Suppose I have two random variables $X$ and $Y$ that are i.i.d. according to a prob. measure $\mu$ whose support is $A$. Take a function $g:A\times A \to \mathbb{R}$ such that $E[g(X,Y)\mid Y \in b(X)]$ exists. Furthermore, we have that $\mu(b(x))>0$ for every $x \in A$.
Suppose we have that for every pair $(x,y) \in A \times A$ we have $y \in b(x) \iff x \in c(y)$, where $\mu(c(y))>0$ for every $y \in A$. Can we show that $$E[g(X,Y)\mid Y \in b(X)] = E[g(X,Y) \mid X \in c(Y)]?$$ The result seems intuitive but I have not been able to prove it.
First Try: It seems logical to try changing the order of integration, i.e. $$ E[g(X,Y)\mid Y \in b(X)] = \int_A \int_{b(x)} g(x,y) \frac{d\mu(y)}{\mu(b(x))} d\mu(x) = \int_A \int_{c(y)} g(x,y) \frac{d\mu(x)}{\mu(b(x))} d\mu(y) \, . $$ However to get from the last integral to $E[g(X,Y) \mid X \in c(Y)]$ I would need to change $\mu(b(x))$ for $\mu(c(y))$, which is not straightforward.