Timeline for Image of probability measures under measurable mappings
Current License: CC BY-SA 4.0
22 events
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| May 16, 2020 at 8:06 | comment | added | Robert Furber | The new version looks fine to me. | |
| May 15, 2020 at 6:29 | comment | added | Dmitri Pavlov | @RobertFurber: Thanks a lot for pointing this out! Fortunately, the wrong half of Lemma 5.10 was not used anywhere, so I threw it away and added the missing reduction to the σ-finite case. I posted a new version here: dmitripavlov.org/equiv.pdf, which will be uploaded to arXiv sometime soon. Let me know if you see anything suspicious left or have any questions. I also added your example (with an acknowledgment in the introduction) as Remark 5.11. | |
| May 14, 2020 at 7:08 | comment | added | Robert Furber | So the join in $\mathcal{P}(X)$ modulo the $\mu$-nullsets is $0$. The good news is: 1. This is essentially the only possible counterexample. 2. If $X$ is compact and $Y$ is strictly localizable, this situation cannot occur and $\mathcal{ML}$ only produces complete Boolean algebra homomorphisms. If you like, I can write a letter to you explaining the proof of this, which you can incorporate into your arxiv paper. | |
| May 14, 2020 at 7:01 | comment | added | Robert Furber | As a counterexample, assume that $X$ is a set such that there exists a countably additive probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons (so $|X|$ exceeds the first real-valued measurable cardinal), and $\nu : \mathcal{P}(X) \rightarrow [0,\infty]$ is the counting measure. Let $f : (X, \mathcal{P}(X), \mu) \rightarrow (X, \mathcal{P}(X), \nu)$ be the identity mapping. The only nullset of $\nu$ is $\emptyset$, so this is a premap of enhanced measurable spaces. The join of all singletons in $X$ is 1, but each singleton has $\mu$-measure zero ... | |
| May 14, 2020 at 6:53 | comment | added | Robert Furber | Unfortunately Lemma 5.10 in your arxiv paper is not correct, unless you assume the nonexistence of real-valued measurable cardinals. The problem is that Lemma 4.37 can only be used to prove that you have a morphism preserving arbitrary suprema for $\sigma$-finite enhanced measurable spaces, not the general localizable case you intend to apply it to there. Counterexample in next comment. | |
| May 14, 2020 at 5:22 | comment | added | Robert Furber | Thanks. That was the fault of the crappiness of the French National Library's website. | |
| May 14, 2020 at 5:06 | comment | added | Dmitri Pavlov | @RobertFurber: Volume 261, pages 4961--4963. Link: gallica.bnf.fr/ark:/12148/bpt6k4027h | |
| May 14, 2020 at 4:52 | comment | added | Robert Furber | That sounds interesting, but a search of the 1965 Comptes Rendus yields no papers by either Ionescu Tulcea. | |
| May 14, 2020 at 1:58 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 14, 2020 at 1:54 | comment | added | Dmitri Pavlov | @RobertFurber: I must also say that the existence of point-set maps of measurable spaces is not due to Fremlin, but rather is a theorem by Cassius Ionescu Tulcea (see his 1965 paper in Comptes Rendus). There were several subsequent improvements, see my paper arxiv.org/abs/2005.05284, where I try to list (hopefully) all contributors. | |
| May 9, 2020 at 20:02 | vote | accept | user2173168 | ||
| May 9, 2020 at 19:54 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 9, 2020 at 19:47 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 9, 2020 at 18:21 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 9, 2020 at 18:13 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 9, 2020 at 18:01 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 9, 2020 at 17:58 | comment | added | Dmitri Pavlov | @RobertFurber: I certainly meant a compact measure, not just a set-theoretic notion. I also added the completeness condition that I left out. | |
| May 9, 2020 at 17:55 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 9, 2020 at 17:15 | comment | added | Robert Furber | And you may call it the "von Neumann-Maharam theorem", but from my perspective that only refers to the existence of an isomorphism of measure algebras as complete Boolean algebras, not of measure spaces. The existence of an isomorphism of measure spaces requires substantial extra work due to Fremlin. | |
| May 9, 2020 at 17:11 | comment | added | Robert Furber | You have also omitted the requirement that $(X,\Sigma,\mu)$ have a lifting. Shelah proved that it is relatively consistent to ZFC that $([0,1],\mathrm{Bo}([0,1]), \lambda)$, $\lambda$ being the Lebesgue measure restricted to Borel sets, have no lifting. Since complete probability spaces always have a lifting, it follows that the Borel restriction of Lebesgue measure is not isomorphic to its completion in this model. (Von Neumann proved, on the other hand, that this space does have a lifting under the continuum hypothesis). | |
| May 9, 2020 at 17:06 | comment | added | Robert Furber | It is not that $\Sigma$ is a compact class, but rather that there exists a compact class $\mathcal{K} \subseteq \Sigma$ such that $\mu$ is inner regular with respect to $\mathcal{K}$ (i.e. that $(X,\Sigma,\mu)$ is compact in the terminology used in Fremlin's Measure Theory volume 3). This cannot be replaced in the general case by any property of $\Sigma$, because there are $\sigma$-algebras that admit both compact and non-compact measures (e.g. the countable-cocountable $\sigma$-algebra), though of course all probability measures on a standard Borel space are compact. | |
| May 9, 2020 at 16:46 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |