Prove that if $P[X\leq Y] =1$ then $E[X]\leq E[Y]$

Question

I would like to prove/disprove that following claim:

Prove that if $P[X\leq Y] =1$ then $E[X]\leq E[Y]$

How can it be done?

Answer 1

If $X$ and $Y$ are random variables defined on a probability space $(\Omega ,\Sigma ,P )$, and $$S=\{\omega\in\Omega\ \text{ such that }Y(\omega)\geq X(\omega)\}$$ Then $P(\Omega\backslash S)=P(X> Y)=0$, $$\begin{split} E(Y)&=\int_{\Omega}Y(\omega)dP(\omega) \,\,\,\text {(by definition)}\\ &= \int_{S}Y(\omega)dP(\omega) + \int_{\Omega\backslash S}Y(\omega)dP(\omega) \\ &= \int_{S}Y(\omega)dP(\omega) + 0 \\ &\geq \int_{S}X(\omega)dP(\omega) \,\,\,\text {(by definition of $S$)} \\ &\geq E(X) \end{split}$$