On a non-empty set $E$, let $(\mathcal{E}_n)$ be an increasing sequence of sigma-algebras, i.e. such that, for every $n \leq m$, $\mathcal{E}_n \subseteq \mathcal{E}_m$. Let us denote by $\mathcal{E}$ its limit, i.e.
$$
\mathcal{E} = \sigma\left(\bigcup_{n\geq 0} \mathcal{E}_n\right)
$$
Is it true that, for each $A\in\mathcal{E}$, there is an increasing sequence $(A_n)$ with $A_n \in \mathcal{E}_n$, $A_n \subseteq A_m$ when $n \leq m$, and $A = \bigcup_{n \ge 0} A_n$?
Increasing sequence of sigma-algebras
analysismeasure-theoryset-theory
Best Answer
It is not true.
Define partitions $\mathcal P_n:=\left\{\left[k\cdot2^{-n},(k+1)\cdot2^{-n}\right)\mid k\in\mathbb Z\right\}$ on $\mathbb R$.
Let $\mathcal E_n$ denote the collection of subsets of $\mathbb R$ that can be written as a union of elements of $\mathcal P_n$.
Then $(\mathcal E_n)_n$ is an increasing sequence of $\sigma$-algebras.
Every singleton is an element of $\sigma(\bigcup_{n=0}^{\infty}\mathcal E_n)$.
This because $\{x\}=\bigcap_{n=0}^{\infty}P_n(x)$ where $P_n(x)$ denotes the unique element of $\mathcal P_n$ that contains $x$ as element.
However, there is no way to write $\{x\}=\bigcup_{n=1}^{\infty} A_n$ where $A_n\in\mathcal E_n$.
This because the $\mathcal E_n$ do not contain singletons.