Timeline for How probability-rich is the $\sigma$-algebra generated by a sequence of sets? (Sierpiński's theorem on non-atomic measures without using the AoC.)
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| May 6 at 19:44 | vote | accept | Iosif Pinelis | ||
| May 6 at 13:24 | answer | added | Michael Greinecker | timeline score: 5 | |
| May 5 at 20:11 | comment | added | Michael Greinecker | I'll try my modest abilities tomorrow. | |
| May 5 at 18:10 | comment | added | Iosif Pinelis | @MichaelGreinecker : Great! Since I have no expertise in mathematical logic and you seem to have it, would you mind summarizing the previous discussion in a formal answer, perhaps supplying some details and/or references (which might also be of interest to other users)? | |
| May 5 at 17:58 | comment | added | Michael Greinecker | Yes, you are really just picking one maximal probability atom of the other to get the purely atomic part. | |
| May 5 at 17:41 | comment | added | Iosif Pinelis | @MichaelGreinecker : Do you think DC is enough for Step (i)? | |
| May 5 at 17:38 | comment | added | Iosif Pinelis | @MichaelGreinecker : Here it was just shown how the closedness of the range of a purely atomic probability measure reduces to the sequential compactness. | |
| May 5 at 17:33 | comment | added | Michael Greinecker | I see. That is basically the sequential compactness of $\{0,1\}^\mathbb{N}$, which does follow from DC. | |
| May 5 at 17:27 | comment | added | Iosif Pinelis | Previous comment continued: Using the diagonal choice argument, we get $c_{n_k,i}\to c_i$ as $k\to\infty$ for some strictly increasing sequence $(n_k)$ of natural numbers, some $c_i$'s in $\{0,1\}$, and all $i$, whence $p=\lim_{k\to\infty}P(A_{n_k})=\sum_{1\le i<m}c_i P(C_i)=P(\bigcup_{i\in I,c_i=1}C_i)$, so that $p$ is in the range of $P_a$. | |
| May 5 at 17:27 | comment | added | Iosif Pinelis | @MichaelGreinecker : Let $(C_i)_{i\in I}$ be a family of distinct (and hence wlog pairwise disjoint) atoms of the purely atomic probability measure $P_a$ such that $\bigcup_{i\in I}C_i=\Omega$, so that $I$ is at most countable and hence wlog $I$ the set of all integers in the set $[1,m]$ for some $m\in\{1,\dots,\infty\}$. Suppose $P_a(A_n)\to p$. For each $n$ and some $c_{n,i}$'s in $\{0,1\}$, $P(A_n)=\sum_{1\le i<m}c_{n,i}P(C_i)$. | |
| May 5 at 17:07 | comment | added | Michael Greinecker | In that case, I simply don't understand what the diagonal choice argument is. | |
| May 5 at 16:24 | comment | added | Iosif Pinelis | Previous comment continued: We know that DC is enough for Steps (ii) and (iii). If DC is enough for Step 1 (which I hope it is), then DC is enough for the highlighted result. | |
| May 5 at 16:24 | comment | added | Iosif Pinelis | @MichaelGreinecker : If I am not mistaken, we can deduce the closedness of the range of a probability measure $P$ from Sierpiński's theorem using the following steps. (i) Represent $P$ as $(1-t)P_a+tP_{na}$, where $t\in[0,1]$, $P_a$ is a purely atomic probability measure, and $P_{na}$ is a non-atomic probability measure. (ii) The range of $P_{na}$ is $[0,1]$, by Sierpiński's theorem. (iii) The range of $P_a$ is closed, by the diagonal choice argument. So, by (i)--(iii), the range of $P$ is closed, and hence the highlighted result follows. | |
| May 5 at 15:48 | comment | added | Michael Greinecker | I'm not sure how the argument works, but if you mean the kind of argument that constructs a convergent subsequence from a sequence of subsequences, that does work with DC. | |
| May 5 at 14:37 | comment | added | Iosif Pinelis | @MichaelGreinecker : Thank you for your comment. Now the highlighted result should follow, using the diagonal choice argument, right? And for the diagonal choice argument, DC is enough, right? | |
| May 5 at 14:23 | comment | added | Michael Greinecker | At least concerning the linked result of Sierpiński, there is a DC-only proof at mathoverflow.net/a/225689/35357 | |
| May 5 at 14:00 | comment | added | Joel David Hamkins | I was imagining that one would find the desired set by choosing things successively in a countable sequence, so as to get close to $p$, as desired. Such an argument would be a proof from DC, but I don't yet have a proof. I'm not sure which way it will go. | |
| May 5 at 13:56 | comment | added | Iosif Pinelis | @JoelDavidHamkins : Thank you for your comment. To my eye, untrained in mathematical logic, DC looks like "sure, how can it be otherwise?". :-) On the other hand, I guess the highlighted statement cannot proved without DC -- what is your opinion/intuition (or more than that?) about this? Anyhow, I would accept a proof using only DC (hopefully I would be able to understand it). | |
| May 5 at 13:37 | comment | added | Joel David Hamkins | Oftentimes with measure and probability, one seeks to eliminate the axiom of choice, but keep dependent choice (and hence also countable choice), since DC enables an adequate theory of measure. But is that what you would allow here? It seems likely one could answer positively with DC. | |
| May 5 at 12:56 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 5 at 12:50 | history | edited | Iosif Pinelis |
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| May 5 at 12:44 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 5 at 12:37 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 5 at 12:13 | history | asked | Iosif Pinelis | CC BY-SA 4.0 |