Timeline for A non-trivial probability measure on $2^{\mathbb R}$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2018 at 16:04 | comment | added | Robert Furber | @TomLaGatta The tensor product $\sigma$-algebra is equal to the Baire $\sigma$-algebra. In general, on a compact Hausdorff space, Baire probability measures extend to Radon probability measures on the Borel sets (i.e. inner regular with respect to compact sets) and Radon probability measures on the Borel sets are equal iff they agree on Baire sets. | |
| Nov 27, 2013 at 15:57 | comment | added | Tom LaGatta | > This generator has the cardinality $c$ of the continuum and since the generated $\sigma$-algebra can be obtained in $\omega_1$ (transfinite) induction steps the cardinality of $\mathcal A$ is also $c$. On the other hand, if $\mathcal T$ is a topology with $\mathcal A=\sigma(\mathcal T)$ then $\mathcal T$ separates points (this should follow from the "good sets principle"), in particular, for two distinct points of $\Omega$ the closures of the corresponding singletons are distinct. Hence $\mathcal T$ has at least $|\Omega|=2^c$ elements. | |
| Nov 27, 2013 at 15:56 | comment | added | Tom LaGatta | > The underlying space is $\Omega= 2^{\mathbb R}$, that is the space of all indicator functions, and the $\sigma$-algebra is $\mathcal A = \bigotimes_{\mathbb R} \mathcal P$ where $\mathcal P$ is the power set of the two element set. It is generated by the system $\mathcal E$ of the "basic open sets" of the product topology (prescribed values in a finite number of points). | |
| Nov 27, 2013 at 15:56 | comment | added | Tom LaGatta | @Qiaochu: thanks! I think you've resolved the issue. It's nice to see that there is an abundance of measures on this space, and not a dearth. For the record, I'll paste in the definition of tensor-product $\sigma$-algebra from Jochen Wengenroth's answer, along with the argument why it cannot be generated by a topology. mathoverflow.net/questions/87838/… | |
| Nov 26, 2013 at 6:22 | comment | added | Qiaochu Yuan | @Tom: if I understand correctly, and it's quite possible I haven't, the tensor product $\sigma$-algebra is strictly contained in the Borel $\sigma$-algebra of the product topology, so any measure I construct on the latter restricts to a measure on the former. (I admit I had initially completely forgotten about this distinction because I didn't know what you meant by "tensor product $\sigma$-algebra" and couldn't find a definition online.) | |
| Nov 26, 2013 at 6:06 | comment | added | Tom LaGatta | Qiaochu & @petrelharp, it dawns on me that I might have accepted this answer too soon. The motivation was to equip $2^{\mathbb R}$ with its tensor-product $\sigma$-algebra, which is not generated by a topology. Your answers use the Borel $\sigma$-algebra of the product topology, which seems to not actually answer the question. Can you speak to this point, particularly in light of Joseph Van Name's answer to this question? | |
| Nov 22, 2013 at 21:55 | comment | added | Qiaochu Yuan | @Tom: yes, it seems there's no obstruction to just constructing the corresponding product measure. | |
| Nov 22, 2013 at 21:53 | comment | added | Tom LaGatta | Thanks Qiaochu & @petrelharp. I think your two answers sufficiently resolve the question. Let $p : \mathbb R \to [0,1]$ be an arbitrary function (possibly non-measurable). Does $2^{\mathbb R}$ always support a product measure where the $t$th marginal is Bernoulli with probability $p(t)$? | |
| Nov 22, 2013 at 21:48 | vote | accept | Tom LaGatta | ||
| Nov 22, 2013 at 20:06 | comment | added | petrelharp | Yep; this follows from Kolmogorov's extension theorem. For a less trivial example, let the probability of the coin at $t$ be $1/(1+\exp(-B_t))$, where $B_t$ is a Brownian motion (run in both directions from $B_0=0$), say. | |
| Nov 22, 2013 at 1:12 | comment | added | Asaf Karagila♦ | Even if we assume the time line is homeomorphic to $\Bbb R$ somehow, flipping a coin $\aleph_0$ many times is an infinite task, but one can achieve that with immortality. Flipping $2^{\aleph_0}$ many coins will certainly amount to a condensation point in which an infinite amount of energy is released. I think we just figured out the big bang. :-) | |
| Nov 22, 2013 at 0:34 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |