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Let $f_n \in [0, 1]$ and suppose if we want to show $$ \lim_{n \to \infty} f_n = 1 $$ almost surely, is it enough to show $$ \lim_{n \to \infty} \mathbb{P}\{ f_n = 1 \} = 1? $$ If not, what if we add the assumption that $f_n$ is increasing?

Attempt: I am trying to use Borel-Cantelli to see if this is plausible, but it seems like in the end we will need a rate for the convergence of $\lim_{n \to \infty} \mathbb{P}\{ f_n = 1 \} = 1$ and this is not given: this leads me to think this is not enough to show a.s. convergence. If not, perhaps there is an easy example showing this is not true, but I am not sure how to find one.

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No, it isn't enough. Let $\mathbb{P}$ be Lebesgue measure on the unit interval and $$ f_n = 1 - \chi \left( \frac{n-2^k}{2^k}, \frac{n-2^k + 1}{2^k} \right)$$ for $k\geq 0$ and $2^k \geq n < 2^{k+1}$. Then your condition holds but this does not converge almost surely. For your question about monotonicity, see: Monotone increasing sequence of random variable that converge in probability implies convergence almost surely

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  • $\begingroup$ Thank you for the answer. Could you clarify what is $k$ here and is $f_n$ dependent on $k$ as well? How then, is $f_n$ defined to be only dependent on $n$? $\endgroup$
    – Partial T
    Apr 16 at 0:34
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    $\begingroup$ @PartialT woops, missed part of my definition. Fixed it now. $\endgroup$ Apr 16 at 5:28

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