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

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 $\mathbb{E}(X)=\infty$.
