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?