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 $[0,1]$.