Does $X\le Y$ imply $\mathbb{E}[X|Z]\le \mathbb{E}[Y|Z]$

Let $ X, Y, Z$ be random variables. If $ X\le Y$ , I wonder if $ $ \mathbb{E}[X|Z=z]\le \mathbb{E}[Y|Z=z]$ $ holds.

I have a contradiction between math and intuition.

First, math. I think math shows that this inequality is correct. Since $ X\le Y$ , we have $ X-Y\le 0$ . Then, $ $ \mathbb{E}[X-Y|Z=z]=\sum_{x,y} (x-y)p(X=x,Y=y|Z=z)$ $ Since $ x-y\le0$ and $ p(x,y|z)\ge0$ , it is clear that $ \mathbb{E}[X-Y|Z=z]\le0$ which implies $ \mathbb{E}[X|Z=z]\le \mathbb{E}[Y|Z=z]$ .

Second, intuition. Intuitively, the inequality is incorrect. For example, it is possible that when $ Z=z$ , the possibility to have $ Y=y$ is zero for all $ y$ . And it is possible that $ \mathbb{E}[X|Z=z]$ is positive. In this case, the inequality does not hold.

Is the intuition wrong? or the math wrong? I searched a bit and found a note saying that the inequality holds almost surely. I don’t understand why it is almost surely.