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Let $a_1,\dots,a_m$ be all the distinct values of $\xi$, and let $b_1,\dots,b_n$ be all the distinct values of $\eta$. Let $p_{ij}:=P(\xi=a_i,\eta=b_j)$, $r_i:=P(\xi=a_i)=\sum_j p_{ij}$, $c_j:=P(\eta=b_j)=\sum_i p_{ij}$. Let us interpret the condition "any value of these values are accepted with a non-zero probability" as the condition that $p_{ij}>0$ for all $i\in[m]:=\{1,\dots,m\}$ and $j\in[n]$. Then the conditions $E(\xi|\eta)\ge\eta$ and $E(\eta|\xi)\ge\xi$ translate into \begin{equation} \frac1{c_j}\,\sum_i a_i p_{ij}\ge b_j\quad\text{and}\quad \sum_j b_j p_{kj}\ge a_k r_k=:A_k \end{equation} for all $k\in[m]$ and $j\in[n]$, whence, for \begin{equation} M_{ik}:=\frac1{r_i}\sum_j\frac{p_{kj}p_{ij}}{c_j}, \end{equation} \begin{equation} \sum_i A_i M_{ik}=\sum_i a_i \sum_j\frac{p_{kj}p_{ij}}{c_j} =\sum_j\frac{p_{kj}}{c_j}\,\sum_ia_i p_{ij}\ge \sum_j b_j p_{kj}\ge A_k. \tag{*} \end{equation} The matrix $M=[M_{ik}]$ is stochastic, i.e., $\sum_k M_{ik}=1$, and so, $\sum_k\sum_i A_i M_{ik}=\sum_i A_i$. Hence, all the inequalities in (*) are equalities, so that \begin{equation} \sum_i A_i M_{ik}=A_k \end{equation} for all $k$. That is, $A=[A_1,\dots,A_m]$ is an invariant measure for the stochastic matrix $M$ with all strictly positive entries. So, the vector $A$ is uniquely determined up to a constant factor. On the other hand, it is straightforward to check that the formula $A_i=r_i$ for all $i$ defines an invariant measure for $M$. So, $a_i=a$ for some real $a>0$ and all $i$. Similarly, $b_j=b$ for some real $b>0$ and all $j$. Now the conditions $E(\xi|\eta)\ge\eta$ and $E(\eta|\xi)\ge\xi$ imply $a\ge b\ge a$, so that $\xi=\eta$.