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Possible Duplicate:
Continuous Strictly Positive Measures on Countable Boolean Algebras

Let B be the Boolean Algebra of sentences in the language of set theory modulo provable equivalence in ZFC. That is to say, for each sentence $\sigma$ in the language of set theory, let $[\sigma]$ be the set of sentences in the language of set theory provably equivalent to $\sigma$. The set of equivalence classes $[\sigma]$ under the obvious operations is the Boolean Algebra in which we are interested.

Is it possible for there to be a countably additive probability measure $\mu$ on $B$ such that $\mu$ assigns every non-zero element of $B$ strictly positive probability? Must there be such a probability measure $\mu$? (Note: I added the condition of countable additivity when some commenters correctly pointed out that every countable Boolean algebra admits a strictly positive probability measure.)

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  • $\begingroup$ Is this close to what you want - mathoverflow.net/questions/99808/… $\endgroup$ Jul 24, 2012 at 2:44
  • $\begingroup$ By the way, Dana Scott's latest work is about probabilistic models of set theory. It might be relevant to what you are trying to do. $\endgroup$ Jul 24, 2012 at 5:07
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    $\begingroup$ I don't understand Qiaochu's comment. Yes, the (Lindenbaum-Tarski) algebra of sentences in the language of set theory modulo equivalence in ZFC is countable and therefore supports a strictly positive probability measure. But there are uncountable Boolean algebras with such a measure, the prime example being the measurable subsets of $[0,1]$ modulo the measure 0 sets. $\endgroup$ Jul 24, 2012 at 6:23
  • $\begingroup$ To Dorais and Qioachu - I suppose I should have made it more explicit, but I want it to be a countably additive probability measure. I don't think every countable Boolean algebra necessarily supports a countably additive, strictly positive countable measure. But it is my fault for not making this condition clear. $\endgroup$
    – idiot
    Jul 24, 2012 at 11:13
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    $\begingroup$ Then the answer is no, there is no countably additive measure on the atomless countable Boolean algebra, see mathoverflow.net/questions/100724 . $\endgroup$ Jul 24, 2012 at 11:44

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This Boolean algebra is atomless (assuming ZFC is consistent). Namely, let $\varphi$ be a sentence whose equivalence class is $\not=0$. Then ZFC+$\varphi$ is consistent. By the incompleteness theorem, ZFC+$\varphi$+$\neg$Con(ZFC+$\varphi$) is consistent as well. It follows that the equivalence classes of $\varphi\wedge\neg\mbox{Con(ZFC}+\varphi$) and $\varphi\wedge\mbox{Con(ZFC}+\varphi$) are both below the one of $\varphi$ and both positive.

Luckily, there is just one countable atomless Boolean algebra, namely the algebra of clopen subsets of the Cantor space ($2^\omega$).
This algebra sits inside the measure algebra of the Cantor space (with the product measure coming from the measure that assigns $1/2$ to the singletons $\{0\}$ and $\{1\}$). So there is your countably additive measure (countably additive with respect to sums that actually exist in the Boolean algebra).

Edit: As Emil Jerabek points out, this is actually not the case. The problem is that the countable BA is not a complete subalgebra of the measure algebra, i.e., there are infinite subsets of the countable BA that have a sup in the countable BA but that sup does not agree with the sup taken in the big measure algebra.

So, it seems this argument only gives a finitely additive measure on this BA.

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    $\begingroup$ This answer is wrong, as the embedding of the algebra in the measure algebra on the Cantor space does not preserve existing countable joins and meets. See mathoverflow.net/questions/100724. $\endgroup$ Jul 24, 2012 at 11:53

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