When is the union of $\sigma$-algebras atomless? Suppose that we are given a probability space $(\Omega, \mathcal{F}, \mathsf P)$ and an increasing sequence of $$\mathcal{F}_1\subset \ldots \subset\mathcal{F}_n\subset \mathcal{F}_{n+1} \subset \ldots \subset \mathcal{F}$$
of $\sigma$-algebras. Assume that $\mathsf P|_{\mathcal{F}_n}$ is atomless for each $n$. Is so $\mathsf P$ restricted to the $\sigma$-algebra generated by the union of $\mathcal{F}_n$ ($n\in \mathbb{N}$)?

An atom is an indivisible set in the sigma algebra with positive measure.
Consider $\mathcal{F}_n$ a sigma algebra of subsets of the space $[0,1]^2$.
The construction is a bit artificial. so I will divide it into steps.
1) consider a the sets $[0,1] \cup (n,n+1)\backslash E_n$ as the basic sets of your sigma algebra $\mathcal{F}_n$ ($E_n  \subset [n-1,n+1]$ is a denumerable set)
Note that there are no atoms in $\mathcal{F}_n$
2) set P([0,1]) = 1 
Note that $[0,1]\in \sigma(\mathcal{F}_1 \cup \mathcal{F}_2)$ $[0,1]$ is indivisible and has positive measure. Therefore it is an Atom.
