$\mathbb E[ |X| ] < \infty \iff \forall \epsilon : \mathbb E[ |X / \epsilon | ] < \infty$

$\newcommand{\dd}{\mathop{}\!\mathrm{d}}$Recall that $\mathbb E|X|=\int_0^\infty\mathbb P(|X|>t)\dd t$. Then since $t\mapsto\mathbb P(|X|>t)$ is a decreasing function, we can bound this integral via upper and lower Riemann sums as $$\epsilon\cdot\sum_{n=1}^\infty\mathbb P(|X|>n\epsilon)\leq\int_0^\infty\mathbb P(|X|>t)\dd t\leq\epsilon\cdot\sum_{n=0}^\infty\mathbb P(|X|>n\epsilon).$$ So it turns out that $\mathbb E|X|<\infty\iff\sum\mathbb P(|X|>n\epsilon)<\infty$.

Using the monotone convergence theorem, for any $\epsilon>0$, $$ \mathsf{E}|X|=\lim_{M\to\infty}\mathsf{E}[|X|\wedge M]=\epsilon\lim_{M\to\infty}\mathsf{E}[|X/\epsilon|\wedge M/\epsilon]=\epsilon\,\mathsf{E}|X/\epsilon|. $$ Thus, $$ \mathsf{E}|X|<\infty \Leftrightarrow \mathsf{E}|X/\epsilon|<\infty \Leftrightarrow \sum_{n}\mathsf{P}(|X/\epsilon|\ge n)<\infty. $$