Timeline for Finitely additive measure over integers [duplicate]
Current License: CC BY-SA 3.0
14 events
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| Jun 10, 2015 at 7:49 | history | closed |
Nate Eldredge Asaf Karagila♦ Johannes Hahn Joonas Ilmavirta Alex Degtyarev |
Duplicate of How to construct a continuous finite additive measure on the natural numbers | |
| Jun 9, 2015 at 22:30 | review | Close votes | |||
| Jun 10, 2015 at 7:49 | |||||
| Jun 9, 2015 at 22:15 | comment | added | Nate Eldredge | @AsafKaragila: Yes, thanks for finding that. Clinton's comment on Stefan's answer sketches (without needing functional analysis) why the existence of such a measure is inconsistent with BP, and as I mentioned, we know from Shelah that ZF+DC+BP is consistent. So that resolves the question at hand. | |
| Jun 9, 2015 at 20:36 | comment | added | Goldstern | A finitely additive probability measure on the reals (still not good enough for an answer): $\mu(A)= \sum_{n\in A\cap \{1,2,\ldots \}} \frac 1{2^n}$. | |
| Jun 9, 2015 at 20:30 | comment | added | Nate Eldredge | Yes, I believe this is consistent. It's consistent with ZF+DC that every set of reals has the Baire property (Solovay / Shelah), and as I recall, under these axioms, you can prove that $\ell^1(\mathbb{N})$ is reflexive. A (strictly) finitely additive measure on $\mathbb{N}$ would give you a continuous linear functional in $(\ell^\infty)^* \setminus \ell^1$, contradicting reflexivity. Unless someone else beats me to it, I will post an answer with details when I have a chance. | |
| Jun 9, 2015 at 20:11 | comment | added | Logica | The context in which I ask this question is de Finetti's subjective probability theory, where he takes that personal probabilities should be finitely additive and there is no need to impose the stronger condition of countable additivity. I was trying to understand how much set theory he needs at the foundational level for his conviction. | |
| Jun 9, 2015 at 20:04 | comment | added | Christian Remling | You probably meant to ask about the existence of a finitely additive measure that is NOT $\sigma$-additive. | |
| Jun 9, 2015 at 20:04 | comment | added | Logica | Thanks, this is very nice. The next question is this. WITHOUT AC, whether or not it can be shown that there does not exists any nontrivial FINITELY additive measure defined over all subsets of the reals? Is there a result showing that ZF + {nonexistence of finitely additive measure over all subsets of reals} is consistent? | |
| Jun 9, 2015 at 19:54 | comment | added | Goldstern | $\mu(A) = \sum_{n\in A } \frac1{2^n}$ for all $A\subseteq \{1,2,3,4,\ldots\}$. | |
| Jun 9, 2015 at 19:51 | comment | added | Logica | Yes, finitely additive probability measure. | |
| Jun 9, 2015 at 19:46 | comment | added | Goldstern | The counting measure is finitely additive. Do you mean a probability measure? Do you mean a measure that assigns 0 to each finite set? | |
| Jun 9, 2015 at 19:38 | history | edited | Logica | CC BY-SA 3.0 |
added 8 characters in body
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| Jun 9, 2015 at 19:32 | review | First posts | |||
| Jun 9, 2015 at 19:40 | |||||
| Jun 9, 2015 at 19:30 | history | asked | Logica | CC BY-SA 3.0 |