Can we approximate a sigma-algebra by an increasing sequence of finite sigma-algebras?

Question

$\newcommand{\N}{\mathbb N}$ $\newcommand{\mc}{\mathcal}$ $\newcommand{\set}[1]{\{#1\}}$ Definition. We say that a measure space $(X, \mc F, \mu)$ has a countable basis if there is a collection $\set{E_n}_{n\in \N}$ of measurable subsets of $X$ such that for all $\varepsilon>0$ and for all $E\in \mc F$, there is $n$ such that $\mu(E_n\Delta E)< \varepsilon$.

Question. With notation as in the above definition, is it true that if $\mc G$ denotes the $\sigma$-algebra generated by $\set{E_n}_{n\in \N}$, then for all $F\in \mc F$, there is $G\in \mc G$ such that $\mu(F\Delta G)=0$?

Answer 1

Let $\mu (E_{n_{k}} \Delta E)<\frac 1 {2^{k}}$ for each $k$. Then $\sum_k \mu (E_{n_{k}} \Delta E)<\infty $ so $\mu ( \lim \sup_k (E_{n_{k}} \Delta E))=0$. This implies $\mu (F\Delta E)=0$ where $F=\lim \sup E_{n_{k}}$.

Answer 2

Another way to see the result is the following:

Let $E\in \mathcal F$ be fixed. By hypothesis, we can find a sequence $n_1<n_2< n_3<\cdots$ such that $\chi_{E_{n_k}}\to \chi_E$ in $L^1$ as $k\to \infty$.

As is well-known, we may pass to a subsequence and assume that the convergence occurs pointwise a.e.

But then we have $\lim_{k\to \infty} \chi_{E_{n_k}} = \limsup_{k\to \infty}\chi_{E_{n_k}}\pmod{\mu}$. The latter is in $\mathcal G$ and we are done.