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We know that, with Axiom of Choice (AC), it can be shown that there exists a finitely additive uniform distribution defined for all subsets of the integers (see, e.g., Hrbacek and Jech 1999, Ch. 11).

I was wondering, WITHOUT AC, whether or not there exists any finitely additive measure defined over $\mathcal{P}(\mathbb{N})$? Is there a result showing that ZF + {nonexistence of finitely additive measure over $\mathcal{P}(\mathbb{N})$} is consistent? Thanks!

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    $\begingroup$ The counting measure is finitely additive. Do you mean a probability measure? Do you mean a measure that assigns 0 to each finite set? $\endgroup$
    – Goldstern
    Commented Jun 9, 2015 at 19:46
  • $\begingroup$ Yes, finitely additive probability measure. $\endgroup$
    – Logica
    Commented Jun 9, 2015 at 19:51
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    $\begingroup$ $\mu(A) = \sum_{n\in A } \frac1{2^n}$ for all $A\subseteq \{1,2,3,4,\ldots\}$. $\endgroup$
    – Goldstern
    Commented Jun 9, 2015 at 19:54
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    $\begingroup$ You probably meant to ask about the existence of a finitely additive measure that is NOT $\sigma$-additive. $\endgroup$
    – Christian Remling
    Commented Jun 9, 2015 at 20:04
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    $\begingroup$ @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. $\endgroup$
    – Nate Eldredge
    Commented Jun 9, 2015 at 22:15

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