I understand from Is projection of a measurable subset in product $\sigma$-algebra onto a component space measurable? that if $A\in\mathcal{G}\otimes\mathcal{F}$ and $\pi_1(A)=\{x:\exists y, (x,y)\in A\}$. Then $\pi_1(A)$ is not necessarily a member of $\mathcal{G}$. However, I am curious about when I can make this inference.
To give more background to my question: Let $(\Omega, \mathcal{F})$ be the Cantor space equipped with the Borel sigma-algebra (generated by sets of sequences that agree on finitely many indices). Let $X_i$ be the projection map of the i-th coordinate, and $\mathcal{T}$ the tail-sigma-field $\bigcap_n\mathcal{G}_n$, with $\mathcal{G}_n=\sigma\{X_{n+1}, X_{n+2}, \dots\}$. We know that the atoms of $\mathcal{T}$ are sets of sequences that only agree on finitely many indices. Let $A$ be an arbitrary union of the atoms of $\mathcal{T}$. I want to verify the following: if $A\in\mathcal{F}$, then $A\in\mathcal{T}$ (equivalently, $A\in \sigma\{X_n, X_{n+1}, \dots\}$ for all $n$). The strategy that I have come up with so far is to let $\pi_n(A)=\{y: \exists x\in \{0,1\}^n, (x,y)\in A\}$. Observe that $A=\pi_n(A)$. I want to show that $\pi_n(A)\in\mathcal{G}_n$, which then implies $A\in\mathcal{G}_n$.
