**EDIT**: The definition of a Suslin measurable set I wrote here is incorrect. It should be that $\mathcal{S}$ contains the [*field*](https://en.wikipedia.org/wiki/Field_of_sets) (or algebra) of open subsets of ${}^\omega\omega$ (or, in other words, it contains all open and closed subsets of ${}^\omega\omega$). I do not intend to re-ask this question, so the question and answers here refer to the unchanged definition below.

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Recall that a *Suslin scheme* is a set of subsets of reals ${}^\omega\omega$ of the form:
$$
\langle X_s : s \in {}^{<\omega}\omega\rangle
$$
and that the *Suslin operation* $\mathcal{A}$ is an operation that takes a Suslin scheme $\mathcal{X} := \langle X_s : s \in {}^{<\omega}\omega\rangle$ and yield:
$$
\mathcal{A}(\mathcal{X}) := \bigcap_{a \in {}^\omega\omega}\bigcup_{n \in \omega} X_{a\upharpoonright n}
$$
The set of all **Suslin measurable sets**, call it $\mathcal{S}$, is the smallest set of subsets of reals such that:

- $\mathcal{S}$ contains all open subsets of ${}^\omega\omega$.
- $\mathcal{S}$ is closed under the Suslin operation (i.e. if $\mathcal{X}$ is a Suslin scheme in which $X_s \in \mathcal{S}$ for all $s$, then $\mathcal{A}(\mathcal{X}) \in \mathcal{S}$).

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A result of Nikodym says that the set of Baire subsets of reals is closed under the Suslin operation (Corollary 4.8 of Todorcevic's *Introduction to Ramsey spaces*). Thus, every Suslin measurable subset of reals has the Baire property. The questions I have are:

1. Can we prove, in $\mathsf{ZFC}$ and $\mathsf{ZF}$, that there exists a Baire subset of reals that is not Suslin measurable?

2. If it is not provable in $\mathsf{ZF}$, is there a well-known model of set theory in which every subset of reals is Suslin measurable?

(Side question: There seems to be very little literature that discusses Suslin measurable sets. Is there another term for such sets?)

**EDIT**: To clarify, a subset $X$ of real is Baire (or has the Baire property) if $X = U \, \triangle \, M$, where $U$ is open and $M$ is meagre.