Metropolis–Hastings flow is reversible and invariant

The acceptance function satisfies $0\le \alpha \le 1$. For each $x$, $0\le a(x):=\int \alpha (x,y)Q(x,dy)\le 1$ because $Q(x,\cdot )$ is a probability measure. The function $a$ is measurable by [F5], so $r(x)=1-a(x)$ is measurable and takes values in $[0,1]$.

For fixed $x$, the accepted term $A\mapsto \int {\mathbf{1}}_A(y)\alpha (x,y)Q(x,dy)$ is a measure by [F9], since the section $y\mapsto \alpha (x,y)$ is measurable by [F8]. The rejection term $A\mapsto r(x){\mathbf{1}}_A(x)$ is a measure: for disjoint sets at most one can contain $x$. Their sum is a measure by [F10], and its total mass is $a(x)+r(x)=1$. For each fixed $A$, the accepted term is measurable in $x$ by [F5], and the rejection term is measurable since $r$ and ${\mathbf{1}}_A$ are measurable. Thus [F2] applies pointwise and $P$ is a probability kernel.

Define $\delta (x)=(x,x)$. For every measurable rectangle $A\times B$, ${\delta }^{-1}(A\times B)=A\cap B\in \mathcal{X}$. Since these rectangles generate $\mathcal{X}\otimes \mathcal{X}$ by [F7], $\delta$ is measurable. By [F9], $\zeta (D):={\int }_{D}rd\pi$ is a measure on $(\mathsf{X},\mathcal{X})$; it is finite because $0\le r\le 1$ and $\pi (\mathsf{X})=1$. Define $R(C):=\zeta ({\delta }^{-1}C)=\int {\mathbf{1}}_C(x,x)r(x)\pi (dx)$. It vanishes at the empty set, and preimages of disjoint sets under $\delta$ are disjoint; countable additivity of $\zeta$ therefore makes $R$ a finite measure on the product sigma-algebra.

Step 2.1 has established that $P$ is a probability kernel, so [F3] now supplies the joint measure $J_P$. Applying the same joint-law result to the input kernel $Q$ gives the proposal flow $\mu$ and its iterated-integral identity. Thus, for every product-measurable $C$, $J_P(C)=\int \text{\hspace{0.17em}⁣}\int {\mathbf{1}}_C(x,y)\alpha (x,y)Q(x,dy)\pi (dx)+R(C)={\int }_C\alpha d\mu +R(C)=\eta (C)+R(C),$ where the last equality uses $\alpha \mu =\eta$ from [F4].

The coordinate swap $\tau (x,y)=(y,x)$ is measurable by [F7] and is an involution. Define ${\eta }^{\mathsf{T}}(C):=\eta ({\tau }^{-1}C)$. This is a measure: measurability of $\tau$ makes the preimages measurable, and preimages preserve the empty set and disjoint countable unions, so [F4, F13] give the measure properties. Since $\eta$ is the greatest measure dominated by both $\mu$ and ${\mu }^{\mathsf{T}}$, its transpose is also dominated by both: for every $C$, ${\eta }^{\mathsf{T}}(C)\le {\mu }^{\mathsf{T}}(C)$ and ${\eta }^{\mathsf{T}}(C)\le \mu (C)$. Maximality gives ${\eta }^{\mathsf{T}}\le \eta$; transposing this inequality and using ${\tau }^2=\mathrm{i}\mathrm{d}$ gives $\eta \le {\eta }^{\mathsf{T}}$. Hence ${\eta }^{\mathsf{T}}=\eta$.

Since $\tau \circ \delta =\delta$, step 2.2 gives $R(C)=R({\tau }^{-1}C)$ for every product-measurable $C$. Step 4.1 gives the same symmetry for $\eta$, so the decomposition in step 3.1 yields $J_P(C)=J_P({\tau }^{-1}C)$ for every $C$. Thus the joint flow is symmetric, so $P$ is reversible by [F11].

Steps 2.1 and 5.1 prove that $P$ is a probability kernel reversible with respect to $\pi$. The general-state result [F12] applies on this measurable space and gives $\int P(x,A)\pi (dx)=\pi (A)$ for every $A\in \mathcal{X}$. Thus $\pi$ is invariant in the sense of [F6].