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If I can show for some event $A_n$, $$ \sum_n P(A_n) < \infty. $$ Then by First Borel-Cantelli Lemma, I get $P(A_n \; i.o.) = 0$; But I still confused about how this infinitely often connects to almost sure convergence.

Is it true that the complement $\{A_n i.o. \}^c$ happens almost surely?

For example, suppose $P( \{ X_n = C \; i.o.\}) =0$ for some constant $C$, where $X_n$ is random variable. Can I conclude $X_n \neq C$ almost surely for free?

Thank you

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    $\begingroup$ In your second example, you probably mean $\{X_n = C\}$ or something like that, because the "infinitely often" part is useless otherwise. In that case, you can conclude that almost surely, $X_n = C$ only for finitely many $n$. $\endgroup$ – PhoemueX Oct 21 '14 at 4:50
  • $\begingroup$ Thanks! @PhoemueX , I revised the question. So there is no connection to say the complement event $X_n \neq C$ almost surely? $\endgroup$ – Fianra Oct 21 '14 at 7:17
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You are using almost surely here in two senses. First, in the $i.o.$ case, you are referring to a sequence of events, whereas in your second question, you are asking a pointwise question.

In short: you cannot conclude $X_n \neq C, a.s.$ as this would imply $P(X_n=C)=0\;\;\forall n$.

However, i think you mean not the above event $\left(\{X_n \neq C\}\right)$but the event $\{X_n\neq C\;\; i.o\}$, correct? If so, then we know that that tail $\sigma-$algebra does not contain $\{X=C\}$, thus, the tail $\sigma-$algebra will be $\sigma\left(\{X\neq C\}\right)$ which implies that the probability must be $1$ for this event.

In English, if an event does not happen infinitely often, then it will eventually cease to happen, hence its complement will always happen.

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