I'm trying to find a proof of this lemma but, since (I guess) its pretty elementary I havent been able to find it in any of the probability theory books I've searched. So, I would really really appreciate if someone could tell me where to find one or provide me with one(or goods hints) I've tried defining a monotone class but I couldnt prove that in fact it was a monotone class).
Lemma:
Let $(\Omega,\mathcal{F},\mathbb{P})$ a probability space and $\mathcal{A_{1}},\mathcal{A_{2}} \subset \mathcal{F}$ two independent algebras, i.e, for all $A_{1} \in \mathcal{A_{1}}$ and $A_{2}\in \mathcal{A_{2}}$ we have that $\mathbb{P}(A_{1}\cap A_{2})=\mathbb{P}(A_{1})\mathbb{P}(A_{2})$. Then, $\sigma(\mathcal{A_{1}})$ and $\sigma(\mathcal{A_{2}})$ are independent $\sigma$-algebras.
Thanks so much in advance <3
Best Answer
First define $$\mathscr{D}_1 = \{A_1 \in \sigma(\mathscr{A}_1): \forall A_2 \in \mathscr{A}_2, P(A_1 \cap A_2) = P(A_1)P(A_2) \}.$$
By condition, $\mathscr{A}_1 \subset \mathscr{D}_1$. Also note
As an algebra, $\mathscr{A}_1$ is a $\pi$-system (thus the original condition can be weakened).
$\mathscr{D}_1$ is a $\lambda$-system, by the countable additivity property of probability measures.
Thus by Dynkin's $\pi$-$\lambda$ theorem, we have $\sigma(\mathscr{A}_1) \subset \mathscr{D}_1$ (i.e., $\sigma(\mathscr{A}_1) = \mathscr{D}_1$).
Next define $$\mathscr{D}_2 = \{A_2 \in \sigma(\mathscr{A}_2): \forall A_1 \in \sigma(\mathscr{A}_1), P(A_1 \cap A_2) = P(A_1)P(A_2) \}.$$ By what was shown above, we know $\mathscr{A}_2 \subset \mathscr{D}_2$. Using the similar trick as in Step 1, one can show that $\sigma(\mathscr{A}_2) =\mathscr{D}_2$, which proves your statement.