# Can a countably generated $\sigma$-algebra be generated by a countable partition?

Let $$\mathcal A := (A_n)$$ be a countable collection of subsets of $$X$$. Let $$B_0 := A_0$$ and $$B_{n+1} :=A_{n+1} \setminus \bigcup_{j=0}^n A_j \quad \forall n.$$

Then $$\mathcal B := (B_n)$$ is pairwise disjoint and $$\mathcal B \subset \sigma(\mathcal A)$$. I think we can not recover $$\mathcal A$$ from $$\mathcal B$$, so I guess it is possible that $$\sigma(\mathcal B) \subsetneq \sigma(\mathcal A)$$. However, I could not come up with a counter-example. Could you please provide one such example?

• with $\sigma$ you mean the generated sigma algebra? Commented Oct 18, 2022 at 11:34
• Well $\sigma(\mathcal A)$ is a $\sigma$-algebra containing $\mathcal B$ so by definition $\sigma(\mathcal B)\subset\sigma(\mathcal A)$ Commented Oct 18, 2022 at 11:35

The Borel $$\sigma-$$ algebra of $$\mathbb R$$ is countably generated: Look at intervals with rational end-points. It is not generated by a countable partition. This is because if it is generated by countbaae a partition $$(P_i)$$ then any mesurable real function on it would be constant on each $$P_i$$ , so it would take only countable number of values.
Take $$A_0=\{0,1\}$$, $$A_1=\{1,2\}$$, $$A_n = \{n+1\}$$ for every $$n>1$$. The generated sets $$B_i$$ are $$B_0=\{0,1\}$$, $$B_1=\{2\}$$, $$B_n = \{n+1\}$$ for every $$n>1$$. You can check by yourself that $$\{1\}$$ is in the sigma algebra of the $$A_i$$, but not in the one of the $$B_i$$.