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 $E(g(X,Y) |Y=t) = \int_{\mathbb{R^2}} g(x,y)\frac{f_{X,Y}(x,y)}{f_Y(t)} dA$?

When dealing with conditional expectations, things like this get slightly trickier than you might anticipate, because these objects are defined formally in a way where you can vary them on a set of probability measure zero.  So, for continuous random variables, you can arbitrarily change the conditional expectation function at any countable number of points and it is still considered to be a valid conditional expectation function.  (This aspect of conditional probability gets into the field of measure theory.)  So the long and short of it is that the two conditional expectations of interest to you can actually be different on a countable set of values, and it will not cause any problem mathematically (though it is not as easily interpretable).
So, technically speaking, your conjecture is false, but let's weaken it a little bit and try to show that these two conditional expectation functions are equal almost surely (i.e., except on a set of values of $Y=t$ occurring with zero probability).  To do this, let's start by creating some simpler notation for the two conditional expectation functions of interest:
$$H(t) \equiv \mathbb{E}(g(X,Y) | Y=t)
\quad \quad \quad \quad \quad 
H_*(t) \equiv \mathbb{E}(g(X,t) | Y=t).$$
To show the relationship between these functions, we will start by observing that conditional expectation is defined formally by the following integration requirement:
$$\mathbb{E}(g(X,Y)) = \int \limits_\mathscr{Y} H(t) f_Y(t) \ dt.$$
Moreover, the marginal expectation can also be written as:
$$\mathbb{E}(g(X,Y)) = \int \limits_{\mathscr{X} \times \mathscr{Y}} g(x,t) \cdot f_{X,Y}(x,t) \ dx \ dt.$$
Putting these together and breaking down the joint integral into an iterated integral gives:$^\dagger$
$$\begin{align}
\int \limits_\mathscr{Y} H(t) f_Y(t) \ dt
&= \mathbb{E}(g(X,Y)) \\[6pt]
&= \int \limits_{\mathscr{X} \times \mathscr{Y}} g(x,t) \cdot f_{X,Y}(x,t) \ dx \ dt \\[6pt]
&= \int \limits_\mathscr{Y} \Bigg[ \int \limits_{\mathscr{X}} g(x,t) \cdot f_{X|Y}(x|t) \ dx \Bigg] f_{Y}(t) \ dt \\[6pt]
&= \int \limits_\mathscr{Y} H_*(t) f_{Y}(t) \ dt. \\[6pt]
\end{align}$$
Since this must hold for all possible density functions $f_Y$, this means that we must have $H(t) = H_*(t)$ for all $t \in \mathscr{Y}_* \subseteq \mathscr{Y}$ where $\mathbb{P}(Y \notin \mathscr{Y}_*) = 0$.  That is, the two conditional expectation functions must be equal except possibly on a set of $Y$-values with probability zero.

$^\dagger$ For simplicity I'm assuming that the range of the random variables is independent.  If this is not the case then the same basic working holds, but with some greater care applied to the range of integration in the iterated integral.
