Does finiteness of expectation of $X$ imply $X$ is finite almost surely? Formally,
let $\left(\Omega, \mathscr{F}, \mathbb{P}\right)$ be a probability triple. Let $X$ be a nonnegative random variable such that $\mathbb{E} \left(X\right) < \infty$, then is it true that $\mathbb{P} \left(X < \infty \right) = 1$? Is this something trivial? I just do not know how to begin the proof
 A: Yes.
$$\mathbb{E}(X)=\int_\Omega X \, d\mathbb{P}\geq \int_{X^{-1}(\infty)} X \, d\mathbb{P}\geq \int_{X^{-1}(\infty)} t \, d\mathbb{P}=t\mathbb{P}(X=\infty)$$
for every $t\in\mathbb{R}^+.$
If $\mathbb{P}(X=\infty)>0$, then taking $t\rightarrow \infty$ shows $\mathbb{E}(X)=\infty$.
A: Similar to Sean G's answer:
Markov's inequality implies $$t P(X \ge t) \le E[X] < \infty$$
for any $t \ge 0$. As $t \to \infty$, the left-hand side remains finite, so we must have $$0 = \lim_{t \to \infty} P(X \ge t) = P(X = \infty).$$
A: Alternative look (in accordance with the comment of Kavi).

Formally if $X$ is a non-negative random variable then $\{0\leq X<\infty\}=\Omega$ so that $P(X<\infty)=1$ is immediate.
If $X$ is a measurable function $\Omega\to[0,+\infty]$ then $X\geq X\mathbf1_{X=+\infty}$ so that: $$\mathbb EX\geq\mathbb EX\mathbf1_{X=+\infty}=+\infty\cdot\mathbb P(X=+\infty)\tag1$$Here $[0,\infty]$ is equipped with a multiplication $\cdot$ determined by:

*

*$x\cdot y=x\times y$ if $x,y<+\infty$.

*$x\cdot y=0$ if $x=0$ or $y=0$.

*$x\cdot y=+\infty$ otherwise.

So $(1)$ tells us that $EX<\infty$ can only be true if $\mathbb P(X=+\infty)=0$ or equivalently $\mathbb P(X<+\infty)=1$.
The answers of Sean and angryavian actually assure/confirm that this multiplication works fine.
