Timeline for Why do we want maps to be measurable (in countably-additive setting)
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 9, 2013 at 7:18 | history | bounty ended | SBF | ||
| Aug 8, 2013 at 23:01 | comment | added | Alexander Pruss | Oh, and yes, you're right: the maximal measure wouldn't be unique in general. But my above "proof" that the measure "extends somewhat" was wrong. I think what you need to do is to basically force $\mu(A)$ to take some value between the lower measure $\sup_{B\subseteq A, B\in F} \mu(B)$ and the upper measure $\inf_{B\supseteq A, B\in F} \mu(B)$. But I haven't checked the details here (I've gone through the analogous proof in the countably additive case some time ago). | |
| Aug 8, 2013 at 7:15 | comment | added | SBF | Dear @Alexander: thank you - it's indeed a very interesting topic! | |
| Aug 7, 2013 at 23:14 | comment | added | Alexander Pruss | One of the best descriptions of why you want conglomerability is here: Kadane, J.B., Schervish, M.J., and Seidenfeld, T. 1996. “Reasoning to a foregone conclusion”, Journal of the American Statistical Association, 91, 1228–35. | |
| Aug 7, 2013 at 22:31 | comment | added | Alexander Pruss | One classic paper is here: tinyurl.com/conglomerability And there are some philosophical references in this paper of mine: onlinelibrary.wiley.com/doi/10.1002/tht3.13/abstract | |
| Aug 7, 2013 at 6:43 | comment | added | SBF | Interesting, let me try work out this procedure - I guess if exists the maximal measures will not be unique. I never heard of conglomerability before - shall I look in the work by de Finetti for this concept? | |
| Aug 6, 2013 at 17:46 | comment | added | Alexander Pruss | You also want countable additivity to avoid the problems of nonconglomerability. | |
| Aug 6, 2013 at 15:38 | comment | added | Alexander Pruss | The answer to your extension question seems affirmative. (1) Any f.a. measure $\mu$ defined on a proper subalgebra $F$ of the powerset extends somewhat. Take a subset $A$ not in $F$. Any set in the algebra gen. by $A$ and $F$ is of the form $(A\cap B)\cup (A^c\cap C)$ for $B,C\in F$. Let the measure of that subset be $\mu(C)$ (exercise: well-defined). So we have an extension from $F$ to the algebra generated by $F$ and $A$. (2) Define an ordering on all f.a. measures by $\mu\le\nu$ iff $\nu$ extends $\mu$. By Zorn (exercise), there is a maximal measure. By (1), it is defined on the powerset. | |
| Aug 6, 2013 at 15:26 | comment | added | Alexander Pruss | Banach-Tarski shows that you can't always extend a f.a. measure to the powerset while keeping isometric invariance on $\mathbb R^n$ ($n\ge 3$). But if you don't require isometric invariance, you can have finitely additive measures over the whole powerset. In fact, given AC, you can have a translation (but not rotation) invariant f.a. measure on all of $2^{\mathbb R^n}$, normalizing the unit cube: the translation group is Abelian, hence amenable, and so we can apply Tarski's Theorem. | |
| Jul 29, 2013 at 8:10 | comment | added | SBF | Very nice, and thanks. Indeed loosing limit theorems is a strong argument. Regarding finite additivity: from what little I've seen, f.a. measures were initially defined over algebras (thus having a domain of definition somewhat smaller that c.a. measures), and still were related to the measurability. At the same time, the book I mentioned seemed to use f.a. measures defined over the powerset. Is the true that any f.a. measure can be extended from the algebra it is initially defined over to the powerset? I guess, your comment on Banach-Tarski says the answer is "no", am I right? | |
| Jul 26, 2013 at 16:09 | history | answered | Alexander Pruss | CC BY-SA 3.0 |