probability of infinite intersection of events given a sequence of mutually independent events $\{A_n\}_{n \in \mathbb{N}}$, I am trying to prove that: $$P\left(\bigcap_{i=n}^{\infty}A_i\right) = \lim \limits_{n \to \infty} \prod_{i=n}^{n}P(A_i)$$
It is easy to prove by induction that $P(\bigcap_{i=n}^{m}A_i) = \prod_{i=n}^{m}P(A_i)$ for all $m\in \mathbb{N}$. Now, my problem is due to the limit. In particular:


*

*Can I write $\bigcap_{i=n}^{\infty}A_i =  \lim \limits_{m \to \infty} \bigcap_{i=n}^{m}A_i$ ?  If yes, is this the traditional way to interpret intersections -
unions of countable sets?

*Can I write $P (\lim \limits_{m \to \infty} \bigcap_{i=n}^{m}A_i) = \lim \limits_{m \to \infty} P(\bigcap_{i=n}^{m}A_i)$ ? If yes, why and does this property hold in any measurable space? For example, we know that countable additivity is an axiom of measures, but there are no axioms for the one I wrote. So in case it is correct there must be a way to prove it. Maybe applying De Morgan's laws?
Thanks a lot.
 A: In general, if $E_1\supseteq E_2\supseteq\dots$ is a series of nested events, then $P(\bigcap_{i=1}^\infty E_i)=\lim_{n\to\infty}P(E_i)$. 
Proof: Let $F_i=E_i-E_{i+1}$ for $i\ge 1$. Then $F_i$ are disjoint and have union $E_1-\bigcap_{i=1}^\infty E_i$. Therefore,
\begin{align}
P\left(E_1-\bigcap_{i=1}^\infty E_i\right)
  &=\sum_{i=1}^\infty P(F_i)
\\P(E_1)-P\left(\bigcap_{i=1}^\infty E_i\right)&=\lim_{n\to \infty}\sum_{i=1}^n P(F_i)
\\&=\lim_{n\to \infty}\sum_{i=1}^n P(E_i)-P(E_{i+1})
\\&\hspace{-.4cm}\stackrel{\text{telescope}}=\lim_{n\to\infty} P(E_1)-P(E_{n+1})
\end{align}
THe result then follows by subtracting each side from $P(E_1)$.   $\square$
Apply this to the nested series $E_{m}=\bigcap_{i=0}^mA_{n+i}$.
A: Since $(\cap_{i=m}^n A_i)_{n\geq m}$ is a decreasing sequence of sets with intersection $\cap _{i=m}^\infty A_i$, it follows by measure continuity from above that
$$
P(\cap_{i=m}^\infty A_i)=\lim_{n\to \infty}P(\cap_{i=m}^n A_i)=\lim_{n\to \infty}\prod_{i=m}^n P(A_i)=\prod_{i=m}^\infty P(A_i)
$$
where we use independence in the second equality and definition of infinite product in the last equality.
