# If $(\mathcal F_1,\mathcal F_2,\mathcal F_3)$ is independent, is $\mathcal F_1\vee\mathcal F_2$ independent of $\mathcal F_3$?

Let $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space and $$\mathcal F_i\subseteq\mathcal A$$ be a $$\sigma$$-algebra on $$\Omega$$.

If $$(\mathcal F_1,\mathcal F_2,\mathcal F_3)$$ is independent, does it follow that $$\mathcal F_1\vee\mathcal F_2:=\sigma(\mathcal F_1\cup\mathcal F_2)$$ is independent of $$\mathcal F_3$$?

In the special case, where $$\mathcal F_i=\sigma(X_i)$$ for some random variable $$X_i$$ taking values in a measurable space $$(E_i,\mathcal E_i)$$, we've got $$\mathcal F_1\vee\mathcal F_2=(X_1,X_2)^{-1}(\mathcal E_1\otimes\mathcal E_2)=(X_1,X_2)^{-1}(\sigma(\mathcal G_1\times\mathcal G_2)\tag1,$$ where $$\mathcal G_i\subseteq\mathcal E_i$$ is arbitrary with $$\sigma(\mathcal G_i)=\mathcal E_i)$$. From $$(1)$$ the desired claim immediately follows.

So, I wondered whether the same holds in general or why it breaks down.

• Yes; you already received a proof below. Note also that much more is true: see this answer of mine for the general theorem. Jan 3, 2022 at 14:18
• @peek-a-boo Thank you for your comment. One of the things I actually used to know, but forgot. Note that your Theorem 1 is even an equivalence (and you only need that $\mathcal P_i\cup\{\emptyset\}$ is a $\pi$-system). Jan 3, 2022 at 15:32

Sets of the form $$A \cap B$$ with $$A \in \mathcal F_1, B \in \mathcal F_2$$ form a $$\pi$$ system which generates $$\mathcal F_1 \vee \mathcal F_2$$. The equation $$P(C\cap D) =P(C)P(D)$$ holds if $$C$$ is above type and $$D \in \mathcal F_3$$. Apply Dymkin's $$\pi -\lambda$$ Theorem (with $$D \in \mathcal F_3$$ fixed) to finish.