Consider three random variables $X,Y,Z$. Suppose that $Z$ has finite support $\mathcal{Z}$. Also, assume that $$\mathbb{E}( Y|Z=z)=0 \quad \text{ for some $z\in \mathcal{Z}$}$$ Claim: $\mathbb{E}( XY|Z=z)=0$
I can prove the claim when $X$ and $Y$ are discrete. Does the same hold with continuous random variables? Am I applying simply the law of iterated expectations?
When $X$ and $Y$ are discrete: $$ \mathbb{E}( XY|Z=z)=\sum_{x} \sum_y xy \Pr(X=x, Y=y|Z=z)\\=\sum_{x} \sum_y xy \Big[\Pr(X=x| Y=y,Z=z)*\Pr(Y=y|Z=z)\Big]\\=\sum_{x} x \underbrace{\Big[\sum_y y \Pr(Y=y|Z=z)\Big]}_{=\mathbb{E}(Y|Z=z)=0}*\Pr(X=x| Y=y,Z=z)=0 $$
Comment which follows answer below: my proof for the discrete case is wrong because I cannot isolate $\Big[\sum_y y \Pr(Y=y|Z=z)\Big]$ as done above!
The conditioning on $Z$ isn't what's important here. We can rephrase the question by replacing $X|Z=z$ with $X$ and $Y|Z=z$ with $Y$. Then the question is: if $\mathbf{E}Y = 0$ is $\mathbf{E} (XY) = 0$ as well?
And the answer is no: if $X = Y$ then $\mathbf{E}(Y^2)$ is never zero unless $Y = 0$ almost surely. For instance if $Y$ is a standard normal then $\mathbf{E}(Y) = 0$ and $\mathbf{E}(Y^2) = 1$. (This happens in the discrete case as well.)
But if $X$ and $Y$ are independent then $$\mathbf{E}(XY) = \mathbf{E}_X \mathbf{E}_Y (XY) = \mathbf{E}_X (X \mathbf{E}_Y Y) = 0.$$