Timeline for Is there a standard way of defining the integral of an extended real function with respect to a finitely additive probability measure?
Current License: CC BY-SA 4.0
13 events
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| Oct 3, 2020 at 6:30 | answer | added | Martin Väth | timeline score: 3 | |
| Oct 3, 2020 at 2:03 | comment | added | aduh | @LSpice Yes, a general construction that works for every (probability) measure. | |
| Oct 3, 2020 at 1:12 | comment | added | LSpice | Are you looking for a construction that, for every measure $\mu$, produces a set $\Phi_\mu$ that includes some unbounded functions? Certainly this can be done for some measures $\mu$. | |
| Oct 2, 2020 at 22:13 | history | edited | aduh | CC BY-SA 4.0 |
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| Oct 2, 2020 at 22:10 | comment | added | aduh | @user95282 No that's not what I mean. I will try to reword it again. Maybe a better way of asking the question is: Is there a notion of integrability satisfying (1)-(3) that includes some unbounded functions? | |
| Oct 2, 2020 at 13:59 | comment | added | user95282 | @aduh Do you mean that $\Phi$ is the set of all functions from $X$ to $\mathbb{R} \cup \{ - \infty, \infty \}$? In that case the expression within the left hand side of (1) is sometimes also of the form $\infty-\infty$. | |
| Sep 30, 2020 at 21:53 | history | edited | aduh | CC BY-SA 4.0 |
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| Sep 30, 2020 at 21:52 | comment | added | aduh | @user95282 Yes, I will edit, thanks. | |
| Sep 30, 2020 at 15:13 | comment | added | user95282 | @aduh Do you also require $\int (-f)d\mu = -\int f d\mu$ ? | |
| Sep 29, 2020 at 21:05 | comment | added | aduh | @GeraldEdgar Right, I shouldn't have said for all functions. I edited a bit. The question is, Is there a notion of integrability that doesn't require the function to be boudned? | |
| Sep 29, 2020 at 21:04 | history | edited | aduh | CC BY-SA 4.0 |
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| Sep 29, 2020 at 13:28 | comment | added | Gerald Edgar | Doing it for all functions involves $\infty - \infty$ calculations. Even for the countably additive case of counting measure on $\mathbb N$, we do not evaluate all series $\sum a_n$. (Equivalent to a probability measure by using $\mu(\{n\}) = 2^{-n}$.) | |
| Sep 29, 2020 at 8:57 | history | asked | aduh | CC BY-SA 4.0 |