 | comment | added | James E Hanson | What about subsets of a countable set? You're not going to get a non-trivial countably additive measure there. Similarly if you require a fairly natural homogeneity condition (every subset of non-maximal cardinality has measure zero) then you'd have problems with countable cofinality too. | |
| May 1, 2019 at 12:04 | comment | added | Michael Greinecker | I think you are right. | |
| May 1, 2019 at 11:56 | comment | added | arsmath | Isn't the first uncountable ordinal a measurable cardinal under AD? Or does measurable cardinal mean something difference in the absence of choice? | |
| May 1, 2019 at 11:39 | comment | added | Michael Greinecker | You cannot have a nontrivial measure on the power set of the first uncountable ordinal. I think you need at most countable choice to prove that. | |
| May 1, 2019 at 11:31 | history | asked | arsmath | CC BY-SA 4.0 |