Let $(\mathcal{F}_n)_{n\geq0}$ be a filtration, i.e. $\mathcal{F}_0 \subset \mathcal{F}_1 \subset \dots $ and define the join of the sigma-algebras (filtration) $\mathcal{F}_\infty := \bigvee_{n\geq 0} \mathcal{F}_n$.
My question is whether the following is true: If $A \in \mathcal{F}_\infty$, then there exists a $n \geq 0$ such that $A \in \mathcal{F}_n$.
Certainly not. On $[0,1)$ let $\mathcal F_n$ be the sigma algebra generated by the partition $\{[\frac {i-1} {2^{n}},\frac i {2^{n}}): 1\leq i <2^{n}\}$. Then the set of all rational numbers is in $\mathcal F_{\infty}$ but not in any of the $\mathcal F_n$'s.