91Ó°ÊÓ

Define\({{\rm B}_1} = {\rm A}_1^c\)and\({{\rm B}_i} = \frac{{{\rm A}_i^c}}{{{\rm A}_{_{i - 1}}^c}}\,for\,i = 2,3...\)

Then\({{\rm B}_1},{{\rm B}_2}...\)are mutually exclusive, and\(\bigcap\limits_{i = 1}^n {{{\rm B}_i}} = \bigcap\limits_{i = 1}^n {{A_i}} \)

Similarly,\(\bigcap\limits_{i = 1}^\infty {{{\rm B}_i}} = \bigcap\limits_{i = 1}^\infty {{{\rm A}_i}} \).

As \(A_1^c \subset A_2^c \subset ...\)we have \({\rm P}\left( {\bigcap\limits_{n = 1}^\infty {{{\rm A}_n}} } \right) = \mathop {lim}\limits_{n \to \infty } \left( {{{\rm A}_n}} \right)\)

\(\begin{aligned}{}{\rm P}\left( {\mathop {lim}\limits_{n \to \infty } \left( {{{\rm A}_n}} \right)} \right) &= {\rm P}\left( {\bigcap\limits_{i = 1}^\infty {{{\rm A}_i}} } \right)\\ &= {\rm P}\left( {\bigcap\limits_{i = 1}^\infty {{{\rm B}_i}} } \right)\\ &= \sum\limits_{i = 1}^\infty {{\rm P}\left( {{{\rm B}_i}} \right)} \end{aligned}\)

\(\begin{aligned}{}{\rm P}\left( {\mathop {lim}\limits_{n \to \infty } \left( {{{\rm A}_n}} \right)} \right) &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{\rm P}\left( {{{\rm B}_i}} \right)} \\ &= \mathop {\lim }\limits_{n \to \infty } {\rm P}\left( {\bigcap\limits_{i = 1}^n {{{\rm B}_i}} } \right)\\ &= \mathop {\lim }\limits_{n \to \infty } {\rm P}\left( {\bigcap\limits_{i = 1}^n {{A_i}} } \right)\end{aligned}\)

Therefore, \({\rm P}\left( {\bigcap\limits_{i = 1}^\infty {{{\rm A}_i}} } \right) = \mathop {\lim }\limits_{n \to \infty } {\rm P}\left( {{A_n}} \right)\)

Hence, the given condition is proved