If $\mathbb{E} |X| < \infty$, then by applying
R2100: Triangle inequality for signed basic expectation, one has
\begin{equation}
\mathbb{E} X
\leq |\mathbb{E} X|
\leq \mathbb{E} |X|
< \infty
\end{equation}
On the other hand, suppose that $\mathbb{E} X \in (-\infty, \infty)$. By the definition of a
D5102: Basic expectation, we have $\mathbb{E} X = \mathbb{E} X^+ - \mathbb{E} X^-$ and thus
\begin{equation}
- \infty
< \mathbb{E} X^+ - X^-
< \infty
\end{equation}
Since $\infty - \infty$ (or $- \infty + \infty$) is not an allowed operation, then this implies that $\mathbb{E} X^+, \mathbb{E} X^- \in (-\infty, \infty)$ so that, in particular, $\mathbb{E} X^+ + \mathbb{E} X^- < \infty$. Result
R3583: Random basic number absolutisation equals sum of positive and negative parts provides the partition $|X| = X^+ + X^-$ and, taking expectations on both sides, we conclude
\begin{equation}
\mathbb{E} |X|
= \mathbb{E} X^+ + \mathbb{E} X^-
< \infty
\end{equation}
$\square$
