Suppose that $X$ and $Y$ are independent standard Gaussian random variables, i.e, $X,Y \sim N(0,1)$. It is a well known fact that the ratio $X/Y$ follows a Cauchy distribution and does not have a mean. Suppose the ratio is denoted by $Z$ as follows: $$ Z \triangleq \frac{X}{Y}. $$ Can we say that $E[Z|Y]=0$? Or does conditional expectation of $Z$ given $Y$ does not exist at all?
All the references that I have come across assume that expectation of the random variable is well-defined.
We can't, really, no; since $Z$ doesn't have finite expectation, conditional expectation isn't technically well-defined.
The main issue is that it depends dramatically on how you define $E[Z | Y]$. For instance, if you define $E[Z | Y = y] = \lim_{h \to 0^+} E[Z | Y \in [y - h,y+h]]$ for $y \neq 0$, then you'd indeed have $E[Z | Y = y] = 0$. As it stands, conditional expectation isn't defined this way, and only makes sense for random variables with finite expectation (more precisely, random variables in $L^1$).