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The Lemma: If the occurrences $A_n, n=1,2,3...$ are independent and $\sum_{n=1}^{\infty}A_n=\infty $ then $P(A^*)=1.$

$$P(A^*)=1-P(\bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty}A_k^c)=1-\lim_{n \to \infty}P( \bigcap_{k=n}^{\infty}A_k^c)$$- because $\bigcap_{k=n}^{\infty}A_k^c$ is a rising set..

$$=1- \lim_{n \to \infty} \lim_{m \to \infty}P(\bigcap_{k=n}^{m}A_k^c)$$-because $\bigcap_{k=n}^{m}A_k^c$ is a falling set $m > n $ it says this makes no sense how its possible..

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One of your questions: show $$ \bigcap_{k=1}^\infty A_k^c \subseteq \bigcap_{k=2}^\infty A_k^c \subseteq \bigcap_{k=3}^\infty A_k^c \subseteq \cdots $$
Can you do this?

The second one $$ \bigcap_{k=7}^8 A_k^c \supseteq \bigcap_{k=7}^9 A_k^c \supseteq \bigcap_{k=7}^{10} A_k^c \supseteq \cdots $$ Can you do this?

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  • $\begingroup$ This I understand, that would make the sets rising $\endgroup$
    – Bozo Vulicevic
    Jun 3, 2015 at 22:25
  • $\begingroup$ How is the other one falling I do not understand $\endgroup$
    – Bozo Vulicevic
    Jun 3, 2015 at 22:25
  • $\begingroup$ Very well, thank you sir.. $\endgroup$
    – Bozo Vulicevic
    Jun 3, 2015 at 22:38

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