$E(X)>E(Y)$ always imply that $P[X>Y]>0$

Question

Prove or disprove- "If $E(X)>E(Y)$ then $P[X>Y]>0$

What I attempted:-

I am using contradiction. Suppose $E(X)>E(Y)$. We assume that $P[X>Y]=0$

Now, we have \begin{equation} \begin{aligned} &P[X>Y]=0\\ \Rightarrow & P[(X-Y)>0]=0 \\ \Rightarrow & P[Z>0]=0 \qquad \mbox{where}\quad Z=X-Y \end{aligned} \end{equation} The last equation imply that \begin{equation} \begin{aligned} & Z\le 0 \\ \Rightarrow & E(Z)\le 0 \\ \Rightarrow & E(X-Y)\le 0 \\ \Rightarrow & E(X)\le E(Y)\\ \end{aligned} \end{equation} which is a contradiction. Therefore we must have $P[X>Y]>0$

Answer 1

Seems fine.

Just that I would write $$Z \le 0 ~ W.P. 1$$

rather than just $Z \le 0$. We can still have null set that take positive values but it wouldn't influece $E[Z]$.