[proofplan] The proof is the detailed-balance argument for a Metropolis-Hastings kernel with the Langevin proposal density. First we check that the proposal and acceptance functions are measurable and that the rejection probability defines a legitimate atom at the current state. The off-diagonal part satisfies detailed balance because $\pi(x)q_h(x,y)\alpha(x,y)$ is exactly the symmetric minimum of $\pi(x)q_h(x,y)$ and $\pi(y)q_h(y,x)$. The diagonal rejection mass is then handled separately, where symmetry follows from the fact that it is supported on $x=y$. Integrating the detailed-balance identity over Borel sets gives reversibility, and taking one set to be the whole space gives invariance. [/proofplan]

[step:Verify that the Metropolis-adjusted transition kernel is well defined] Since $\pi: \mathbb{R}^d \to (0,\infty)$ is $C^1$ and strictly positive, the function $\log \pi: \mathbb{R}^d \to \mathbb{R}$ is $C^1$, and its gradient $\nabla\log\pi: \mathbb{R}^d \to \mathbb{R}^d$ is continuous. Hence $q_h: \mathbb{R}^d \times \mathbb{R}^d \to (0,\infty)$ is continuous, so it is Borel measurable. For fixed $x \in \mathbb{R}^d$, the map $q_h(x,\cdot): \mathbb{R}^d \to (0,\infty)$ is the Gaussian density with mean $x+\frac{h}{2}\nabla\log\pi(x)$ and covariance matrix $hI_d$. Therefore \begin{align*} \int_{\mathbb{R}^d} q_h(x,y)\,d\mathcal{L}^d(y) = 1. \end{align*} The ratio \begin{align*} (x,y) \mapsto \frac{\pi(y)q_h(y,x)}{\pi(x)q_h(x,y)} \end{align*} is positive and continuous, so $\alpha$ is Borel measurable and satisfies $0 < \alpha(x,y) \leq 1$ for all $x,y \in \mathbb{R}^d$. The rejection probability $r: \mathbb{R}^d \to [0,1]$ is defined in the theorem statement by \begin{align*} r(x) := 1-\int_{\mathbb{R}^d}q_h(x,z)\alpha(x,z)\,d\mathcal{L}^d(z). \end{align*} Since $0 \leq \alpha(x,z) \leq 1$ and $q_h(x,\cdot)$ integrates to $1$, we have $0 \leq r(x) \leq 1$. The function $r$ is Borel measurable by the standard parameterized-integral measurability lemma for nonnegative Borel functions, proved by first checking indicators of Borel rectangles and then using the [monotone class theorem](/theorems/4925) together with Tonelli's theorem. Thus, for every $x \in \mathbb{R}^d$, the formula \begin{align*} K(x,A) = \int_A q_h(x,y)\alpha(x,y)\,d\mathcal{L}^d(y) + r(x)\delta_x(A) \end{align*} defines a probability measure on the Borel subsets of $\mathbb{R}^d$. Conversely, for every Borel set $A \subset \mathbb{R}^d$, the map $x \mapsto \int_A q_h(x,y)\alpha(x,y)\,d\mathcal{L}^d(y)$ is Borel measurable by the same monotone-class and Tonelli parameterized-integral argument, and $x \mapsto \delta_x(A)=\mathbb{1}_A(x)$ is Borel measurable, where $\mathbb{1}_A: \mathbb{R}^d \to \{0,1\}$ denotes the indicator function of $A$. Hence $x \mapsto K(x,A)$ is Borel measurable, so $K$ is a Markov transition kernel. [/step]

[step:Show that the accepted move density satisfies pointwise detailed balance]Define $m: \mathbb{R}^d \times \mathbb{R}^d \to [0,\infty)$ by \begin{align*} m(x,y) := \pi(x)q_h(x,y)\alpha(x,y). \end{align*} Using the definition of $\alpha$, for every $x,y \in \mathbb{R}^d$, \begin{align*} m(x,y) = \pi(x)q_h(x,y)\min\left\{1,\frac{\pi(y)q_h(y,x)}{\pi(x)q_h(x,y)}\right\}. \end{align*} Because $\pi(x)q_h(x,y) > 0$, multiplying the minimum by $\pi(x)q_h(x,y)$ gives \begin{align*} m(x,y) = \min\{\pi(x)q_h(x,y),\pi(y)q_h(y,x)\}. \end{align*} The right-hand side is unchanged when $x$ and $y$ are interchanged. Hence \begin{align*} m(x,y) = m(y,x) \end{align*} for all $x,y \in \mathbb{R}^d$.[/step]

[guided]The accepted part of the transition kernel has density $q_h(x,y)\alpha(x,y)$ with respect to $d\mathcal{L}^d(y)$. To check detailed balance for this part, we multiply by the target density at the starting point and define \begin{align*} m(x,y) := \pi(x)q_h(x,y)\alpha(x,y). \end{align*} This is the density of the measure “start at $x$ under $\Pi$, then make an accepted proposal to $y$.” Substitute the definition of $\alpha$. For every $x,y \in \mathbb{R}^d$, \begin{align*} m(x,y) = \pi(x)q_h(x,y)\min\left\{1,\frac{\pi(y)q_h(y,x)}{\pi(x)q_h(x,y)}\right\}. \end{align*} The denominator is nonzero because $\pi$ is strictly positive and $q_h$ is strictly positive. Therefore multiplication by the positive number $\pi(x)q_h(x,y)$ gives \begin{align*} m(x,y) = \min\{\pi(x)q_h(x,y),\pi(y)q_h(y,x)\}. \end{align*} This is the core Metropolis-Hastings cancellation: even though $q_h(x,y)$ need not equal $q_h(y,x)$, the acceptance probability converts the accepted-flow density into the smaller of the two directional proposal flows. The expression \begin{align*} \min\{\pi(x)q_h(x,y),\pi(y)q_h(y,x)\} \end{align*} is symmetric in $x$ and $y$, so \begin{align*} m(x,y) = m(y,x) \end{align*} for all $x,y \in \mathbb{R}^d$.[/guided]

[step:Integrate the accepted move symmetry over Borel rectangles] Let $A,B \subset \mathbb{R}^d$ be Borel sets. Since $m$ is nonnegative and Borel measurable, Tonelli's theorem applies to the product measure $\mathcal{L}^d \otimes \mathcal{L}^d$ on $\mathbb{R}^d \times \mathbb{R}^d$. The accepted part of the left-hand detailed-balance measure is \begin{align*} \int_A \int_B q_h(x,y)\alpha(x,y)\,d\mathcal{L}^d(y)\,\pi(x)\,d\mathcal{L}^d(x) = \int_{A \times B} m(x,y)\,d(\mathcal{L}^d \otimes \mathcal{L}^d)(x,y). \end{align*} Using the symmetry $m(x,y)=m(y,x)$, define the coordinate swap map $S: \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}^d \times \mathbb{R}^d$ by $S(x,y)=(y,x)$. Applying this change of variables gives \begin{align*} \int_{B \times A} m(y,x)\,d(\mathcal{L}^d \otimes \mathcal{L}^d)(y,x), \end{align*} because $S^{-1}(B \times A)=A \times B$ and the product [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^d \otimes \mathcal{L}^d$ is invariant under $S$. Therefore \begin{align*} \int_A \int_B q_h(x,y)\alpha(x,y)\,d\mathcal{L}^d(y)\,\pi(x)\,d\mathcal{L}^d(x) = \int_B \int_A q_h(y,x)\alpha(y,x)\,d\mathcal{L}^d(x)\,\pi(y)\,d\mathcal{L}^d(y). \end{align*} [/step]

[step:Handle the rejection atom on the diagonal] The rejection part of $K$ contributes, for Borel sets $A,B \subset \mathbb{R}^d$, \begin{align*} \int_A r(x)\delta_x(B)\pi(x)\,d\mathcal{L}^d(x). \end{align*} Let $\mathbb{1}_B: \mathbb{R}^d \to \{0,1\}$ denote the indicator function of the Borel set $B$. Since $\delta_x(B)=\mathbb{1}_B(x)$, this term is \begin{align*} \int_A r(x)\mathbb{1}_B(x)\pi(x)\,d\mathcal{L}^d(x) = \int_{A \cap B} r(x)\pi(x)\,d\mathcal{L}^d(x). \end{align*} The same calculation with $A$ and $B$ interchanged gives \begin{align*} \int_B r(y)\delta_y(A)\pi(y)\,d\mathcal{L}^d(y) = \int_{A \cap B} r(y)\pi(y)\,d\mathcal{L}^d(y). \end{align*} These two quantities are equal because they are the same integral over $A \cap B$. Thus the diagonal rejection mass satisfies detailed balance. [/step]

[step:Conclude reversibility and invariance] Combining the accepted-move identity with the diagonal rejection identity, for all Borel sets $A,B \subset \mathbb{R}^d$ we obtain \begin{align*} \int_A K(x,B)\pi(x)\,d\mathcal{L}^d(x) = \int_B K(y,A)\pi(y)\,d\mathcal{L}^d(y). \end{align*} Since $d\Pi(x)=\pi(x)\,d\mathcal{L}^d(x)$, this is exactly \begin{align*} \int_A K(x,B)\,d\Pi(x) = \int_B K(y,A)\,d\Pi(y). \end{align*} Hence $K$ is reversible with respect to $\Pi$. To prove invariance, take $A=\mathbb{R}^d$. Since $K(y,\mathbb{R}^d)=1$ for every $y \in \mathbb{R}^d$, reversibility gives \begin{align*} \int_{\mathbb{R}^d} K(x,B)\,d\Pi(x) = \int_B K(y,\mathbb{R}^d)\,d\Pi(y). \end{align*} Therefore \begin{align*} \int_{\mathbb{R}^d} K(x,B)\,d\Pi(x) = \int_B 1\,d\Pi(y) = \Pi(B). \end{align*} This proves that $\Pi$ is invariant for the Metropolis-adjusted Langevin transition kernel. [/step]