[proofplan] We prove invariance first for a single coordinate update by testing against an arbitrary bounded measurable function. The defining property of the regular conditional distribution disintegrates $\pi$ into the marginal law of the deleted-coordinate projection $\operatorname{pr}_{-i}$ and the conditional kernel $\pi_i(x_{-i},\cdot)$. Since the Gibbs update resamples exactly from this same conditional law while keeping $x_{-i}$ fixed, the iterated integral after the update is the original integral under $\pi$. Random-scan invariance then follows from linearity, and systematic-scan invariance follows by repeated application of the one-coordinate invariance. [/proofplan]

[step:Disintegrate $\pi$ along the deleted coordinate projection]Fix $i\in\{1,\dots,d\}$. Define the marginal probability measure \begin{align*} \pi_{-i}:=\pi\circ \operatorname{pr}_{-i}^{-1} \end{align*} on $(E_{-i},\mathcal{E}_{-i})$. Let \begin{align*} I_i:E_{-i}\times E_i &\to E \end{align*} be the coordinate insertion map sending $(x_{-i},y_i)$ to the unique point of $E$ whose deleted-coordinate vector is $x_{-i}$ and whose $i$-th coordinate is $y_i$. Equivalently, if $x\in E$ satisfies $\operatorname{pr}_{-i}(x)=x_{-i}$, then $I_i(x_{-i},y_i)=x_{(i,y_i)}$. The map $I_i$ is measurable because it is a coordinate map between finite products of measurable spaces. For $A\in\mathcal{E}$, the set $I_i^{-1}(A)\in\mathcal{E}_{-i}\otimes\mathcal{E}_i$ represents the event that the pair consisting of the remaining coordinates and the $i$-th coordinate reconstructs a point in $A$. Since $\pi_i$ is the regular conditional distribution of $\operatorname{pr}_i$ given $\operatorname{pr}_{-i}$ under $\pi$, this gives the displayed identity first for indicator functions $f=\mathbb{1}_A$. The class of bounded $\mathcal{E}$-[measurable functions](/page/Measurable%20Functions) for which it holds is closed under bounded pointwise limits and linear combinations, so the [monotone class theorem](/theorems/4925) extends the identity to every bounded $\mathcal{E}$-measurable function $f:E\to\mathbb{R}$: \begin{align*} \int_E f(x)\,d\pi(x)=\int_{E_{-i}}\int_{E_i} f(I_i(x_{-i},y_i))\,d\pi_i(x_{-i},\cdot)(y_i)\,d\pi_{-i}(x_{-i}). \end{align*}[/step]

[guided]The purpose of this step is to write the joint law $\pi$ as “marginal law of the other coordinates” followed by “conditional law of the updated coordinate.” We first name the marginal law precisely. The projection $\operatorname{pr}_{-i}:E\to E_{-i}$ deletes the $i$-th coordinate, so the law of the remaining coordinates under $\pi$ is the pushforward probability measure \begin{align*} \pi_{-i}:=\pi\circ \operatorname{pr}_{-i}^{-1}. \end{align*} We also need notation for rebuilding a full point of $E$ after choosing a new $i$-th coordinate. Define \begin{align*} I_i:E_{-i}\times E_i &\to E \end{align*} to be the coordinate insertion map: $I_i(x_{-i},y_i)$ is the point whose non-$i$ coordinates are $x_{-i}$ and whose $i$-th coordinate is $y_i$. This map is measurable because each of its coordinate projections is one of the coordinate projections on $E_{-i}\times E_i$. Now apply the defining property of the regular conditional distribution $\pi_i$. For an indicator function $f=\mathbb{1}_A$ with $A\in\mathcal{E}$, the relevant event in $E_{-i}\times E_i$ is $I_i^{-1}(A)$, which is measurable because $I_i$ is measurable. This event says precisely that the pair made from the deleted-coordinate vector and the $i$-th coordinate reconstructs a point of $A$. The regular conditional distribution property gives the integral identity for these indicators. By the monotone class theorem, the same identity extends from indicators to all bounded $\mathcal{E}$-measurable functions $f:E\to\mathbb{R}$. Therefore conditioning first on $x_{-i}$ and then integrating the conditional law of the $i$-th coordinate gives \begin{align*} \int_E f(x)\,d\pi(x)=\int_{E_{-i}}\int_{E_i} f(I_i(x_{-i},y_i))\,d\pi_i(x_{-i},\cdot)(y_i)\,d\pi_{-i}(x_{-i}). \end{align*} This formula is exactly the bridge between the conditional distribution hypothesis and the Gibbs update kernel: both use the same conditional measure $\pi_i(x_{-i},\cdot)$.[/guided]

[step:Show that one Gibbs coordinate update preserves $\pi$]Let $f:E\to\mathbb{R}$ be bounded and $\mathcal{E}$-measurable. The action of $K_i$ on $f$ is the bounded measurable function \begin{align*} K_i f:E &\to \mathbb{R} \end{align*} defined by \begin{align*} K_i f(x):=\int_E f(z)\,dK_i(x,\cdot)(z). \end{align*} For each fixed $x\in E$, the map $A\mapsto K_i(x,A)$ is the pushforward of the probability measure $\pi_i(\operatorname{pr}_{-i}(x),\cdot)$ under the measurable map $y_i\mapsto I_i(\operatorname{pr}_{-i}(x),y_i)$, hence is a probability measure on $(E,\mathcal{E})$. For each fixed $A\in\mathcal{E}$, the map $x\mapsto K_i(x,A)$ is $\mathcal{E}$-measurable by the Markov-kernel measurability of $\pi_i$ and the measurability of $(x,y_i)\mapsto I_i(\operatorname{pr}_{-i}(x),y_i)$. Thus $K_i$ is a Markov kernel. By the definition of $K_i$, this equals \begin{align*} K_i f(x)=\int_{E_i} f(I_i(\operatorname{pr}_{-i}(x),y_i))\,d\pi_i(\operatorname{pr}_{-i}(x),\cdot)(y_i). \end{align*} Hence $K_i f(x)$ depends on $x$ only through $\operatorname{pr}_{-i}(x)$. Define \begin{align*} G_i:E_{-i} &\to \mathbb{R} \end{align*} by \begin{align*} G_i(x_{-i}):=\int_{E_i} f(I_i(x_{-i},y_i))\,d\pi_i(x_{-i},\cdot)(y_i). \end{align*} The Markov-kernel measurability of $\pi_i$ and bounded measurability of $f\circ I_i$ imply that $G_i$ is bounded and $\mathcal{E}_{-i}$-measurable. Then $K_i f=G_i\circ \operatorname{pr}_{-i}$, so $K_i f$ is bounded and $\mathcal{E}$-measurable, and the definition of the pushforward measure $\pi_{-i}$ gives \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_{E_{-i}}G_i(x_{-i})\,d\pi_{-i}(x_{-i}). \end{align*} Substituting the definition of $G_i$ and using the disintegration identity from the previous step, \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_{E_{-i}}\int_{E_i} f(I_i(x_{-i},y_i))\,d\pi_i(x_{-i},\cdot)(y_i)\,d\pi_{-i}(x_{-i}). \end{align*} Therefore \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_E f(x)\,d\pi(x). \end{align*} Since this holds for every bounded $\mathcal{E}$-measurable $f:E\to\mathbb{R}$, the measures $\pi K_i$ and $\pi$ agree. Thus $\pi K_i=\pi$.[/step]

[guided]Fix a bounded $\mathcal{E}$-measurable [test function](/page/Test%20Function) $f:E\to\mathbb{R}$. Proving $\pi K_i=\pi$ is equivalent to proving equality of integrals against every such $f$: \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_E f(x)\,d\pi(x). \end{align*} The function obtained after applying the kernel is \begin{align*} K_i f:E &\to \mathbb{R} \end{align*} with \begin{align*} K_i f(x):=\int_E f(z)\,dK_i(x,\cdot)(z). \end{align*} Using the definition of the Gibbs update kernel, the measure $K_i(x,\cdot)$ is obtained by keeping $\operatorname{pr}_{-i}(x)$ fixed and sampling a fresh $i$-th coordinate $y_i$ from $\pi_i(\operatorname{pr}_{-i}(x),\cdot)$. Therefore \begin{align*} K_i f(x)=\int_{E_i} f(I_i(\operatorname{pr}_{-i}(x),y_i))\,d\pi_i(\operatorname{pr}_{-i}(x),\cdot)(y_i). \end{align*} This expression depends on $x$ only through the deleted-coordinate vector $\operatorname{pr}_{-i}(x)$. To make that dependence explicit, define \begin{align*} G_i:E_{-i} &\to \mathbb{R} \end{align*} by \begin{align*} G_i(x_{-i}):=\int_{E_i} f(I_i(x_{-i},y_i))\,d\pi_i(x_{-i},\cdot)(y_i). \end{align*} The Markov-kernel measurability of $\pi_i$ and bounded measurability of $f\circ I_i$ imply that $G_i$ is bounded and $\mathcal{E}_{-i}$-measurable. We have $K_i f=G_i\circ \operatorname{pr}_{-i}$. Now integrate with respect to $\pi$. Since $\pi_{-i}$ is the pushforward of $\pi$ by $\operatorname{pr}_{-i}$, the pushforward identity gives \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_E G_i(\operatorname{pr}_{-i}(x))\,d\pi(x). \end{align*} Thus \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_{E_{-i}}G_i(x_{-i})\,d\pi_{-i}(x_{-i}). \end{align*} Substitute the definition of $G_i$: \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_{E_{-i}}\int_{E_i} f(I_i(x_{-i},y_i))\,d\pi_i(x_{-i},\cdot)(y_i)\,d\pi_{-i}(x_{-i}). \end{align*} The right-hand side is exactly the disintegration formula for $\int_E f\,d\pi$. Hence \begin{align*} \int_E K_i f(x)\,d\pi(x)=\int_E f(x)\,d\pi(x). \end{align*} Because bounded measurable test functions determine probability measures on $(E,\mathcal{E})$, this proves $\pi K_i=\pi$.[/guided]

[step:Use linearity to prove random-scan invariance] Let \begin{align*} K:=\sum_{i=1}^d p_iK_i. \end{align*} For every bounded $\mathcal{E}$-measurable function $f:E\to\mathbb{R}$, \begin{align*} Kf(x)=\sum_{i=1}^d p_iK_i f(x) \end{align*} for every $x\in E$. Therefore, using finiteness of the sum and the already proved identity $\pi K_i=\pi$, \begin{align*} \int_E Kf(x)\,d\pi(x)=\sum_{i=1}^d p_i\int_E K_i f(x)\,d\pi(x). \end{align*} Thus \begin{align*} \int_E Kf(x)\,d\pi(x)=\sum_{i=1}^d p_i\int_E f(x)\,d\pi(x). \end{align*} Since $\sum_{i=1}^d p_i=1$, \begin{align*} \int_E Kf(x)\,d\pi(x)=\int_E f(x)\,d\pi(x). \end{align*} Hence $\pi K=\pi$. [/step]

[step:Apply one-coordinate invariance repeatedly for the systematic scan] Let \begin{align*} K_{\mathrm{sys}}:=K_1K_2\cdots K_d. \end{align*} We use the standard convention that the product of kernels is defined by associating the measure action from left to right: \begin{align*} \pi K_{\mathrm{sys}}:=(((\pi K_1)K_2)\cdots K_d). \end{align*} Since $\pi K_i=\pi$ for every $i\in\{1,\dots,d\}$, repeated substitution gives \begin{align*} \pi K_{\mathrm{sys}}=(((\pi K_1)K_2)\cdots K_d)=((\pi K_2)\cdots K_d). \end{align*} Continuing through the finite product yields \begin{align*} \pi K_{\mathrm{sys}}=\pi. \end{align*} Thus $\pi$ is invariant for the systematic-scan Gibbs kernel. This completes the proof. [/step]