Timeline for Can invariant means be really considered as the generalization of the uniform measure?
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| Aug 24, 2013 at 19:01 | comment | added | Alexander Pruss | It's hard to say. Invariant means are typically non-unique, while uniqueness seems important to my intuitive concept of uniform measure. Also, they tend only to be finitely additive, which rather changes things from the uniform case. (Merely finitely additive probabilities have some rather counterintuitive consequences, mainly due to nonconglomerability.) I have been exploring invariant means as a generalization of uniform measure, but I think how good a generalization they are is up for grabs. | |
| Jun 6, 2012 at 19:47 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
added 63 characters in body
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| Jun 6, 2012 at 19:38 | comment | added | Valerio Capraro | I am not against non-ergodic argument, but in its present form I don't think this is very convincing. Why a mean value defined using an invariant mean is better than a mean value defined used a non-invariant measure? What are properties that definitely force the use of an invariant mean? Concerning the limit of every bounded sequence, I think this is a property of ultrafilters (and not, in general, invariant means), which are one of the least invariant things. Am I misunderstanding anything? | |
| Jun 5, 2012 at 15:01 | comment | added | Yulia Kuznetsova | I can suggest a non-ergodic argument. A finite invariant measure allows you to assign the mean value to every integrable function. An invariant mean assigns a mean value to every bounded function. For example, on Z you get a "limit at infinity" of every bounded sequence. This was the initial motivation of S. Banach, as far as I know. | |
| Jun 5, 2012 at 13:37 | history | edited | Valerio Capraro | CC BY-SA 3.0 |
edited title
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| Jun 5, 2012 at 13:13 | history | asked | Valerio Capraro | CC BY-SA 3.0 |