If $X, Y \in L^1(P)$ such that $P(X \le t) < P(Y \le t) \implies E(Y) < E(X)$? I have to decide if this statement is either true or false, my intuition tells me that it is true, but at the moment of demonstrating it I encounter some technical difficulties. Here is what I tried at the moment.
$X, Y$ are random variables $\in L^1(P)$, so their mean is bounded because functions that belong to this space are those  $f$ measurable such that $f: \Omega \to \mathbb{K} : \int_{\Omega} |f|\, dP < \infty.$
If $P(X \le t) < P(Y \le t)\quad \forall t \in \mathbb{R}\implies P(X \le t) - P(Y \le t) < 0\quad \forall t \in \mathbb{R} \implies P(X -Y \le t)<0 \quad \forall t \in \mathbb{R} \implies X < Y \text{ a.s.} \implies \int_{\Omega} X\, dP < \int_{\Omega} Y \,dP \implies E(X) < E(Y) < \infty$.
But I do not know if there is any step wrong, or if I have not even used the hypothesis $X, Y \in L^1(P)$.
 A: Your reasoning is not correct, e.g. $\mathbb{P}(X \leq t)-\mathbb{P}(Y \leq t)<0$ does not imply $\mathbb{P}(X-Y \leq t) < 0$.

If $X,Y \geq 0$ and $\mathbb{P}(X \leq t)<\mathbb{P}(Y \leq t)$ for all $t>0$ (...or, equivalently, $\mathbb{P}(X>t)> \mathbb{P}(Y>t)$ for all $t>0$), then (as @GabrielRomon pointed out)
$$\mathbb{E}Y= \int_0^{\infty} \mathbb{P}(Y>t) \, dt \leq \int_0^{\infty} \mathbb{P}(X > t) \, dt = \mathbb{E}X. \tag{1}$$
In fact, the inequality is strict because otherwise it would follow that $\mathbb{P}(X \geq t) =\mathbb{P}(Y \geq t)$ for Lebesgue almost every $t$, which is not possible. Note that we may replace in $(1)$ the probabilities $\mathbb{P}(Y >t)$ (resp. $\mathbb{P}(X >t)$) by $\mathbb{P}(Y \geq t)$ (resp. $\mathbb{P}(X \geq t)$) since there at most countably many $t$ with $\mathbb{P}(Y=t)>0$ (resp. $\mathbb{P}(X=t)>0$).
It remains to consider the case that $X,Y \in L^1$ are not necessarily non-negative. Write $X=X^+-X^-$ for the positive part $X^+ = \max\{0,X\}$ and negative part $X^- = \max\{0,-X\}$, respectively. For $t>0$ we have
$$\mathbb{P}(X^+>t) = \mathbb{P}(X>t) > \mathbb{P}(Y>t) = \mathbb{P}(Y^+>t),$$
and so, by the first part of this answer,
$$\mathbb{E}Y^+ < \mathbb{E}X^+.$$
On the other hand, we have for $t>0$
$$\mathbb{P}(X^-  \geq t) = \mathbb{P}(X \leq -t) < \mathbb{P}(Y \leq -t) = \mathbb{P}(Y^- \geq t).$$
Hence, by the first part of this answer, $\mathbb{E}X^- < \mathbb{E}Y^-$.  In conclusion, $$\mathbb{E}(Y) = \underbrace{\mathbb{E}Y^+}_{< \mathbb{E}X^+}- \underbrace{\mathbb{E}Y^-}_{>\mathbb{E}X^-}<\mathbb{E}(X^+)-\mathbb{E}(X^-) = \mathbb{E}(X).$$
