Let $ \{A_{n}\} $ be a sequence of measurable sets , **then show that $ \\ \ P(\bigcap_{n=1}^{\infty} A_{n}) \geq 1-\sum_{n=1}^{\infty} P(A_{n}^{c})**. $ $$ $$ We know that if $ \{A_{n}\} $ is measurable , then $ P(\cap _{n=1}^{\infty} A_{n} )$ is also measurable. Now, $ \begin {align} \cap A_{n} \subset A_{1} \\ \cap A_{n} \subset A_{2}\\ \cap A_{n} \subset A_{3} \\ ... so \ on \\ or, P(\cap A_{n}) \leq P(A_{1}=1-P(A_{n}) \\ P(\cap A_{n}) \leq P(A_{2})=1-P(A_{2}) \\ and \ so \ on . \end{align}. $ . Any help is appreciating .
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The inequality doesn't seem right. The left hand side is non-negative, while the right hand side might very well be negative. – Quang Hoang Apr 07 '17 at 05:15
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Do you mean $\le$ or $\ge$? – gt6989b Apr 07 '17 at 05:16
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Let $A_1 = A_2=\emptyset$, then the left hand side is $0$ and the right hand side is negative. – copper.hat Apr 07 '17 at 05:21
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it is my mistake the actual inequality , is $ P( \bigcap A_{n}) \geq 1- \sum P(A_{n}) $ – MAS Apr 07 '17 at 05:26
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Hint: $$1-P\left(\bigcap A_n\right) = P\left(\left(\bigcap A_n\right)^c\right)= P\left(\bigcup A_n^c\right).$$
Quang Hoang
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@mabmath Sure. See proofs of this fact: http://math.stackexchange.com/q/270393/413376 – NCh Apr 07 '17 at 14:34