When comparing two sub-$\sigma$-algebras on a probability space $(\Omega,\Sigma,\pi)$, say $\mathcal{X}$ and $\mathcal{Y}$, say that $\mathcal{X}$ is strictly coarser than $\mathcal{Y}$ if the completion of $\mathcal{X}$ does not contain $\mathcal{Y}$. Here completion always refers to the restriction of $\pi$. Do there exist probability spaces $(\Omega,\Sigma,\pi)$ satisfying the following property?

  • For any countably-generated sub-$\sigma$-algebra $\mathcal{X}$ strictly coarser than $\Sigma$, and containing a set of interior measure (strictly between $0$ and $1$), there exists an atomless sub-$\sigma$-algebra $\mathcal{U}\subseteq\Sigma$, independent of $\mathcal{X}$.

Furthermore, does this imply the following property?

  • For every $\mathcal{X}$ as above, there exists an independent sub-$\sigma$-algebra $\mathcal{U}\subseteq\Sigma$ independent of $\mathcal{X}$ such that the completion of $\mathcal{X}\vee\mathcal{U}$ contains $\Sigma$.

I know the first can't hold in $[0,1]$ equipped with the Borel measure from Ramachandran (1979), although it does hold in that case when $X$ is restricted to be generated by a countable partition of $[0,1]$.


1 Answer 1


The answer to the first question is yes. There is a class of probability spaces known under various names such as superatomless, saturated, nowhere countably-generated, $\aleph_1$-atomless, and a couple of other names that have exactly this property. Note that the restriction to sub-$\sigma$-algebras admitting sets of interior measure is superfluous. Otherwise, the $\sigma$-algebra $\Sigma$ would be trivially independent of it. This paper might be a good entry to the topic.

A typical example of such a probability space would be the independent product measure on uncountably many copies of the unit interval endowed with the uniform distribution. It is the canonical example in a special sense. If you take a probability space $(\Omega,\Sigma,\pi)$ and identify two measurable sets $A$ and $B$ with $\pi(A\Delta B)=0$, you obtain the so-called measure algebra. There is a fundamental theorem due to Dorothy Maharam that for every probability space, the measure algebra is a countable weighted sum of product measures obtained from $[0,1].$ I think one might also be able to use this to prove that the second part of the question holds true.

  • $\begingroup$ Nice example. My intuition is that the second part is probably false, and I am wondering if one of the examples from mathoverflow.net/questions/54033/… could be adapted. $\endgroup$ Mar 25 at 14:21
  • $\begingroup$ @NateEldredge In the example, the problem is that there are not enough independent events. On the level of measure algebras, the second question has a positive answer. This follows from results on factor spaces in volume 3 of Fremlin's magnus opus on measure theory. If there are problems, they should have to do with managing null sets. $\endgroup$ Mar 25 at 14:39
  • $\begingroup$ Regarding the second question, there are complete strict Sigma-sub-algebra (not countably generated) that do not have independent none trivial events. Or is the claim that there are none? $\endgroup$ Mar 25 at 18:07
  • $\begingroup$ @RabeeTourky The question asks only about the case with $\mathcal{X}$ being countably generated. $\endgroup$ Mar 25 at 18:54
  • $\begingroup$ Yes I know. The question is closely related. Take a maximal U independent of X. Must the completion of U contain Sigma? it would be odd intuitively, and that is equivalent to my question. So if your conjecture is correct. The second question becomes, there is U independent of X whose completion contains all information Sigma $\endgroup$ Mar 25 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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