The common-density Metropolis–Hastings ratio

Put $\mathrm{\Lambda }=\lambda \otimes \lambda$. The lift $(x,y)\mapsto p(x)$ is product-measurable because the preimage of each Borel set $B$ is the rectangle $p^{-1}(B)\times \mathsf{X}$. Each section $C_x$ is measurable: for fixed $x$, the class of all $D\subseteq {\mathsf{X}}^2$ with $D_x\in \mathcal{X}$ is a sigma-algebra containing every measurable rectangle, so it contains $\mathcal{X}\otimes \mathcal{X}$. Since $q$ and ${\mathbf{1}}_C$ are product-measurable, [F6] makes their product measurable, and [F5] makes $H(x):={\int }_{\mathsf{X}}{\mathbf{1}}_C(x,y)q(x,y)\lambda (dy)=Q(x,C_x)$ measurable, where the equality is the given density identity for $Q(x,\cdot )$. The density identity for $\pi$ and [F10], first for simple functions and then for $H$ by [F9] and monotone convergence, give $\mu (C)={\int }_{\mathsf{X}}H(x)\pi (dx)={\int }_{\mathsf{X}}H(x)p(x)\lambda (dx)\mathrm{.}$ Substituting the expression for $H$ and applying Tonelli [F5] yields $\mu (C)={\int }_{\mathsf{X}}\text{\hspace{0.17em}⁣}{\int }_{\mathsf{X}}{\mathbf{1}}_C(x,y)p(x)q(x,y)\lambda (dy)\lambda (dx)={\int }_{C}md\mathrm{\Lambda }\mathrm{.}$ Thus $m$ is a density of the forward probability flow with respect to $\mathrm{\Lambda }$.

Let $\tau (x,y)=(y,x)$ and define ${\mathrm{\Lambda }}^{\mathsf{T}}(C)=\mathrm{\Lambda }({\tau }^{-1}C)$. By [F7], $\tau$ is measurable, so ${\mathrm{\Lambda }}^{\mathsf{T}}$ is a measure. On a rectangle $A\times B$, ${\mathrm{\Lambda }}^{\mathsf{T}}(A\times B)=\lambda (B)\lambda (A)=\lambda (A)\lambda (B)=\mathrm{\Lambda }(A\times B),$ using the product-measure convention $0\cdot \mathrm{\infty }=0$. Choose increasing measurable $X_n\uparrow \mathsf{X}$ with $\lambda (X_n)<\mathrm{\infty }$. The rectangles $X_n\times X_n$ increase to ${\mathsf{X}}^2$ and have equal finite measure for ${\mathrm{\Lambda }}^{\mathsf{T}}$ and $\mathrm{\Lambda }$. Measurable rectangles form a generating pi-system, so [F3]–[F4] give ${\mathrm{\Lambda }}^{\mathsf{T}}=\mathrm{\Lambda }$. For every nonnegative product-measurable $u$, the identity ${\int }_{{\tau }^{-1}C}ud\mathrm{\Lambda }={\int }_{C}u\circ \tau d\mathrm{\Lambda }$ holds first for indicators from ${\mathrm{\Lambda }}^{\mathsf{T}}=\mathrm{\Lambda }$, then for simple functions by finite addition, and for all such $u$ by [F9]. Applying it to $u=m$ yields ${\mu }^{\mathsf{T}}(C)=\mu ({\tau }^{-1}C)={\int }_{C}m(y,x)d\mathrm{\Lambda }\mathrm{.}$ Hence $m^{\mathsf{T}}(x,y)=p(y)q(y,x)$ is the reverse-flow density.

Apply [F8] to the finite probability measures $\mu ,{\mu }^{\mathsf{T}}$ and the sigma-finite reference measure $\mathrm{\Lambda }$. Their greatest common submeasure $\eta =\mu \wedge {\mu }^{\mathsf{T}}$ has density $h=\min \{m,m^{\mathsf{T}}\}$, so $\eta (C)={\int }_{C}hd\mathrm{\Lambda }(C\in \mathcal{X}\otimes \mathcal{X})\mathrm{.}$

Define ${\alpha }_{\mathrm{c}\mathrm{d}}=h\mathrm{/}m$ on $\{m>0\}$ and ${\alpha }_{\mathrm{c}\mathrm{d}}=1$ on $\{m=0\}$. It takes values in $[0,1]$. For a real $c<0$, $\{{\alpha }_{\mathrm{c}\mathrm{d}}>c\}={\mathsf{X}}^2$; for $0\le c<1$, $\{{\alpha }_{\mathrm{c}\mathrm{d}}>c\}=\{m=0\}\cup (\{m>0\}\cap \{h-cm>0\});$ for $c\ge 1$ the set is empty. Each set is measurable by [F6], so ${\alpha }_{\mathrm{c}\mathrm{d}}$ is jointly measurable. By its definition, ${\alpha }_{\mathrm{c}\mathrm{d}}m=h$ everywhere. For any $C$, choose nonnegative simple $s_n\uparrow {\alpha }_{\mathrm{c}\mathrm{d}}{\mathbf{1}}_C$. The simple density identity [F10] gives $\int s_{n}d\mu =\int s_{n}md\mathrm{\Lambda }$; monotone convergence on both sides then gives ${\int }_C{\alpha }_{\mathrm{c}\mathrm{d}}d\mu ={\int }_C{\alpha }_{\mathrm{c}\mathrm{d}}md\mathrm{\Lambda }={\int }_{C}hd\mathrm{\Lambda }=\eta (C)\mathrm{.}$ Thus ${\alpha }_{\mathrm{c}\mathrm{d}}$ is a version of $d\eta \mathrm{/}d\mu$.

On $\{m>0\}$, algebra gives ${\alpha }_{\mathrm{c}\mathrm{d}}=\frac{\min \{m,m^{\mathsf{T}}\}}{m}=\min \{1,\frac{p(y)q(y,x)}{p(x)q(x,y)}\}\mathrm{.}$ On $\{m=0\}$ its specified value is $1$, which is permissible because $\mu (\{m=0\})={\int }_{\{m=0\}}md\mathrm{\Lambda }=0$. The algorithm's chosen derivative and ${\alpha }_{\mathrm{c}\mathrm{d}}$ are two versions of the same Radon–Nikodym derivative, so [F11] makes them equal $\mu$-almost everywhere. This proves the common-density ratio and the forward-zero convention without asserting pointwise uniqueness on null sets.