This is from Kallenberg.
Let $(\xi, \eta) \stackrel{d}{=} (\tilde{\xi}, \tilde{\eta})$, with $\xi \in L^1$. Then $E[ \xi | \eta ] \stackrel{d}{=} E[ \tilde{\xi} | \tilde{\eta} ]$.
The hint is to let $E[ \xi | \eta] = f(\eta)$ and first argue that $E[\tilde{\xi} | \tilde{\eta}] = f(\tilde{\eta})$. Supposing this, then the result should follow.
My idea was to try to show for all measurable $g$, we have $E[ g(f(\eta)) ] = E[ g(f(\tilde{\eta})) ]$. I can see this for $g$ identity, but I'm having trouble seeing this for arbitrary measurable $g$.
Thanks!
For $g= 1_B$ for $B$ a Borel set, $E[g(f(\eta))] = E[g(f(\tilde \eta))]$ says $P(\eta \in f^{-1}(B)) = P(\tilde \eta \in f^{-1}(B))$. The result then follows by approximating an arbitrary $g$ by simple functions.