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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?

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  • $\begingroup$ with $\sigma$ you mean the generated sigma algebra? $\endgroup$
    – Exodd
    Commented Oct 18, 2022 at 11:34
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    $\begingroup$ Well $\sigma(\mathcal A)$ is a $\sigma$-algebra containing $\mathcal B$ so by definition $\sigma(\mathcal B)\subset\sigma(\mathcal A)$ $\endgroup$
    – maxjw91
    Commented Oct 18, 2022 at 11:35

2 Answers 2

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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.

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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 $B_i$.

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