claim : let $\{X_n, \, n \geq 1\}$ be a monotonic sequence of random variables

$X_n\rightarrow0\quad \text{in probability} \implies X_n\rightarrow 0\quad \text{a.s}$

proof attempt :

we have for every $\epsilon > 0$ $$\lim_{n \to +\infty}\mathbb{P}(|X_n| \geq \epsilon) = 0$$

which intuitively means to me that : (since it holds for every $\epsilon, \, $ and $\epsilon > 0$ and $|X_n |\geq 0$)

$$\lim_{n \to +\infty}\mathbb{P}(|X_n| > 0) =\lim_{n \to +\infty}\mathbb{P}(|X_n| \neq 0) = 0$$

I'm not sure concerning this step.... anyways

by the monotone convergence theorem : $$\mathbb{P}(\omega \mid\lim_{n \to +\infty}|X_n(w)| \neq 0) = 0$$

then :

$$\mathbb{P}(\omega \mid\lim_{n \to +\infty}|X_n(w)| = 0) = \mathbb{P}(\omega \mid\lim_{n \to +\infty} X_n(w) = 0) = 1$$

thus the almost sure convergence, I need some feedback concerning this attempt at proof, thanks !

  • 1
    $\begingroup$ "I'm not sure concerning this step" Indeed, you do not justify it, and it happens to be wrong, as very simple examples show. To prove the result, note that if $(X_n)$ is monotonous, $X_n\to X$ almost surely hence $X_n\to X$ in probability. By the almost sure uniqueness of the limit in probability, $X=0$ almost surely, end of the proof. $\endgroup$ – Did Jan 18 at 14:04

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