I Is this conditional expectation identity true?

[psie]

TL;DR Summary

I am working an exercise to show a conditional expectation identity, but I'm not sure if it is true at all.

I'm working through an exercise to prove various identities of the conditional expectation. One of the identities I need to show is the following $$E(f(X,Y)\mid Y=y)=E(f(X,y)\mid Y=y).$$ But I am a little concerned about this identity from things I've read elsewhere. I am paraphrasing from Breiman's book Probability:

Proposition 4.36: Let $X$ and $Y$ be random variables. Let $f(X,Y)$ be a real valued, random variable such that the expectation of the absolute value of $f(X,Y)$ is finite. If $Q(\cdot\mid Y=y)$ is a regular conditional distribution for $X$ given $Y=y$, then, $$E(f(X,Y)\mid Y=y)=\int f(x,y) \, dQ(\cdot \mid Y=y)\quad \text{a.s. with respect to the law of }Y.\tag{4.37}$$

Again, paraphrasing Breiman in section 4.3, page 80:

It is tempting to replace the right-hand side of (4.37) by $E(f(X,y)\mid Y=y)$. But this object cannot be defined through the standard definition of conditional expectation (4.18) [defined below].

Definition 4.18: $E(X\mid Y=y)$ is any random variable on $\mathbb R$, where $Q(B)=P(Y \in B)$, satisfying $$ \int_B E(X\mid Y=y) dQ = \int_{Y \in B} X dP,$$ for all Borel sets $B$.

Thus, does $E(f(X,Y)\mid Y=y)= E(f(X,y)\mid Y=y)$ make sense? Breiman seems to suggest that $E(f(X,y)\mid Y=y)$ is not well defined, so I'm not sure how to proceed.

 

[psie]

I think this pdf clarified things a bit.