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Clement Yung
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A Baire subset of reals that is not Suslin measurable

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: $$ \mathcal{A}(\mathcal{X}) := \bigcap_{a \in {}^\omega\omega}\bigcup_{n \in \omega} X_s $$ 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 $\mathcal{A}(\mathcal{X}) \in \mathcal{S}$).

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.

Clement Yung
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