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Let $\Omega$ be a set and $\Sigma$ be a $\sigma$-algebra of subsets of $\Omega$. Let $N$ be a collection of measurable subsets of $\Sigma$.

Question: What conditions on $\Sigma$ and $N$ guarantee that there exists a non-atomic probability measure $\mu:\Sigma\to [0,1]$ such that for any $E\in \Sigma$ if $\mu(E)=0$, then $E\in N$ ?

Edited to make question coherent.

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    The condition on $N$ seems strange, because you can always choose $E'=\Omega$, at least the way that it is written now.2012-10-27
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    Thanks Lukas. Brain wasn't fully engaged.2012-10-27
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    @Lukas: Either of you could write that up as an answer so the question doesn't remain unanswered.2012-10-27
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    @joriki I edited the question in response to Lukas' question. It was a silly question initially.2012-10-27
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    +1, it's a very interesting question. The title seems misleading though -- the condition in the body prescribes the sets of non-zero measure and doesn't force any set to have zero measure, right?2012-10-27
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    You are right joriki. But I'm not sure of a title that's more descriptive.2012-10-27
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    "Characterizing $\sigma$-filters of positive measure for arbitrary atomless probability spaces"?2012-11-26
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    @MichaelGreinecker: You might be interested in Kelley's criterion ([original article here](http://dx.doi.org/10.2140/pjm.1959.9.1165)) covered in several books (e.g. Fremlin vol 3, ch. 39) as well as Maharam's "control measure problem" which generated some excitement in the past decade due to its [negative resolution by Talagrand](http://arxiv.org/abs/math/0601689) in '06. I don't have the time to go digging any further, but this should give some pointers. As an aside: there was also [this MO thread](http://mathoverflow.net/questions/110972/) by the OP but I didn't read it closely.2012-11-26
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    @commenter Thank you, but results about measure algebras tell us very little about the ideal of null sets that gets crunched to a single element.2012-11-26
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    @Michael: I don't understand. Take a $\sigma$-ideal $J$ included in $N$ and consider $\mathfrak{A} = \Sigma/J$. Every property of $\mathfrak{A}$ *is* a property of how $J$ sits inside $\Sigma$.2012-11-26
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    @commenter Thank you, I get it now.2012-11-26

1 Answers 1

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Thanks Michael Greinecker and commenter.

The main practical problem for me in applying commenter's idea was that the weak $\sigma$-distributive property in Maharam's 1947 paper, in Kelley's paper, and in Todorcevic's amazing paper of 2004 on measure algebras may not hold if we choose an arbitrary $\sigma$-ideal $J$ in $N$ (and it certainly has no clear meaning for what I am doing).

In the end, the best fit for my work was Ryll-Nardzewski's result published in the addendum section of Kelley and not Kelley's result with the distributive property.

1) There exists a sequence $B_n$ of families of subsets of $\Sigma$ such that $(\Sigma\setminus N)\subseteq \bigcup_{n} B_n$.

2) Each $B_n$ has a positive intersection number (as in Kelley).

3) Each $B_n$ is open for increasing sequences; (if $E_m\uparrow E\in B_n$, then eventually $E_m\in B_n$).

The final condition (3) of Nardzewski guarantees that $\Sigma\setminus \bigcup_{n} B_n$ is a $\sigma$-ideal.

Condition (2) guarantees that there is a finitely additive (positive) probability measure $\nu_n$ on $\Sigma$ that is bounded away from zero on $B_n$.

Condition (3) tells us that from $\nu_n$ we can define a countably additive probability measure $\mu_n$ that also measures elements of $B_n$ positively.

Letting $\mu= \sum_{n=1}^\infty 2^{-n} \mu_n$, we have the required measure.

For the converse suppose that $\mu$ is the required measure, letting $B_n=\{ \mu>1/n\}$ we see that (1), (2), and (3); hold.

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    Nice. Glad to see that my suggestion helped even if it was in an unexpected way...2013-01-21