Counterexample for a probability space and two independent family of sets I am having difficulties solving the following task:
Find a counterexample for a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and two nontrivial, independent family of sets $\xi_1,\xi_2 \subset\mathcal{A}$, so that $\sigma(\xi_1)$ and $\sigma(\xi_2)$ are not independent.
Hope someone can help with this task. Thanks in advance.
 A: Let $\Omega= \{ 1, 2, 3, 4 \}$, $\mathcal{A}= 2^{\Omega}$ and $\mathbb{P}$ be define by  $\mathbb{P}(\{1\}) = \mathbb{P}(\{2\})=\mathbb{P}(\{3\}) = \mathbb{P}(\{4\})=\frac{1}{4}$ .
Now let $\xi_1=\{\emptyset, \{1, 2\}, \{2, 3\}\}$ and  $\xi_2=\{\emptyset, \{1, 3\}\}$. It is easy to see that they are independent. In fact:   
If $A=\emptyset$ or $B=\emptyset$, we clearly have 
$$ \mathbb{P}(A \cap B) = \mathbb{P}(\emptyset)= 0 = \mathbb{P}(A)\mathbb{P}(B)$$
And we have 
$$ \mathbb{P}(\{1, 2\}\cap \{1, 3\}) = \mathbb{P}(\{1\}) = \frac{1}{4} = \frac{1}{2}\cdot\frac{1}{2}= \mathbb{P}(\{1, 2\})\mathbb{P}(\{1, 3\}) $$
In a similar way:
$$ \mathbb{P}(\{2,3\}\cap \{1, 3\}) = \mathbb{P}(\{3\}) = \frac{1}{4} = \frac{1}{2}\cdot\frac{1}{2}= \mathbb{P}(\{2,3\})\mathbb{P}(\{1, 3\}) $$
So, for any $A \in \xi_1$ and any $B \in \xi_2$, $A$ is independent of $B$. So $\xi_1$ and $\xi_2$ are independent.
Now, note that $\{2\} \in \sigma(\xi_1)$, and $\{1, 3\} \in \sigma(\xi_2)$, but  $\{2\}$ and $\{1, 3\}$ are not independent. In fact:
$$ \mathbb{P}(\{2\}\cap \{1, 3\}) = \mathbb{P}(\emptyset)=0 \neq  \frac{1}{4}\cdot\frac{1}{2}= \mathbb{P}(\{2\})\mathbb{P}(\{1, 3\}) $$
So $\sigma(\xi_1)$ and $\sigma(\xi_2)$ are not independent.
