TITLE: Does finiteness of expectation of $X$ imply $X$ is finite almost surely?
QUESTION [6 upvotes]: 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

REPLY [3 votes]: 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.