[proofplan] The proof separates the delayed-acceptance transition into its off-diagonal proposal part and its diagonal rejection part. The central calculation is a pointwise symmetry identity for the off-diagonal flow $\gamma(x)q(x,y)\alpha_1(x,y)\alpha_2(x,y)$, with proposal-zero cases handled separately. Once this flow is symmetric under exchanging $x$ and $y$, integration over rectangles gives detailed balance for the proposal part, while the rejection part is supported on the diagonal and is automatically symmetric. Finally, integrating detailed balance in one variable gives invariance. [/proofplan]

[step:Verify that the displayed formula defines a Markov kernel] The functions $\alpha_1$ and $\alpha_2$ are $\mathcal E\otimes\mathcal E$-measurable because they are obtained from measurable positive functions and the measurable operation $(u,v)\mapsto \min\{u,v\}$, with $\alpha_1$ defined piecewise on the measurable set \begin{align*} \{(x,y)\in E\times E:q(x,y)q(y,x)>0\}. \end{align*} Moreover $0\leq \alpha_1\leq 1$ and $0\leq \alpha_2\leq 1$, so for every $x\in E$, \begin{align*} 0\leq \int_E q(x,z)\alpha_1(x,z)\alpha_2(x,z)\,d\lambda(z)\leq \int_E q(x,z)\,d\lambda(z)=1. \end{align*} Thus the rejection coefficient is nonnegative. For fixed $x\in E$, the map $A\mapsto P(x,A)$ is the sum of the finite measure with density $y\mapsto q(x,y)\alpha_1(x,y)\alpha_2(x,y)$ with respect to $\lambda$ and the nonnegative multiple of the Dirac measure at $x$. Hence it is a probability measure because \begin{align*} P(x,E)=\int_E q(x,y)\alpha_1(x,y)\alpha_2(x,y)\,d\lambda(y)+1-\int_E q(x,z)\alpha_1(x,z)\alpha_2(x,z)\,d\lambda(z)=1. \end{align*} For fixed $A\in\mathcal E$, $\sigma$-finiteness of $\lambda$ and Tonelli's measurability theorem imply that \begin{align*} x\mapsto \int_A q(x,y)\alpha_1(x,y)\alpha_2(x,y)\,d\lambda(y) \end{align*} is $\mathcal E$-measurable; the diagonal term $x\mapsto \mathbb 1_A(x)$ is also measurable. Therefore $x\mapsto P(x,A)$ is measurable, and $P$ is a Markov kernel. [/step]

[step:Prove the pointwise symmetry of the accepted proposal flow]Define the accepted proposal flow density $h:E\times E\to[0,\infty)$ by \begin{align*} h(x,y):=\gamma(x)q(x,y)\alpha_1(x,y)\alpha_2(x,y). \end{align*} We prove that \begin{align*} h(x,y)=h(y,x) \end{align*} for every $(x,y)\in E\times E$. If $q(x,y)q(y,x)=0$, then by definition $\alpha_1(x,y)=0$ and $\alpha_1(y,x)=0$, since the same product appears after interchanging $x$ and $y$. Hence $h(x,y)=0=h(y,x)$. Assume now that $q(x,y)q(y,x)>0$. Let \begin{align*} S:=\{(u,v)\in E\times E:q(u,v)q(v,u)>0\}. \end{align*} Define the auxiliary maps $A:S\to(0,\infty)$ and $B:S\to(0,\infty)$ by \begin{align*} A(u,v):=\frac{\widetilde\gamma(v)q(v,u)}{\widetilde\gamma(u)q(u,v)} \end{align*} and \begin{align*} B(u,v):=\frac{\gamma(v)\widetilde\gamma(u)}{\gamma(u)\widetilde\gamma(v)}. \end{align*} Then \begin{align*} A(x,y)B(x,y)=\frac{\gamma(y)q(y,x)}{\gamma(x)q(x,y)}. \end{align*} Also $A(y,x)=A(x,y)^{-1}$ and $B(y,x)=B(x,y)^{-1}$. For every positive number $a$, the identity \begin{align*} \min\{1,a\}=a\min\{1,a^{-1}\} \end{align*} holds by checking the two cases $a\leq 1$ and $a\geq 1$. Applying this identity first with $a=A(x,y)$ and then with $a=B(x,y)$ gives \begin{align*} \alpha_1(x,y)\alpha_2(x,y)=A(x,y)B(x,y)\alpha_1(y,x)\alpha_2(y,x). \end{align*} Multiplying by $\gamma(x)q(x,y)$ yields \begin{align*} \gamma(x)q(x,y)\alpha_1(x,y)\alpha_2(x,y)=\gamma(y)q(y,x)\alpha_1(y,x)\alpha_2(y,x). \end{align*} Thus $h(x,y)=h(y,x)$ in the positive proposal case as well.[/step]

[guided]The accepted proposal flow from $x$ to $y$ is the exact target weight at $x$, multiplied by the proposal density from $x$ to $y$, multiplied by both acceptance probabilities. We therefore define \begin{align*} h(x,y):=\gamma(x)q(x,y)\alpha_1(x,y)\alpha_2(x,y). \end{align*} The goal is to prove that this quantity is unchanged when $x$ and $y$ are interchanged. First consider the case $q(x,y)q(y,x)=0$. The first-stage acceptance probability was defined with this case separated precisely to avoid division by zero. Since the product $q(x,y)q(y,x)$ is symmetric in $x$ and $y$, the same zero-product condition holds after swapping the two variables. Hence \begin{align*} \alpha_1(x,y)=0 \end{align*} and \begin{align*} \alpha_1(y,x)=0. \end{align*} Therefore \begin{align*} h(x,y)=0=h(y,x). \end{align*} Now assume $q(x,y)q(y,x)>0$. This ensures every quotient below is well-defined and positive, since $\gamma$ and $\widetilde\gamma$ take values in $(0,\infty)$. Let \begin{align*} S:=\{(u,v)\in E\times E:q(u,v)q(v,u)>0\}. \end{align*} Define the auxiliary maps $A:S\to(0,\infty)$ and $B:S\to(0,\infty)$ by \begin{align*} A(u,v):=\frac{\widetilde\gamma(v)q(v,u)}{\widetilde\gamma(u)q(u,v)} \end{align*} and \begin{align*} B(u,v):=\frac{\gamma(v)\widetilde\gamma(u)}{\gamma(u)\widetilde\gamma(v)}. \end{align*} The first ratio $A(x,y)$ is the surrogate Metropolis-Hastings ratio, while $B(x,y)$ is the correction from the surrogate density back to the exact density. Their product cancels the surrogate terms: \begin{align*} A(x,y)B(x,y)=\frac{\widetilde\gamma(y)q(y,x)}{\widetilde\gamma(x)q(x,y)}\frac{\gamma(y)\widetilde\gamma(x)}{\gamma(x)\widetilde\gamma(y)}=\frac{\gamma(y)q(y,x)}{\gamma(x)q(x,y)}. \end{align*} After swapping $x$ and $y$, the two ratios invert: \begin{align*} A(y,x)=A(x,y)^{-1} \end{align*} and \begin{align*} B(y,x)=B(x,y)^{-1}. \end{align*} The elementary minimum identity we need is \begin{align*} \min\{1,a\}=a\min\{1,a^{-1}\} \end{align*} for every $a>0$. If $0<a\leq 1$, both sides equal $a$. If $a\geq 1$, both sides equal $1$. Applying this identity to $A(x,y)$ gives \begin{align*} \alpha_1(x,y)=A(x,y)\alpha_1(y,x), \end{align*} and applying it to $B(x,y)$ gives \begin{align*} \alpha_2(x,y)=B(x,y)\alpha_2(y,x). \end{align*} Multiplying these two equalities, \begin{align*} \alpha_1(x,y)\alpha_2(x,y)=A(x,y)B(x,y)\alpha_1(y,x)\alpha_2(y,x). \end{align*} Finally multiply by $\gamma(x)q(x,y)$ and substitute the product identity for $A(x,y)B(x,y)$: \begin{align*} \gamma(x)q(x,y)\alpha_1(x,y)\alpha_2(x,y)=\gamma(x)q(x,y)\frac{\gamma(y)q(y,x)}{\gamma(x)q(x,y)}\alpha_1(y,x)\alpha_2(y,x). \end{align*} Cancelling the positive factor $\gamma(x)q(x,y)$ gives \begin{align*} \gamma(x)q(x,y)\alpha_1(x,y)\alpha_2(x,y)=\gamma(y)q(y,x)\alpha_1(y,x)\alpha_2(y,x). \end{align*} This is exactly the desired symmetry $h(x,y)=h(y,x)$.[/guided]

[step:Integrate the symmetric flow to obtain off-diagonal detailed balance] Define the finite measure $\mu_{\mathrm{acc}}$ on $(E\times E,\mathcal E\otimes\mathcal E)$ by \begin{align*} \mu_{\mathrm{acc}}(C):=Z^{-1}\int_E\int_E \mathbb 1_C(x,y)h(x,y)\,d\lambda(y)\,d\lambda(x), \end{align*} for $C\in\mathcal E\otimes\mathcal E$. This measure is finite because \begin{align*} \mu_{\mathrm{acc}}(E\times E)\leq Z^{-1}\int_E\int_E \gamma(x)q(x,y)\,d\lambda(y)\,d\lambda(x)=Z^{-1}\int_E\gamma(x)\,d\lambda(x)=1. \end{align*} The pointwise identity $h(x,y)=h(y,x)$ and Tonelli's theorem imply that $\mu_{\mathrm{acc}}$ is invariant under the coordinate-swap map $(x,y)\mapsto(y,x)$. Therefore, for all $A,B\in\mathcal E$, \begin{align*} \mu_{\mathrm{acc}}(A\times B)=\mu_{\mathrm{acc}}(B\times A). \end{align*} Equivalently, \begin{align*} \int_A\int_B q(x,y)\alpha_1(x,y)\alpha_2(x,y)\,d\lambda(y)\,d\pi(x) = \int_B\int_A q(y,x)\alpha_1(y,x)\alpha_2(y,x)\,d\lambda(x)\,d\pi(y). \end{align*} [/step]

[step:Add the diagonal rejection measure without using a density] Define the measurable rejection probability $\rho:E\to[0,1]$ by \begin{align*} \rho(x):=1-\int_E q(x,z)\alpha_1(x,z)\alpha_2(x,z)\,d\lambda(z). \end{align*} Define the diagonal rejection measure $\mu_{\mathrm{rej}}$ on $E\times E$ by \begin{align*} \mu_{\mathrm{rej}}(C):=\int_E \mathbb 1_C(x,x)\rho(x)\,d\pi(x), \end{align*} for $C\in\mathcal E\otimes\mathcal E$. This formulation treats the diagonal mass as a measure supported on $\{(x,x):x\in E\}$ and does not require any density with respect to $\lambda\otimes\lambda$. For $A,B\in\mathcal E$, \begin{align*} \mu_{\mathrm{rej}}(A\times B)=\int_E \mathbb 1_A(x)\mathbb 1_B(x)\rho(x)\,d\pi(x)=\int_{A\cap B}\rho(x)\,d\pi(x). \end{align*} The same expression is obtained with $A$ and $B$ interchanged, so \begin{align*} \mu_{\mathrm{rej}}(A\times B)=\mu_{\mathrm{rej}}(B\times A). \end{align*} [/step]

[step:Combine accepted and rejected parts to prove reversibility] For $A,B\in\mathcal E$, decompose the joint transition measure as \begin{align*} \int_A P(x,B)\,d\pi(x) = \int_A\int_B q(x,y)\alpha_1(x,y)\alpha_2(x,y)\,d\lambda(y)\,d\pi(x) + \int_{A\cap B}\rho(x)\,d\pi(x). \end{align*} The first term is symmetric in $A$ and $B$ by the accepted-flow calculation, and the second term is symmetric in $A$ and $B$ because it is an integral over $A\cap B$. Hence \begin{align*} \int_A P(x,B)\,d\pi(x)=\int_B P(y,A)\,d\pi(y). \end{align*} Thus $P$ is reversible with respect to $\pi$. [/step]

[step:Integrate detailed balance to conclude invariance] Set $B=E$ in the detailed balance identity. Since $P(y,E)=1$ for every $y\in E$, we obtain, for every $A\in\mathcal E$, \begin{align*} \int_A P(x,E)\,d\pi(x)=\int_E P(y,A)\,d\pi(y). \end{align*} The left-hand side is \begin{align*} \int_A 1\,d\pi(x)=\pi(A). \end{align*} Therefore \begin{align*} \int_E P(y,A)\,d\pi(y)=\pi(A), \end{align*} which is precisely invariance of $\pi$ for $P$. [/step]