Timeline for Hyperreal finitely-additive measure on [0,1) assigning $b-a$ to $[a,b)$ or $(a,b]$ and infinitesimals to singletons
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2012 at 16:51 | vote | accept | Alexander Pruss | ||
| Sep 27, 2012 at 15:56 | comment | added | Joel David Hamkins | I've realized that $\mu$ defined like this can never work, since the values $k/N$ are always in $\mathbb{Q}^\ast$, and so will not be equal exactly to $b-a$ as desired on $[a,b)$ when this difference is irrational. | |
| Sep 27, 2012 at 14:57 | answer | added | David Milovich | timeline score: 7 | |
| Sep 27, 2012 at 13:31 | comment | added | Alexander Pruss | A correspondent tells me that if $\mu$ is the Bernstein-Wattenberg measure (i.e., the $\mu$ in the last comment) then we have translation invariance when you translate by $1/n$ (for a natural $n$). So we're going to have $\mu([a,b))=b-a$ at least when $b-a$ is rational. | |
| Sep 27, 2012 at 13:19 | history | edited | Alexander Pruss | CC BY-SA 3.0 |
edited title
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| Sep 27, 2012 at 2:37 | comment | added | Joel David Hamkins | Kevin, that seems highly relevant. Let's fix such a $b$, and then define $\mu(X)$ to be $k/N$, where $b$ has $N$ members and $k$ is the number of members in $X^\ast\cap b$. This is additive and gives measure $\epsilon=1/N$ to singletons. To work correctly on intervals, however, you need that $b$ is somehow sufficiently uniform. Is this achievable? | |
| Sep 27, 2012 at 1:41 | comment | added | Kevin O'Bryant | I can throw some buzzwords around, but I'm out of my depth here. If you take a nonstandard extension that is an enlargement (or polysaturated) then there is a hyperfinite set $b\subseteq *[0,1]$ with $[0,1] \subseteq b$. This feels relevant, but I'm not sure how exactly. | |
| Sep 26, 2012 at 22:36 | comment | added | Michael Albanese | @Alexander: In the title, it should be $(a, b]$ not $(b, a]$. | |
| Sep 26, 2012 at 21:51 | answer | added | Joel David Hamkins | timeline score: 5 | |
| Sep 26, 2012 at 21:07 | comment | added | Alexander Pruss | Actually, now that I think about it, we don't get the "same one" property from 1, unless it does so trivially because there is no measure satisfying 1. For suppose that $\mu$ satisfies 1 and 2. Let $\nu(A) = \mu(A) - \mathrm{st} \mu(A)$ be the infinitesimal part of $\mu$. Let $\rho(A) = \mu(A) + \nu(A \cap [0,1/2)) + 2\nu(A \cap [1/2,1))$. Then $\rho(A)$ satisfies 1 but not 2. | |
| Sep 26, 2012 at 21:02 | comment | added | Alexander Pruss | I don't see it. Could you elaborate? | |
| Sep 26, 2012 at 18:45 | comment | added | Gerald Edgar | We get the "same one" property of 2 from 1. | |
| Sep 26, 2012 at 17:38 | history | edited | Douglas Zare | CC BY-SA 3.0 |
Corrected typo, tried to improve formatting, added NSA tag
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| Sep 26, 2012 at 15:21 | history | asked | Alexander Pruss | CC BY-SA 3.0 |