[proofplan] The proof is the deterministic Metropolis argument for an involutive proposal. First we prove the pointwise symmetry identity saying that the accepted-move density $e^{-H(z)}\alpha(z)$ is invariant under the involution $\Psi$. Then we verify detailed balance for Borel sets by splitting the kernel into its accepted-move part and its rejection part; the accepted part is symmetric by the measure-preserving property of $\Psi$, and the rejection part is symmetric because it stays on the diagonal. Finally, invariance of the position kernel follows by integrating the phase-space invariance of $M$ and disintegrating $\widetilde{\pi}$ into $\pi(dx)\kappa_x(dp)$. [/proofplan]

[step:Verify measurability of the Metropolized kernels] Since $H: \mathbb{R}^{2d} \to \mathbb{R}$ and $\Psi: \mathbb{R}^{2d} \to \mathbb{R}^{2d}$ are Borel measurable, the composition $H \circ \Psi: \mathbb{R}^{2d} \to \mathbb{R}$ is Borel measurable. Therefore the function $\alpha: \mathbb{R}^{2d} \to [0,1]$ defined by \begin{align*}\alpha(z) := \min\{1,\exp(-H(\Psi(z)) + H(z))\}\end{align*} is Borel measurable. For each Borel set $E \subset \mathbb{R}^{2d}$, the map $z \mapsto \delta_{\Psi(z)}(E)$ is the indicator function $\mathbb{1}_{\Psi^{-1}(E)}(z)$, and $z \mapsto \delta_z(E)$ is the indicator function $\mathbb{1}_E(z)$. Hence $z \mapsto M(z,E)$ is Borel measurable. For each fixed $z \in \mathbb{R}^{2d}$, $M(z,\cdot)$ is a probability measure because it is a convex combination of two Dirac probability measures with weights $\alpha(z)$ and $1-\alpha(z)$. Thus $M$ is a Markov kernel. For a Borel set $C \subset \mathbb{R}^d$, the map $(x,p) \mapsto M((x,p), C \times \mathbb{R}^d)$ is Borel measurable and bounded. The space $\mathbb{R}^d$ with its Borel $\sigma$-algebra is a standard Borel space, and the statement supplies $(\kappa_x)_{x \in \mathbb{R}^d}$ as a Markov kernel realizing the regular conditional distribution. Define the bounded Borel map $h_C: \mathbb{R}^d \times \mathbb{R}^d \to [0,1]$ by \begin{align*}h_C(x,p) := M((x,p), C \times \mathbb{R}^d).\end{align*} For a Markov kernel, integration of a bounded jointly Borel function against the kernel gives a Borel function of the base point; this follows from the defining measurability on indicator functions by the standard monotone-class extension. Applying this property to $h_C$ and $(\kappa_x)_{x \in \mathbb{R}^d}$ shows that the map \begin{align*}x \mapsto \int_{\mathbb{R}^d} h_C(x,p) \, d\kappa_x(p)\end{align*} is Borel measurable. For each fixed $x \in \mathbb{R}^d$, the set function $C \mapsto M((x,p), C \times \mathbb{R}^d)$ is a probability measure for each $p \in \mathbb{R}^d$, and integrating these probability measures against the probability measure $\kappa_x$ preserves total mass and countable additivity by monotone convergence for nonnegative sums. Therefore $P$ is a Markov kernel on $\mathbb{R}^d$. [/step]

[step:Prove the accepted-move density is invariant under the involution]Define $r: \mathbb{R}^{2d} \to (0,\infty)$ by \begin{align*}r(z) := e^{-H(z)}.\end{align*} For every $z \in \mathbb{R}^{2d}$, the acceptance rule gives \begin{align*}r(z)\alpha(z) = e^{-H(z)}\min\{1,e^{-H(\Psi(z)) + H(z)}\}.\end{align*} Thus \begin{align*}r(z)\alpha(z) = \min\{e^{-H(z)},e^{-H(\Psi(z))}\}.\end{align*} Since $\Psi$ is an involution, $\Psi(\Psi(z))=z$, and therefore \begin{align*}r(\Psi(z))\alpha(\Psi(z)) = \min\{e^{-H(\Psi(z))},e^{-H(z)}\}.\end{align*} Hence \begin{align*}r(z)\alpha(z) = r(\Psi(z))\alpha(\Psi(z)).\end{align*} for every $z \in \mathbb{R}^{2d}$.[/step]

[guided]The accepted part of a Metropolis kernel has to be symmetric after weighting by the target density. Here the target density with respect to $\mathcal{L}^{2d}$ is $Z^{-1}e^{-H(z)}$, so the factor that matters is $e^{-H(z)}\alpha(z)$. Define the Borel measurable function $r: \mathbb{R}^{2d} \to (0,\infty)$ by \begin{align*}r(z) := e^{-H(z)}.\end{align*} Using the definition of $\alpha$, we compute \begin{align*}r(z)\alpha(z) = e^{-H(z)}\min\{1,e^{-H(\Psi(z)) + H(z)}\}.\end{align*} Multiplying the two alternatives inside the minimum by $e^{-H(z)}$ gives \begin{align*}r(z)\alpha(z) = \min\{e^{-H(z)},e^{-H(\Psi(z))}\}.\end{align*} This is the key Metropolis symmetry: the accepted flux from $z$ to $\Psi(z)$ is the smaller of the two unnormalised target densities. Now apply the same computation at $\Psi(z)$. Because $\Psi$ is an involution, $\Psi(\Psi(z))=z$, so \begin{align*}r(\Psi(z))\alpha(\Psi(z)) = \min\{e^{-H(\Psi(z))},e^{-H(\Psi(\Psi(z)))}\}.\end{align*} Using $\Psi(\Psi(z))=z$, this becomes \begin{align*}r(\Psi(z))\alpha(\Psi(z)) = \min\{e^{-H(\Psi(z))},e^{-H(z)}\}.\end{align*} The minimum is symmetric in its two arguments, so \begin{align*}r(z)\alpha(z) = r(\Psi(z))\alpha(\Psi(z)).\end{align*} This identity is the whole reason the Metropolis acceptance probability has exactly this form.[/guided]

[step:Show the accepted-move part satisfies detailed balance] Let $\mathcal{B}(\mathbb{R}^{2d})$ denote the Borel $\sigma$-algebra on $\mathbb{R}^{2d}$. Define the accepted-move contribution as the map \begin{align*}I: \mathcal{B}(\mathbb{R}^{2d}) \times \mathcal{B}(\mathbb{R}^{2d}) \to [0,1]\end{align*} given, for Borel sets $A,B \subset \mathbb{R}^{2d}$, by \begin{align*}I(A,B) := \int_A \alpha(z)\delta_{\Psi(z)}(B) \, d\widetilde{\pi}(z).\end{align*} Since $\delta_{\Psi(z)}(B)=\mathbb{1}_{\Psi^{-1}(B)}(z)$, we have \begin{align*}I(A,B) = Z^{-1}\int_{A \cap \Psi^{-1}(B)} r(z)\alpha(z) \, d\mathcal{L}^{2d}(z).\end{align*} Let $g: \mathbb{R}^{2d} \to [0,\infty)$ be defined by \begin{align*} g(z) := r(z)\alpha(z). \end{align*} The previous step gives $g \circ \Psi = g$. We first record the integration-invariance consequence of the measure-preserving involution. If $S \subset \mathbb{R}^{2d}$ is Borel, then $\Psi(S)=\Psi^{-1}(S)$ because $\Psi$ is an involution, and this set is Borel because $\Psi$ is Borel measurable. For a Borel set $E \subset \mathbb{R}^{2d}$, measure preservation gives \begin{align*}\int_{\mathbb{R}^{2d}} \mathbb{1}_E(\Psi(z)) \, d\mathcal{L}^{2d}(z) = \mathcal{L}^{2d}(\Psi^{-1}(E)) = \mathcal{L}^{2d}(E) = \int_{\mathbb{R}^{2d}} \mathbb{1}_E(z) \, d\mathcal{L}^{2d}(z).\end{align*} By linearity this identity holds for nonnegative simple Borel functions, and by the [monotone convergence theorem](/theorems/509) it holds for every nonnegative Borel function $f: \mathbb{R}^{2d} \to [0,\infty]$: \begin{align*}\int_{\mathbb{R}^{2d}} f(\Psi(z)) \, d\mathcal{L}^{2d}(z) = \int_{\mathbb{R}^{2d}} f(w) \, d\mathcal{L}^{2d}(w).\end{align*} Define the nonnegative Borel map $f: \mathbb{R}^{2d} \to [0,\infty]$ by $f(z) := g(z)\mathbb{1}_{\Psi(A \cap \Psi^{-1}(B))}(z)$. Apply the integration-invariance identity to this function $f$. Since $g \circ \Psi = g$ and $\Psi^{-1}(\Psi(A \cap \Psi^{-1}(B)))=A \cap \Psi^{-1}(B)$, we obtain \begin{align*}\int_{A \cap \Psi^{-1}(B)} g(z) \, d\mathcal{L}^{2d}(z) = \int_{\Psi(A \cap \Psi^{-1}(B))} g(w) \, d\mathcal{L}^{2d}(w).\end{align*} Since $\Psi$ is an involution, \begin{align*}\Psi(A \cap \Psi^{-1}(B)) = \Psi(A) \cap B = B \cap \Psi^{-1}(A).\end{align*} Therefore \begin{align*}I(A,B) = Z^{-1}\int_{B \cap \Psi^{-1}(A)} g(w) \, d\mathcal{L}^{2d}(w) = I(B,A).\end{align*} [/step]

[step:Show the rejection part satisfies detailed balance] Let $\mathcal{B}(\mathbb{R}^{2d})$ denote the Borel $\sigma$-algebra on $\mathbb{R}^{2d}$. Define the rejection contribution as the map $R: \mathcal{B}(\mathbb{R}^{2d}) \times \mathcal{B}(\mathbb{R}^{2d}) \to [0,1]$ given, for Borel sets $A,B \subset \mathbb{R}^{2d}$, by $R(A,B) := \int_A (1-\alpha(z))\delta_z(B) \, d\widetilde{\pi}(z)$. Since $\delta_z(B)=\mathbb{1}_B(z)$, we obtain \begin{align*}R(A,B) = \int_{A \cap B} (1-\alpha(z)) \, d\widetilde{\pi}(z).\end{align*} The set $A \cap B$ is symmetric in $A$ and $B$, so \begin{align*}R(A,B) = R(B,A).\end{align*} [/step]

[step:Combine accepted and rejected moves to obtain phase-space reversibility] For Borel sets $A,B \subset \mathbb{R}^{2d}$, the definition of $M$ gives \begin{align*}\int_A M(z,B) \, d\widetilde{\pi}(z) = I(A,B) + R(A,B).\end{align*} The accepted-move symmetry gives $I(A,B)=I(B,A)$, and the rejection symmetry gives $R(A,B)=R(B,A)$. Hence \begin{align*}\int_A M(z,B) \, d\widetilde{\pi}(z) = \int_B M(w,A) \, d\widetilde{\pi}(w).\end{align*} Thus $M$ is reversible with respect to $\widetilde{\pi}$. Taking $A=\mathbb{R}^{2d}$ in the detailed balance identity gives phase-space invariance: \begin{align*}\int_{\mathbb{R}^{2d}} M(z,B) \, d\widetilde{\pi}(z) = \widetilde{\pi}(B).\end{align*} for every Borel set $B \subset \mathbb{R}^{2d}$. [/step]

[step:Project phase-space invariance to obtain invariance of the position kernel] Let $C \subset \mathbb{R}^d$ be a Borel set. Apply phase-space invariance with $B=C \times \mathbb{R}^d$: \begin{align*}\int_{\mathbb{R}^{2d}} M(z,C \times \mathbb{R}^d) \, d\widetilde{\pi}(z) = \widetilde{\pi}(C \times \mathbb{R}^d).\end{align*} Since $\pi$ is the first-coordinate marginal of $\widetilde{\pi}$, \begin{align*}\widetilde{\pi}(C \times \mathbb{R}^d) = \pi(C).\end{align*} By the disintegration property of the regular conditional distribution $(\kappa_x)_{x \in \mathbb{R}^d}$, applied to the bounded Borel function $(x,p) \mapsto M((x,p),C \times \mathbb{R}^d)$, \begin{align*}\int_{\mathbb{R}^{2d}} M(z,C \times \mathbb{R}^d) \, d\widetilde{\pi}(z) = \int_{\mathbb{R}^d}\int_{\mathbb{R}^d} M((x,p),C \times \mathbb{R}^d) \, d\kappa_x(p) \, d\pi(x).\end{align*} Using the definition of $P$, this becomes \begin{align*}\int_{\mathbb{R}^d} P(x,C) \, d\pi(x) = \pi(C).\end{align*} Since $C \subset \mathbb{R}^d$ was arbitrary, $\pi$ is invariant for the position transition kernel $P$. [/step]