Consider an arbitrary sequence of events $\{A_{i}\}_{i \in N}$. Assume there exists $c > 0$ such that, for every $i \in N$, $P(A_{i}) \geq c$. Is it true that $P(A_{i} \text{ infinitelly often}) \geq c$?
Does $P(A_{i} \text{ infinitelly often}) \geq c$ follow from $P(A_{i}) \geq c$?
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probability-theory
2 Answers
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I think yes? Namely, $\{\omega: \omega \in A_i \text{ for infinitely many $A_i$ } \} = \cap_{j \geqslant 0} \cup_{k \geqslant j} A_k$ Now for fixed $j$, the union has probability certainly at least $c$, and pushing $j$ to infinity gives the result.
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Yes. Silly me :)
$P(A_{i} \text{infinitelly often}) := P(\cap_{i=1}^{\infty} \cup_{j=i}^{\infty} A_{j})$.
Define $B_{i} = \cup_{j=i}^{\infty} A_{i}$. Observe that if $i \geq j$, $B_{i} \subset B_{j}$ and, by problem assumption, $P(B_{i}) \geq c$, for every $i \in N^{*}$. Hence, by continuity of probability,
$P(A_{i} \text{infinitelly often}) = P(\cap_{i=1}^{\infty} B_{i}) = P(\lim B_{i}) = \lim P(B_{i}) \geq c$