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Assume that $F_1$ and $F_2$ are two independent sigma fields. We know that union of $F_1$ and $F_2$ is not necessarily a sigma-field. Suppose we define $ \mathcal{A} = \{A \cap B: A\in F_1, B\in F_2\} $. How to show that:

$\sigma(\mathcal{A}) = \sigma(F_1 \cup F_2) $

Thanks,

Here is how I thought about it. I divide the proof into two parts:

1) $\sigma(\mathcal{A}) \subset \sigma(F_1 \cup F_2)$

2) $\sigma(F_1 \cup F_2) \subset \sigma(\mathcal{A})$

I guess, part 1 is easy since $\mathcal{A} \subset (F_1 \cup F_2)$ . Right? How about part 2?

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    Did you follow the hint I gave you in a comment?2012-11-29

1 Answers 1

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Consider the sigma-algebras generated by each class of subsets involved in the double inclusion $ F_1\cup F_2\subseteq\mathcal A\subseteq\sigma(F_1\cup F_2).$

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    The relevance of the *independence* of $F_1$ and $F_2$ is mysterious to me in the context of this question.2012-11-25