I am interested in knowing wether the following statement is true of false.
Let $ (\Omega, \Sigma , \mathbb{P})$ be a probability space and $\mathcal{A}, \mathcal{B} \subseteq \Sigma$ independant subsets. Then $\sigma A(\mathcal{A})$ and $\sigma A(\mathcal{B})$ are independent sigma algebras.
I think the statement is true but haven't been able to finish the proof. What I've done thus far is the following.
First define $\{M \in \Sigma : \forall A \in \mathcal{A} \text{ it holds } A, M \text{ independent}\}$ and I claim that it's a sigma algebra.
Assuming that's true and since it contains $\mathcal{B}$. This implies that it must contain $\Sigma_2 = \sigma A(\mathcal{B})$.
We then define the set $\{M \in \Sigma : \forall B \in \Sigma_2 \text{ it holds } B, M \text{ independent}\}$ Then clearly by the last part this set contains $\mathcal{A}$.
I think here (were we able to probe that the first set is a sigma algebra) we should be able to recycle the argument to show that this new set is also a sigma algebra. This would imply
$$\Sigma_1 \subseteq \{M \in \Sigma : \forall B \in \Sigma_2 \text{ it holds } B, M \text{ independent}\}$$
where $\Sigma_1 = \sigma A(\mathcal{A}) $ so we would be done.
Now I just have to show that $\mathcal{O}=\{M \in \Sigma : \forall A \in \mathcal{A} \text{ it holds } A, M \text{ independent}\}$ is a sigma algebra. But I have been unable to do so. The part that I'm having trouble with is the countable union part.
UPDATE: So as someone in the comments suggested if we restrict the case to only showing that the countable union of pairwise disjoint elements $M_n$ is still in $\mathcal{O}$ we get the result directly since.
\begin{align} \mathbb{P}(A \cap (\cup_{i=1}^\infty M_n)) &= \mathbb{P}( \cup_{i=1}^\infty (A \cap M_n))\\
&= \sum_{i=1}^\infty \mathbb{P}(A \cap M_n)\\
&=\sum_{i=1}^\infty \mathbb{P}(A)\mathbb{P}(M_n)\\
&=\mathbb{P}(A)\mathbb{P}(\cup_{i=1}^\infty M_n)
\end{align}
However I was still unable to show that it's possible to restrict the general case to that case. I tried using induction on the sequence $M_n' = M_n \setminus (\cup_{i=1}^{n-1} M_i')$ but failed.
Best Answer
It is not true. Consider two independent coin flips. Let $\Omega = \{ HH, HT, TH, TT\}$. Let $A = \{HH, TT\}$ and let $\mathcal{A} = \{A\}$.
Let $B_1 = \{HH, HT\}$, $B_2 = \{HT, TT\}$ and $\mathcal{B} = \{B_1, B_2\}$.
Observe that the sets in $\mathcal{A}$ are independent of the sets in $\mathcal{B}$ but their sigma-algebras are not independent. In particular, $A$ is not independent of $B_1 \cup B_2$.