# Spaces with atomless independent $\sigma$-sub-algebras

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 $$[0,1]$$.

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.
• @RabeeTourky The question asks only about the case with $\mathcal{X}$ being countably generated. Mar 25 at 18:54