Is it true that $P(A_n \text{ i.o.})=1 \implies A_n$ occurs a.s.?
I think it's true that $P(A_n \text{ i.o.})=0 \implies A_n^c$ occurs a.s. But how to prove it?
Is it true that $P(A_n \text{ i.o.})=1 \implies A_n$ occurs a.s.?
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probability-theory
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0What is "i.o."? – 2017-01-02
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0I guess it is the abbreviation of 'infinitey often', which is often found in probability literature. Anyway, for each fixed $n$ it is not necessarily that $A_n$ occurs a.s. – 2017-01-02
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0short for 'infinitely often'. $\left\{A_n \text{ i.o.}\right\} = \left\{\cup_{n=1}^\infty \cap_{m=n}^\infty A_n}$ – 2017-01-02
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1"$\implies A_n$ occurs a.s." For which $n$? – 2017-01-02
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0@Did It's a good point. I guess my question is not a proper one. This doubt comes from the answer of another question. It would be great if you can take a look at that one. Thanks. [Why $P(X_n>A \text{i.o.})$ implies $\sup X_n < \infty$?](http://math.stackexchange.com/questions/2080539/why-px-na-texti-o-implies-sup-x-n-infty?noredirect=1#comment4274765_2080539) – 2017-01-02
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0What is your doubt about this (inexplicably downvoted) other answer? – 2017-01-02
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0I think the answer you are linking is saying something wrong (and sort of nonsensical) which is why your question about it seems nonsensical. @Did: I cast that downvote and left a comment, and will happily explain it as much as you like. – 2017-01-02
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0@NateEldredge OK, I looked at it more carefully now and it is true that the answer needs some modifications, but the core is there... – 2017-01-02
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0@Did: Yes, I agree that what that answer probably *meant* to say is correct. What it *actually* says is wrong, and that is what led to this question. – 2017-01-02
1 Answers
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If $X_n\sim Bernoulli(1/2)$ for all $n$ with $X_n\in \{0,1\}$, then $$\mathbb P\{X_n=1,\ i.o.\}$$ but $P\{X_n=1\}=1/2$, and thus $\{X_n=1\}$ doesn't occur a.s.