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

Question

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.

Answer 1

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.