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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:

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


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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The question in the title does not match the question in the body (as a matter of fact, the answer to the question in the title is NO). Please consider editing. – Gautam Shenoy Nov 25 '12 at 17:15
Hi @GautamShenoy, you are right, I didn't know how to write a title. I wanted to have something correct but brief and this is what I came up with... – Sam Nov 25 '12 at 17:17
You are using $A$ too much. – Asaf Karagila Nov 25 '12 at 17:18
The set is actually script A ! Please see the edits! – Sam Nov 25 '12 at 17:19
In fact, $F_1\cup F_2\subset\mathcal A$, not (in general) the other way. – Did Nov 25 '12 at 17:50
up vote 2 down vote accepted

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. – Did Nov 25 '12 at 18:38

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