Let $X$ be a Polish space. Recall that the Suslin operation is the operation $\mathcal{A}$ such that for any Suslin scheme $\{A_s : s \in \omega^{<\omega}\}$ of subsets of $X$, we have: $$ \mathcal{A}\{A_s : s \in \omega^{<\omega}\} := \bigcup_{a \in \omega^\omega} \bigcap_{n<\omega} A_{a\upharpoonright n} $$ The $\sigma$-algebra of Suslin measurable $\mathcal{S}$ subsets of $X$ is the smallest $\sigma$-algebra, containing all open subsets of $X$, such that $\mathcal{S}$ is closed under the Suslin operation. Since every analytic set is the application of the Suslin operation to a Suslin scheme of closed subsets of $X$, $\mathcal{S}$ contains all analytic subsets of $X$.
Is $\mathcal{S}$ exactly the smallest $\sigma$-algebra containing all analytic subsets of $X$?
