$\mathrm E [f(X,Y) \mid Y=y] = \mathrm E [f(X,y)]$? Let $X,Y$ be independent integrable random variables and let $f :\mathbb{R}^2 \to \mathbb{R}$ be integrable.  It makes intuitive sense that 
$$
\mathrm E [f(X,Y) \mid Y=y] = \mathrm E [f(X,y)],
$$
but I'm having trouble showing this.  Does this only hold if $Y$ is discrete?  I'd like to show it rigorously using regular conditional probabilities, so starting out I would have
$$
\mathrm E [f(X,Y) \mid Y=y] = \int_{\Omega} (f \circ (X,Y))(\omega)\, P^Y(\mathrm d\omega \mid y),
$$
where $P^Y(\cdot \mid \cdot)$ is the regular conditional probability of $P$ given $Y$.  I'm really not sure how to proceed form here, so any help is appreciated.
Update:  Added that $X$ and $Y$ should be independent.

To flesh out Did's comment: let $P(X=1)=P(X=-1)=1/2$. Let $Y=-X$ and note that $X,Y$ are not independent. Then define $f(x,y)=x+y$. 
$0=E[X+Y|Y=1]\neq E[X+1]=1.$
