Prove $P(\cap_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$? In class we proved that:
If $A_1$ is in $A_2$ and $A_2$ is in $A_3$ and ..., then
$P(\cup_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$
By declaring the following groups: $B_n=A_n\smallsetminus A_{n-1}$

Now I want to prove that if $A_1$ contains $A_2$ and $A_2$ contains $A_3$ and so on, then:
$P(\cap_{n=1}^{\infty} A_n)=\lim_{n\to\infty} P(A_n)$
What groups should I declare? Using the same $B_n$ doesn't help.
 A: Define the sets $B_n = A_n^c$, where $A^c$ denotes the complementary of $A$. Since $A_1 \supset A_2 \supset \dots$, we have that $A_1^c \subset A_2^c \subset \dots$ and so $B_1 \subset B_2 \subset \dots$.
Notice as well that $\bigcup_{i=1}^\infty B_i = (\bigcap_{i=1}^\infty A_i)^c$. Now apply your first result to these new sets to get:
$$\lim\limits_{n\to \infty} P(B_n) = P(\bigcup_{n=1}^\infty B_n) = P((\bigcap_{i=1}^\infty A_i)^c) = 1 - P(\bigcap_{i=1}^\infty A_i)$$
But we also have that:
$$\lim\limits_{n\to \infty} P(B_n) = \lim\limits_{n\to \infty} (1 - P(A_n)) = 1 -  \lim\limits_{n\to \infty} P(A_n)$$
And so $1 -  \lim\limits_{n\to \infty} P(A_n) = 1 - P(\bigcap_{i=1}^\infty A_i)$, and hence $\lim\limits_{n\to \infty} P(A_n) = P(\bigcap_{i=1}^\infty A_i)$.
A: $...A_3\subseteq A_2\subseteq A_1\implies  P(\cap_{n=1}^\infty A_n)=1-P(\cup_{n=1}^\infty A_n^C)$ where $A_1^C\subseteq A_2^C\subseteq A_3^C...$
$=1-\lim_{n\to\infty} P(A_n^C)=\lim_{n\to\infty}(1-P(A_n^C))=\lim_{n\to\infty} P(A_n)$
A: $
\newcommand{\P}{\mathrm{P}}
\newcommand{\bb}[1]{\left( #1 \right)}
$
$$
\P\bb{\bigcap_{i=1}^\infty A_n} = 1 - \P\bb{\bigcup_{i=1}^\infty A_n^c} = 1 - \lim_{n \to \infty} \P(A_n^c) = \lim_{n \to \infty} \bb{1 - P(A_n^c)} = \lim_{n \to \infty} P(A_n)
$$
