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Let $m=|V|\ge 1$ and put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, counting each edge once. If only the spin at $x$ changes from $-1$ to $+1$, the change in $H$ is $2N(x)$, soThe given opposite transition probabilities have the same ratio. More explicitly,For pairs differing at more than one site both sides vanish, and for equal configurations balance is automatic. This proves detailed balance, hence the chain is reversible with stationary law $\pi$. All single-spin changes have positive probability, so the finite chain is irreducible; it also has positive holding probability and is aperiodic.

Its random-map version is a heat-bath Markov chain. Choose a vertex $I$ uniformly and a uniform $U\in [0,1]$, independently, and set its spin to $+1$ whenand to $-1$ otherwise, leaving all other spins unchanged. This also specifies the holding probabilities omitted from the off-diagonal formulas. The conditional Ising law at $I$ is exactly this Bernoulli choice.

Order spin configurations coordinatewise with $-1<+1$. Since both the neighbour sum and $h$ are increasing, every map defined by a fixed $(I,U)$ preserves this order. For monotone coupling from the past, generate one two-sided independent sequence of such maps $F_n$, $n\in \mathbb{Z}$, and, for a horizon $T$, apply $F_{-T},\dots ,F_{-1}$ to the all-minus and all-plus states at time $-T$. If their time-zero images agree, every initial state has that same image by monotonicity. Return it. Otherwise enlarge $T$, for example by doubling it, retaining all maps already generated at existing negative times. Any earlier starting configuration is also forced to the same output once a horizon coalesces, so the result is consistent under further backward extension.

To prove termination, let $D={\max }_x\mathrm{deg}(x)$ andList the vertices $v_1,\dots ,v_m$. A block of $m$ updates choosing them in that order and having every uniform less than $a$ sets every vertex to $+1$, regardless of the starting configuration. Each choice has probability $1/m$, and every uniform condition has probability $a$. Thus a block is a universal reset with probabilityIndependent disjoint blocks in the past giveOnce such a block occurs, later maps preserve agreement. Therefore the backward search terminates in finite time almost surely, including the doubling implementation.

Finally we prove that the common output $Y$ has law $\pi$. For a deterministic horizon $T$, independently start another chain at time $-T$ with initial law $\pi$ and use the same maps. Its time-zero state $Z_0^{(T)}$ has law $\pi$ by stationarity. On the event that this deterministic horizon coalesces, $Z_0^{(T)}=Y$, so for every event $B\subseteq \Sigma$,Letting deterministic $T\to \infty$ provesThis argument justifies exact sampling without treating a random, data-dependent horizon as a deterministic stationary run.