Independence of increasing limits 
If $\{E_n\}_{n\ge1}$, and $\{F_n\}_{n\ge1}$ are increasing and independent for each $n$, show that their limits are independent.

Here is my attempt: 
Note that $\{E_n \cap F_n\}$ is also increasing. Then,
\begin{align}
    P(E \cap F) &=P\left(  \lim_{n \to \infty}\{ E_n\} \cap \lim_{n \to \infty}\{ F_n\}  \right)\\
     &=P\left(  \bigcup_{n=1}^{\infty} \{ E_n \}\cap  \bigcup_{n=1}^{\infty} \{ F_n \} \right)\\
     &= P\left(  \bigcup_{n=1}^{\infty} \{ E_n \cap F_n\} \right)        \\
     &= P(\lim_{n \to\infty} \{E_n \cap F_n \}) \\
     &=\lim_{n\to\infty}  P(E_n \cap F_n )\\
     &=\lim_{n\to\infty}P(E_n)P(F_n)\\
     &=P(E)P(F)  \hspace{5mm} \blacksquare 
\end{align}

What you did is essentially correct. Here are some remarks.


*

*It is not necessary to write the $\lim$ of a sequence of sets (unless it has been properly defined with the equality of $\limsup$ and $\liminf$ of the corresponding sequence). All we actually need is that if $\left(A_n\right)_{n\geqslant 1}$ is a non-decreasing sequence of sets, then $\lim_{n\to +\infty}\mathbb P\left(A_n\right)=\mathbb P\left(\bigcup_{j=1}^{+\infty}A_j\right)$.

*The place where the assumption that the sequences $\left(E_n\right)_{n\geqslant 1}$ and $\left(F_n\right)_{n\geqslant 1}$ are non-decreasing is used should appear more clearly. It is used to justify the equality
$$
  \left( \bigcup_{j=1}^{+\infty}E_j\right)\cap \left( \bigcup_{k=1}^{+\infty}F_k\right)=  \bigcup_{n=1}^{+\infty}\left(E_n\cap F_n\right).
 $$
The last equality is not true in general (take $E_{2i}=\emptyset=F_{2i+1}$, $E_{2i+1}=F_{2i}=\Omega$ for example.
