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Let $X$ be a random variable and $\left\{X_{n}\right\}$ be a sequence of random variables on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Prove that $X_n \to X$ a.s. implies $X_n \to X$ in probability.

Could you check if my attempt is correct?

It suffices to prove that $$\mathbb P(|X_n-X| > 1/m) \to 0 \quad \text{as} \quad n \to \infty, \quad m \in \mathbb N^*.$$

Notice that $$\{\omega \mid X_n (\omega) \to X (\omega)\} = \bigcap_{m \in \mathbb N^*} \bigcup_{N \in \mathbb N} \bigcap_{n \ge N} \{ \omega | X_n (\omega) - X (\omega) | \le 1/m \}.$$

Because $\mathbb P( \{\omega \mid X_n (\omega) \to X (\omega)\} )=1$, we have $$\mathbb P \left ( \bigcup_{N \in \mathbb N} \bigcap_{n \ge N} \{ \omega \mid | X_n (\omega) - X (\omega) | \le 1/m \} \right ) = 1, \quad m \in \mathbb N^*$$

and thus

$$\begin{aligned} &\mathbb P (\limsup_n \{ \omega \mid | X_n (\omega) - X (\omega) | > 1/m \}) \\ ={} &\mathbb P \left ( \bigcap_{N \in \mathbb N} \bigcup_{n \ge N} \{ \omega \mid | X_n (\omega) - X (\omega) | > 1/m \} \right ) = 0, \quad m \in \mathbb N^*. \end{aligned}$$

It follows that $$\begin{aligned} &0 \\ ={} &\mathbb P (\limsup_n \{ \omega \mid | X_n (\omega) - X (\omega) | > 1/m \}) \\ \ge {} &\limsup_n \mathbb P ( \{ \omega \mid | X_n (\omega) - X (\omega) | > 1/m \} \\ \ge {} &\lim_n \mathbb P ( \{ \omega \mid | X_n (\omega) - X (\omega) | > 1/m \} \\ = {} &\lim_n \mathbb P (| X_n - X | > 1/m). \end{aligned}$$

The result then follows.

Akira
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    Imo some steps need to be justified. Like why $\mathbb P(\bigcup\ldots)=1$ holds from $\mathbb P(X_n\to X)=1$ and why is $\mathbb P(\limsup{\ldots})\ge\limsup_n:\mathbb P({\ldots})$. Otherwise it looks good! (There is a quicker proof of this fact using Fatou lemma/dominated convergence once we know that convergence in probability means $\mathbb E[\min(|X_n-X|,1)]\to0$.) – nejimban Oct 05 '21 at 16:00
  • @nejimban I showed the first fact here. The inequality was proved in my class. – Akira Oct 05 '21 at 16:02
  • You are also using that $\mathbb P(\bigcap A_n)=\lim_{n\to\infty}\mathbb P(A_n)$ for a non-increasing sequence of events, aren't you? – nejimban Oct 05 '21 at 16:03
  • @nejimban I only use $\mathbb P (\bigcap_{m \in \mathbb N^} \bigcup_{N \in \mathbb N} \bigcap_{n \ge N} A_{n,m}) = 1 \implies \mathbb P (\bigcup_{N \in \mathbb N} \bigcap_{n \ge N} A_{n,m}) = 1$ for all $m \in \mathbb N^$. – Akira Oct 05 '21 at 16:08
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    Right, this is obvious. And I guess the $\mathbb P(\limsup{\ldots})\ge\limsup_n:\mathbb P({\ldots})$ bit is (reverse) Fatou lemma. – nejimban Oct 05 '21 at 16:18

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