I found this exercise in the book of Kallenberg -Foundations of modern probability-.
Let $(\xi,\eta)\overset{d}{=}(\tilde{\xi},\tilde{\eta})$, with $\xi \in \mathcal{L}^1$. Then $ E[\xi|\eta] \overset{d}{=} E[\tilde{\xi}|\tilde{\eta}]$.
The author gives the hint: If $ E[\xi|\eta] = f(\eta)$ then $E[\tilde{\xi}|\tilde{\eta}] = f(\tilde{\eta})$ $a.s$.
I seem to be able to conclude assuming the hint, but I'm having troubles proving the hint, so does anyone has a hint for the hint?
Let $B$ be a Borel subset of $\mathbb R$. Then by definition of conditional expectation, $$\mathbb E\left[\xi \mathbf 1\left\{\eta\in B\right\}\right]=\mathbb E\left[\mathbb E\left[\xi\mid \eta\right] \mathbf 1\left\{\eta\in B\right\}\right]=\mathbb E\left[f\left(\eta\right)\mathbf 1\left\{\eta\in B\right\}\right].$$ Using equality in distribution between $\left(\xi,\eta\right)$ and $\left(\widetilde{\xi},\widetilde{\eta}\right)$, we derive that $$\mathbb E\left[\widetilde{\xi} \mathbf 1\left\{\widetilde{\eta}\in B\right\}\right] =\mathbb E\left[f\left(\widetilde{\eta}\right)\mathbf 1\left\{\widetilde{\eta}\in B\right\}\right].$$