We are asked to show:
if a random variable sequence $X_n \to +\infty $ almost surely then $ P(X_n < M \text{ i.o}) = 0 \ \forall \ M \gt 0$
I try show this using $E(X_n) = \infty $
$E(X_n) = \int_{X_n < M} X_n dP + \int_{X_n \ge M} X_n dP= \infty$
$\int_{X_n < M} X_n dP \le M\int_{A_n: (X_n < M)}dP \le M\sum P(A_n)$
Since $M\sum P(A_n) \le \infty \implies P(A_n \text{ i.o}) = 0$ by Borell Cantelli I.
I'm not sure if the above is correct though.
Best Answer
You don't need anything like an integral or an expectation for this.
If $X_n\to \infty$, then with probability one $X_n\ge M$ eventually so with probability zero $X_n<M$ infinitely often.
Formally, let $M > 0$. Then we know that $$\{\omega \in \Omega: \lim_n X_n(\omega) = \infty\}\subseteq \{\omega \in \Omega: X_n(\omega) \ge M \mathrm{\ eventually}\}$$
Hence, $$\mathbb{P}\{\omega \in \Omega: X_n(\omega) \ge M \mathrm{\ eventually}\} = 1.$$ Next, note that $$\{\omega \in \Omega: X_n(\omega) \ge M \mathrm{\ eventually}\}^c = \{\omega \in \Omega: X_n(\omega) < M \mathrm{\ infinitely \ often}\}$$ and thus $$\mathbb{P}\{\omega \in \Omega: X_n(\omega) < M \mathrm{\ infinitely \ often}\} = 1- \mathbb{P}\{\omega \in \Omega: X_n(\omega) \ge M \mathrm{\ eventually}\} = 0$$ as desired.