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?

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*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?

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*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 $[0,1]$.
 A: 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.
