Proving $E(g(X,Y) |Y=t) = E(g(X,t)|Y=t)$ for continuous random variables For continuous random variables $X$ and $Y$, I want to show that $E(g(X,Y) |Y=t) = E(g(X,t)|Y=t)$. This looks simple, but I can't do it.
I know that if $X$ and $Y$ were discrete then $E(g(X,Y) |Y=t) = \sum_{(x,y) \in \mathbb{Z^2}} g(x,y)P(X=x,Y=y|Y=t) = \sum_{(x,t) \in \mathbb{Z^2}} g(x,t)P(X=x|Y=t) = \sum_{x \in \mathbb{Z}} g(x,t)P(X=x|Y=t)$
I want to do something similar for the continuous case, but not sure how.
Should I start with $E(g(X,Y) |Y=t) = \int_{\mathbb{R^2}} g(x,y)\frac{f_{X,Y}(x,y)}{f_Y(t)} dA$?

This equality is nothing more than the particularization for $y=t$ of:
$$\int f(X,Y)dP_{\Omega /Y}=\int f(x,y)dP_{X/Y}$$
Formula that is demonstrated in the thread:
Why does conditional expectation have this property for independent random variables?
So the equality is true for almost all $t$. Furthermore, it is valid in general, not only for continuous random variables.
