I need to show that:
If $(E_n), n\geq 1$ and $(F_n), n\geq 1$ are non-decreasing sequences which converge respectively to $E$ and $F$, and $E_n$ and $F_n$ are independent, then $E$ and $F$ are independent.
I want to know if my proof is logical. Could you people help me out?
To show that, I needed an auxiliary proposition I had to prove.
Let $(E_n), n\geq 1$ and $(F_n), n\geq 1$ such that $ E_n$ and $F_n$ are non-decreasing. Then $ E_n \cap F_n$ is non-decreasing.
Proof: if the sequences are non-decreasing, then $E_n \subset E_{n+1}$ and $F_n \subset F_{n+1}$.
Let $ x \in E_n\cap F_n \Rightarrow x \in E_n \text{ and } x\in F_n $. We have: $$ x \in E_{n+1} \text{ and } x \in F_{n+1} \Rightarrow x \in E_{n+1} \cap F_{n+1}$$ which follows from the assumption that: $$E_n \subset E_{n+1} \text{ and } F_n \subset F_{n+1}$$
So for every arbitrary $n \geq 1$, we have that if $x \in E_n \cap F_n \Rightarrow x \in E_{n+1} \cap F_{n+1} $, which means that $E_n \cap F_n \subset E_{n+1} \cap F_{n+1} $ so the sequence is non-decreasing.
With that, we can show the first statement. If $E_n$ and $F_n$ are independent:
$$P(E_n \cap F_n) = P(E_n) P(F_n) $$
$$ \lim_{n \to \infty} P(E_n \cap F_n) = \lim_{n \to \infty} P(E_n) P(F_n) $$
$$ P(\lim_{n \to \infty} E_n \cap F_n) = P( \lim_{n \to \infty} E_n) P( \lim_{n \to \infty} F_n)$$ due to continuity of probability.
We know by hypothesis $E_n \uparrow E $ and $ F_n \uparrow F$. Now, because I showed that $E_n \cap F_n$ is non-decreasing, then its limit is given by:
$$ P(\lim_{n \to \infty} E_n \cap F_n) = P\left(\bigcup_{n=1}^{\infty} E_n \cap F_n\right) = P\left(\bigcup_{n=1}^{\infty} E_n \cap \bigcup_{n=1}^{\infty} F_n\right) = P(E\cap F) $$
which concludes the statement:
$$ P(E \cap F) = P(E) P(F) $$
Is it right? Thank you so much!
I read your proof all the way through and couldn't see anything wrong with it.