Suppose that $(C_n)_{n\geqslant1}$ is a sequence of events such that $C_{n+1} \subset C_n$ for every $n\geqslant1$. Then how to prove the following probability statement? $$ \lim_{n \to \infty} P(C_n) = P\left( \bigcap_{n=1}^\infty C_n \right).$$
I know how to prove the following:
Suppose that $(C_n)_{n\geqslant1}$ is a sequence of events such that $C_n \subset C_{n+1}$ for every $n\geqslant1$. Then we have $$ \lim_{n \to \infty} P(C_n) = P\left( \bigcup_{n=1}^\infty C_n \right).$$
Best Answer
If $C_{n+1} \subset C_{n}$, then $C_{n}^c \subset C_{n+1}^c$. Thus we have $\lim_{n\to \infty} P(C_n^c) = P(\cup_{n=1}^\infty C_n^c)$, by what you know how to prove.
Since $P(C_n^c) = 1 - P(C_n)$ and $P(\cup_{n=1}^\infty C_n^c) = P((\cap_{n=1}^\infty C_n)^c) = 1 - P(\cap_{n=1}^\infty C_n)$, we get $\lim_{n\to\infty}P(C_n) = P(\cap_{n=1}^\infty C_n)$