# Sufficient Condition on Almost Surely Convergence

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.

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
• 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 $n$? Apr 16 at 0:34