Let $(E,\mathscr E)$ be a measurable space and $(S,2^S)$ be a finite set. Let $ \xi:(E,\mathscr E)\to(S,2^S) $ be a mesaurable function, i.e. $\xi^{-1}(s)\in \mathscr E$ for any $s\in S$. Now, let us denote by $\Omega = E^{\mathbb N_0}$ and by $\mathscr F$ its product $\sigma$-algebra. Also, let $\Sigma = S^{\mathbb N_0}$ and let $\mathscr S$ be the correspondent product $\sigma$-algebra.
Let $\eta:\Omega\to\Sigma$ be the element-wise extension of $\xi$, i.e. $ \eta(\omega_0,\omega_1,\dots) = (\xi(\omega_0),\xi(\omega_1),\dots). $
I wonder if $\mathscr S$ is different from $ \mathscr C = \{A\subseteq \Sigma:\eta^{-1}(A)\in \mathscr F\}. $ It is clear that $\mathscr S\subseteq \mathscr C$ - but can the strict inclusion actually happen?