Baire category theorem for uncountable unions

Question

Any compact Hausdorff space $X$ is a Baire space: if the set $X$ is a meager set (meaning a countable union of nowhere dense subsets, also known as a set of first category), then $X$ is empty.

I am interested in analogues of this theorem for uncountable unions.

Specifically, suppose a compact Hausdorff space $X$ is partitioned into a disjoint family $\{Y_i\}_{i∈I}$ of nowhere dense subsets. To exclude trivial counterexamples such as partitions into singleton subsets, assume that for any subset $J⊂I$ the union $⋃_{i∈J}U_i$ is a set with the Baire property (meaning it is the symmetric difference of an open set and a meager set).

If $I$ is countable, then the condition involving the Baire property is trivially satisfied. Furthermore, any countable collection of nowhere dense subsets can be easily adjusted to a countable disjoint collection of nowhere dense subsets with the same union by replacing $Y_i$ with $Y_i∖⋃_{j<i}Y_j$. Thus, the above assumption is indeed an analogue for uncountable unions of the assumption of the Baire category theorem.

Under what additional conditions on $X$ (if any) can we conclude that $X$ is empty?

If additional assumptions are necessary, I am specifically interested in cases when $X$ is extremally disconnected or even hyperstonean.

I do not want to impose any countability (or cardinality) assumptions on $X$, e.g., being metrizable, separable, first countable, etc., as done (for example) in a related question about partitioning of Polish spaces. I also do not want to impose any cardinality assumptions on $I$, as is done in a related question about Baire spaces for higher cardinalities.

In fact, for hyperstonean spaces the answer is positive if we assume the nonexistence of real-valued-measurable cardinals (see Lemma 438B in Fremlin's Measure Theory, which proves a more general result), which can be seen as evidence in favor of the positive answer to the above question. The question is then whether the large cardinal hypothesis can be removed if we assume $X$ to be compact and Hausdorff, and if necessary, extremally disconnected or hyperstonean.

Answer 1

The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire property whose nullsets are exactly the meagre sets and which is inner regular with respect to compact subsets. Therefore $(X, \mathcal{BP}(X), \mu)$ (where $\newcommand{\BP}{\mathcal{BP}}\BP(X)$ is the $\sigma$-algebra of sets with the Baire property) is a compact semi-finite measure space, so we can apply Fremlin's Lemma 451Q. Specialized to this case, this states that if $(E_i)_{i \in I}$ is a pairwise disjoint family of sets in $\BP(X)$ such that for each $J \subseteq I$ we have $\bigcup_{i \in J}E_i \in \BP(X)$, then $\mu\left(\bigcup_{i \in I}E_i\right) = \sum_{i \in I}\mu(E_i)$. In particular, if each $E_i$ is a meagre set, (so $\mu(E_i) = 0$ for all $i \in I$), then $\bigcup_{i \in I}E_i$ is meagre.

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[Removed wrong suggestion for a different proof.]

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In the absence of the axiom of choice, it is consistent that there is a counterexample to the question for compact Hausdorff spaces. An example is given by the partition of $[0,1]$ into singletons in Shelah's model where all subsets of $\mathbb{R}$ have the Baire property.