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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 $[0,1]$.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented Mar 25, 2022 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. $\endgroup$
    – Michael Greinecker
    Commented Mar 25, 2022 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? $\endgroup$
    – Rabee Tourky
    Commented Mar 25, 2022 at 18:07
  • $\begingroup$ @RabeeTourky The question asks only about the case with $\mathcal{X}$ being countably generated. $\endgroup$
    – Michael Greinecker
    Commented Mar 25, 2022 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 $\endgroup$
    – Rabee Tourky
    Commented Mar 25, 2022 at 19:00

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