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 $\mathcal{F}_n$ ($n\in \mathbb{N}$)?
 A: Actually, something much stronger is true: If $P$ is a measure on a $\sigma$-algebra $\mathcal{F}$ and $\mathcal{G}\subseteq\mathcal{F}$ is a $\sigma$-subalgebra on which $P$ is atomless and $\sigma$-finite, then $P$ is atomless on all of $\mathcal{F}$.  To show this, suppose $A\in \mathcal{F}$ is an atom.  Then for any $B\in \mathcal{G}$, either $P(A\cap B)=0$ or $P(A\cap B)=P(A)$.  Let $\mathcal{I}$ be the collection of all $B\in\mathcal{G}$ such that $P(A\cap B)=0$.  Then $\mathcal{I}$ is a closed under countable unions, and must contain all $B$ such that $P(B)<P(A)$.  But since $P$ is atomless and $\sigma$-finite on $\mathcal{G}$, every element of $\mathcal{G}$ can be written as a union of countably many sets $B\in\mathcal{G}$ such that $P(B)<P(A)$.  It follows that $\mathcal{I}$ is all of $\mathcal{G}$.  But clearly $\Omega\not\in \mathcal{I}$, so this is a contradiction.
A: 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.
