Is it true or false? if $E(X)\geq E(Y)$ then $X\geq Y$ I'm not pretty sure if this is true. We already know that if $X\leq Y$ then $E(X)\leq E(Y)$. But it would be great if someone can show me a counterexample.
 A: False. Let $X$ be the amount of money you spend on a lottery ticket and $Y$ be the amount you win. $E(X) > E(Y)$, but it's possible to win more than you spent.
A: The statement is false, here is a counterexample: let $(S,\mathbb{P}) = ([0,1],\lambda)$ and consider $X = \chi_{[0,2^{-1}]}$, $Y = \chi_{[2^{-1},1]}$. Then $\mathbb{E}(X) \ge \mathbb{E}(Y)$, but $X(\omega) < Y(\omega)$ for $\omega \in (2^{-1},1]$.
A: Yes, we can show that:

If $X< Y $ almost surely, then $\forall \omega\in\Omega :X(\omega)< Y(\omega)$
So: $\mathsf E(X) = \int_\Omega X(\omega)\operatorname{d} P \;<\; \int_\Omega Y(\omega)\operatorname d P = \mathsf E(Y)$
Ergo: $X < Y \text{ a.s.}\implies \mathsf E(X)< \mathsf E(Y)$


However   We cannot say the converse, because the contrapositive is: $$\mathsf E(X)\geq \mathsf E(Y) \implies X\geq Y \text{ sometimes}$$
That is not the result we want.   The ordering of the expectation does not guarantee that the variables will be ordered, almost surely, only that that ordering occurs for some outcomes.
For example: Let $Y$ be the result of a standard cube dice (a d6).  Let $X$ be the result of a standard octahedral dice (a d8), tossed at the same time.  Then $\mathsf E(X)=4.5, \mathsf E(X)=4.5$, but sometimes $X<Y, X=Y,$ or $X>Y$ are all possible (they are non-zero measure events).
$$\mathsf E(X)\geq \mathsf E(Y) \quad\nRightarrow\quad X\geq Y\text{ a.s.}$$
