91Ó°ÊÓ

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

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

Similarly,\(\bigcup\limits_{i = 1}^\infty {{{\rm B}_i}} = \bigcup\limits_{i = 1}^\infty {{A_i}} \)

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

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

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

Therefore

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

Hence, the given situation is proved