My question is a bit vague, hopefully someone can still clarify. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and assume that $\mathcal F$ is countably generated. My question is, does there exists a Theorem asserting that there exists a $\sigma$-algebra $\mathcal E\subseteq\mathcal F$ which is generated by a countable partition and such that $\mathbb P$ on $\mathcal F$ can be well approximated by elements in $\mathcal E$? Related references are here Countably generated versus being generated by a countable partition and here Approximating a $\sigma$-algebra by a generating algebra. What I have in mind is for example that for every $F\in\mathcal F$ and every $\varepsilon>0$ there exists $E\in\mathcal E$ with $\mathbb P(F\Delta E)<\varepsilon.$ I am generally looking for a strategy to prove some property, which holds for $\sigma$-algebras generated by countably partition, for countably generated $\sigma$-algebra…
After the answer of Michael Greinecker, I realized that what I am really looking for is a Theorem stating something similar to what follows: Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space as above. Let $\varepsilon>0.$ Does there exists a $\sigma$-algebra $\mathcal E_\varepsilon$ which is generated by a countable partition and such that for every $F\in\mathcal F$ there exists $E\in\mathcal E$ with $$\mathbb P(F\Delta E)<\varepsilon?$$
Best Answer
Let $(\Omega,\mathcal{F},\mu)$ be an atomless probability space and let $\mathcal{P}=\{P_0,P_1,\ldots\}$ be a countable measurable partition of $\Omega$. Let $A_n$ be a measurable subset of $P_n$ with half the probability of $P_n$ for each $n$. Let $A=\bigcup_n A_n$. Let $B$ be in the $\sigma$-algebra generated by $\mathcal{P}$. There is $I\subseteq\mathbb{N}$ such that $B=\bigcup_{n\in I}P_n$. We have $$A\triangle B=A\backslash B\cup B\backslash A=A\Big\backslash \bigcup_{n\in I}P_n\cup \bigcup_{n\in I}P_n\Big\backslash A=\bigcup_{n\in\mathbb{N}}A_n\Big\backslash \bigcup_{n\in I}P_n\cup \bigcup_{n\in I}P_n\Big\backslash\bigcup_{n\in\mathbb{N}}A_n$$ $$=\bigcup_{n\in\mathbb{N}\backslash I}A_n\cup \bigcup_{n\in I}P_n\backslash A_n.$$ This allows us to write $A\triangle B$ as the countable union of disjoint sets. Now $$\mu(A\triangle B)=\mu\bigg( \bigcup_{n\in\mathbb{N}\backslash I}A_n\cup \bigcup_{n\in I}P_n\backslash A_n\bigg)=\sum_{n\in\mathbb{N}\backslash I}\mu(A_n)+\sum_{n\in I}\mu(P_n/A_n)$$ $$=\sum_{n\in\mathbb{N}\backslash I}\mu(P_n)/2+\sum_{n\in I}\mu(P_n)/2=1/2 \sum_{n\in\mathbb{N}}\mu(P_n)=1/2.$$ Since $B$ was arbitrary, we see that the conjecture does not hold.