# Sigma algebras generated by independent sets of events

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.

• If $M_i$ are pairwise disjoint and independent from each $A\in\mathcal A$, then $P(A\cap\bigcup_i M_i) =P(\bigcup_i(A\cap M_i))=\sum_i P(A\cap M_i) =\sum_i P(A)P(M_i) =P(A)\cdot\sum_i P(M_i) =P(A)\cdot P(\bigcup_i M_i)$. – Berci Apr 9 at 11:49
• Perhaps begin with: For a fixed $A \in \Sigma$, show that $\{B \in \Sigma : \mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B)\}$ is a sigma-algebra. – GEdgar Apr 9 at 12:17