$\mathbf{E}[X|Z=z]=\mathbf{E}[Y|Z=z]$ implies $X=Y$ a.s.?

Question

Suppose that $X,Y,Z$ are real-valued random variables defined on a probability space $(\Omega, \mathscr{F}, P)$ such that for all reals $z$ such that $P(Z=z)>0$ it holds $$ \mathbf{E}[X|Z=z]=\mathbf{E}[Y|Z=z]. $$ What can be said about $X$ and $Y$? In particular, is it true that $X=Y$ a.s. $P\upharpoonright \sigma(Z)$?

Answer 1

Just consider $X, Z$ and $Y, Z$, to be independent, $X$ and $Y$ can have any probability distribution as long as they have equal mean. Then your equality holds for all $z$, but $X$ and $Y$ are not a.s. equal.