Finite maximal coupling and heat-bath stationary law

Put $m(a)=\min \{p(a),q(a)\}$ and $r=1-{\sum }_{a}m(a)=\frac{1}{2}{\sum }_a\mathrm{\mid }p(a)-q(a)\mathrm{\mid }$. Couple the common mass $m(a)$ on the diagonal. If $r>0$, couple the residual masses $p-m$ and $q-m$ by their product divided by $r$; their supports are disjoint, so this part always disagrees. If $r=0$, the diagonal coupling already has the required marginals.

On the finite configuration space $A^{\mathrm{\Lambda }}$, choose a site uniformly and resample its spin from the conditional Gibbs law. The finite-volume Gibbs law is invariant under each such update because it resamples from its own conditional distribution; hence it is stationary for the random-scan mixture.

Let $N=\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }>0$ and fix a target configuration. Since there are finitely many strictly positive conditional probabilities, their minimum $p_{\ast }>0$. In $N$ updates, the probability of selecting a fixed ordering of all sites is $N^{-N}$, and conditional on that ordering the probability of setting every spin to its target is at least $p_{\ast }^N$. Thus every starting state reaches the target with probability at least $\epsilon =N^{-N}{p}_{\ast }^N>0$. The $N$-step kernel has a common component $\epsilon {\delta }_{\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{t}}$, so it contracts total variation by at most $1-\epsilon$; two stationary laws must coincide. If $\mathrm{\Lambda }=\mathrm{\varnothing }$, the configuration space is a singleton and uniqueness is immediate.