Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The condition on $N$ seems strange, because you can always choose $E'=\Omega$, at least the way that it is written now. – Lukas Geyer Oct 27 '12 at 2:18
Thanks Lukas. Brain wasn't fully engaged. – Rabee Tourky Oct 27 '12 at 2:43
@Lukas: Either of you could write that up as an answer so the question doesn't remain unanswered. – joriki Oct 27 '12 at 6:19
@MichaelGreinecker: You might be interested in Kelley's criterion (original article here) 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 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 by the OP but I didn't read it closely. – commenter Nov 26 '12 at 18:47
@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 $\Sigma$. – commenter Nov 26 '12 at 20:38
up vote 2 down vote accepted

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.

share|cite|improve this answer
Nice. Glad to see that my suggestion helped even if it was in an unexpected way... – commenter Jan 21 '13 at 21:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.