Each exact Gibbs block update is reversible and invariant

The full conditional identity [F1] says that the target law in coordinates $(z,u)$ and the iterated law ${\pi }_{-B}(dz)K_B(z,du)$ agree on coordinate rectangles. Both are probability measures, and rectangles generate the finite product sigma-algebra, so [F8] makes them equal on every product-measurable event. Integrating nonnegative simple functions and then increasing limits gives $\int h(x)\pi (dx)=\int {\pi }_{-B}(dz)\int h({\mathrm{splice}}_B(z,u))K_B(z,du)$ for every nonnegative product-measurable $h$.

Let $M$ be the joint law of the current and updated states under initial law $\pi$ and transition $P_B$, which exists by [F3]–[F4]. For arbitrary measurable $A,C\subseteq S$, the update formula and step 1.1 give $M(A\times C)=\int {\pi }_{-B}(dz)\int K_B(z,du){\mathbf{1}}_A({\mathrm{splice}}_B(z,u))\int K_B(z,dv){\mathbf{1}}_C({\mathrm{splice}}_B(z,v))$. All integrands are bounded measurable functions, so the iterated integrals are defined.

For each $z$, the product measure $K_B(z,\cdot )\otimes K_B(z,\cdot )$ is invariant under swapping its two coordinates: the product measure and its swapped image agree on every rectangle because $K_B(z,A)K_B(z,C)=K_B(z,C)K_B(z,A)$, hence agree on the generated product sigma-algebra by [F8]. Applying this symmetry to the two inner factors in step 2.1 yields $M(A\times C)=M(C\times A)$ for arbitrary $A,C$. By [F6], $P_B$ is reversible with respect to $\pi$.

Set $C=S$ in the detailed-balance identity of step 3.1. Since $P_B(x,S)=1$, its left side is $\pi (A)$ and its right side is $\int P_B(y,A)\pi (dy)=(\pi P_B)(A)$. This is invariance by [F7], for every measurable $A$.