How to prove that E[X] < E[Y]?

Question

Consider two random variables $X$ and $Y$ such that $P(X \le Y) = 1$ but $P(X = Y) < 1$, that is $P(X < Y) > 0$. Then, how do we prove that $E[X] < E[Y]$?

My attempt is as below:

Let $A = \{x \in \mathbb{R}: x = X(w) < Y(w) = y \}$, then $$ E[Y] = \int_{\mathbb{R}} x \, dF_{Y}(x) = \int_{\mathbb{R}/A} x \, dF_{Y}(x) + \int_{A} x \, dF_{Y}(x) $$ but not able to get it to the end. And, I am not familiar with the measure theoretic approach.

Answer 1

$$\begin{align} \operatorname{E}[Y - X] &= \operatorname{E}[Y - X \mid Y = X]\Pr[Y = X] + \operatorname{E}[Y - X \mid Y > X]\Pr[Y > X] \\ &= \operatorname{E}[Y - X \mid Y - X > 0]\Pr[Y > X] \\ \end{align}$$ and since $\Pr[Y > X] > 0$ and $\operatorname{E}[Y - X \mid Y - X > 0] > 0$, the result follows.