Proof of Gaussian Smoothing of a Kernel-Weighted Discrete Family with Countable Support: Joint Density, van Trees Regularity, and the Symmetrised Directional Score Bound

Preliminaries. We call The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder the weight lemma and Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging the averaging lemma. Integrals of nonnegative functions are handled with claim 1 of Linearity and Monotonicity of the Lebesgue Integral (linearity, monotonicity), integrable functions with claim 2 there and Integrable Function and the Lebesgue Integral; the monotone convergence theorem (MCT) and the dominated convergence theorem (DCT) are used on the measure spaces $(\mathsf{S},2^{\mathsf{S}},\mathsf{c})$, $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),\lambda_d)$, $(\mathsf{R},\mathcal{R},\rho)$ and the product $(\mathbb{R}^d\times\mathsf{R},\mathcal{B}(\mathbb{R}^d)\otimes\mathcal{R},\lambda_d\otimes\rho)$; on the last, Tonelli and Fubini Theorems (Tonelli) applies, both factors being $\sigma$-finite. Sums, products, absolute values and pointwise limits of measurable real functions are measurable by Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Write $d(\cdot,\cdot)$ for the Euclidean distance; $d(\theta-x,\theta'-x)=d(\theta,\theta')$ for all $\theta,\theta',x$, so translates of sequentially continuous functions are sequentially continuous, and sequentially continuous functions on $\mathbb{R}^d$ are measurable by claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets.

(P1) Countable sums. Since $\mathsf{S}$ is countable, there is a sequence $(x_k)_{k\in\mathbb{N}}$ in $\mathsf{S}$ whose set of values is $\mathsf{S}$ (a constant sequence if $\mathsf{S}$ is a single point; otherwise enumerate, repeating values if $\mathsf{S}$ is finite); put $F_k=\{x_1,\dots,x_k\}$, a nondecreasing sequence of finite sets with union $\mathsf{S}$, so that every finite subset of $\mathsf{S}$ is contained in some $F_k$. For $h:\mathsf{S}\to[0,\infty]$ the finite partial sums $\sum_{x\in F_k}h(x)=\int h\mathbf{1}_{F_k}\,d\mathsf{c}$ are nondecreasing in $k$ and cofinal among all finite partial sums, hence $\sum_{x\in\mathsf{S}}h(x)=\sup_k\sum_{x\in F_k}h(x)$; for $\mathsf{c}$-integrable real $h$, DCT with the dominating function $|h|$ gives $\int h\,d\mathsf{c}=\lim_k\sum_{x\in F_k}h(x)$. Consequently a function on $\mathsf{R}$ (or on $\mathbb{R}^d\times\mathsf{R}$) of the form $r\mapsto\sum_{x\in\mathsf{S}}h(x,r)$ with each $h(x,\cdot)$ measurable and nonnegative (resp. with $h(\cdot,r)$ $\mathsf{c}$-integrable for every $r$) is measurable, as a supremum (measurable by the MCT, which asserts measurability of the supremum) resp. a pointwise limit (claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions) of the measurable finite sums $\sum_{x\in F_k}h(x,\cdot)$; and for such nonnegative $h$, MCT on $(\mathsf{R},\rho)$ applied to the finite partial sums gives $\int\sum_{x\in\mathsf{S}}h(x,r)\,\rho(dr)=\sum_{x\in\mathsf{S}}\int h(x,r)\,\rho(dr)$, and likewise on $(\mathbb{R}^d,\lambda_d)$ and on the product.

(P2) Reindexing along a move. Fix $j$ and let $\sigma_j:\mathsf{S}\to\mathsf{S}$, $\sigma_j(x)=x+a_j$, which is injective with image $\mathsf{S}_j^{+}=\{y\in\mathsf{S}:y-a_j\in\mathsf{S}\}$. Let $H:\mathsf{S}\to[0,\infty)$ vanish off $\mathsf{S}_j^{+}$. For a finite $F\subseteq\mathsf{S}$, reindexing along the bijection $\sigma_j:F\to\sigma_j(F)$ (claim 2 of Properties of a Sum over a Finite Index Set) gives $\sum_{x\in F}H(x+a_j)=\sum_{y\in\sigma_j(F)}H(y)$; conversely, for a finite $F'\subseteq\mathsf{S}$, dropping the vanishing terms off $\mathsf{S}_j^{+}$ (claims 3 and 4 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set) gives $\sum_{y\in F'}H(y)=\sum_{y\in F'\cap\mathsf{S}_j^{+}}H(y)=\sum_{x\in\sigma_j^{-1}(F')}H(x+a_j)$. So the two families of finite partial sums coincide, and

$$\sum_{x\in\mathsf{S}}H(x+a_j)=\sum_{y\in\mathsf{S}}H(y).\tag{R}$$

(P3) Translations in $\theta$. For $b\in\mathbb{R}^d$ the map $\mathsf{T}_b(\theta,r)=(\theta-b,r)$ is measurable on the product: the preimage of a rectangle $A\times B$ ($A\in\mathcal{B}(\mathbb{R}^d)$, $B\in\mathcal{R}$) is $(A+b)\times B$, with $A+b$ Borel by claim 1 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$, and such rectangles generate the product $\sigma$-algebra. For measurable $F\ge0$ on the product, Tonelli and claim 2 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$ applied to each section give $\int F\circ\mathsf{T}_b\,d(\lambda_d\otimes\rho)=\int F\,d(\lambda_d\otimes\rho)$, and $\int_{\mathbb{R}^d}F(\theta-b,r)\,\lambda_d(d\theta)=\int_{\mathbb{R}^d}F(\theta,r)\,\lambda_d(d\theta)$ for every $r$. In particular $\int\varphi_\eta(\theta-x)\,\lambda_d(d\theta)=1$ (claim 1 of the weight lemma).

(P4) Cauchy-Schwarz on $\mathsf{S}$. For $\alpha:\mathsf{S}\to\mathbb{R}$ and $\beta:\mathsf{S}\to[0,\infty)$ with $\sum_x\beta(x)<\infty$, $\alpha=0$ where $\beta=0$, and $\sum_x\alpha(x)^{2}/\beta(x)<\infty$ (terms with $\beta(x)=0$ read as $0$), claim 1 of the averaging lemma on $(\mathsf{S},\mathsf{c})$ gives that $\alpha$ is $\mathsf{c}$-integrable and $(\sum_x\alpha(x))^{2}\le(\sum_x\beta(x))(\sum_x\alpha(x)^{2}/\beta(x))$.

(P5) Sequential continuity implies continuity. Let $G:\mathbb{R}^d\to\mathbb{R}$ be sequentially continuous and $\theta\in\mathbb{R}^d$. If $G$ were not continuous at $\theta$, there would be $\varepsilon>0$ such that for every natural $k$ some $\theta^{k}$ has $d(\theta^{k},\theta)<1/k$ and $|G(\theta^{k})-G(\theta)|\ge\varepsilon$ (using Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field to pass between squared and unsquared inequalities); then $d(\theta^{k},\theta)\to0$ while $G(\theta^{k})\not\to G(\theta)$, a contradiction.

(P6) Vanishing off $\mathsf{R}_{+}$. If $r\notin \mathsf{R}_{+}$ then $\sum_xp(x)f(x,r)=0$, so $f(x,r)=0$ for every $x\in\mathsf{S}$ (each $p(x)>0$); hence every function below of the form $\sum_xp(x)f(x,r)(\cdots)$ vanishes at such $r$, in particular $g$, $g_j$, $\bar g$, $\partial_ig$, $A_1$ and $B_j$.

Step 1: claim 1. $\mathsf{S}$ is nonempty, since otherwise the sum of $p$ over $\mathbb{R}^d$ would be $0$ rather than $1$. By (K2) and $\sum_xp(x)=1$, $\bar f(r)\le\sum_xp(x)M(r)=M(r)<\infty$ for every $r$. For each $x$, $(\theta,r)\mapsto p(x)\varphi_\eta(\theta-x)f(x,r)$ is measurable on the product, being the product of the compositions of the measurable maps $\theta\mapsto\varphi_\eta(\theta-x)$ and $r\mapsto f(x,r)$ with the coordinate projections (whose preimages are rectangles); it is nonnegative and at most $c_\eta p(x)f(x,r)$ (claim 1 of the weight lemma). By (P1), $\bar f$ is $\mathcal{R}$-measurable, $g$ is measurable on the product, and $\mathsf{R}_{+}=\{\bar f>0\}\in\mathcal{R}$; monotonicity gives $0\le g\le c_\eta\bar f(r)<\infty$, so $g$ is finite. For each $x$ and $j$, the maps $r\mapsto p(x)f(x,r)$ and $r\mapsto p_j(x)f_j(x,r)$ are $\mathcal{R}$-measurable (the latter is either $0$ or $p(x-a_j)f(x-a_j,\cdot)$), so $r\mapsto\Phi_w(p(x)f(x,r),p_1(x)f_1(x,r),\dots,p_n(x)f_n(x,r))$ is measurable by claim 2(a) of the averaging lemma and Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable, and $\mathsf{J}^{\mathrm{sym}}$ is well defined. For fixed $r$, MCT on $(\mathbb{R}^d,\lambda_d)$ applied to the finite partial sums, together with (P3) and linearity, gives $\int g(\theta,r)\,\lambda_d(d\theta)=\sup_k\sum_{x\in F_k}p(x)f(x,r)=\bar f(r)$. Tonelli and (P1) then give $\int g\,d(\lambda_d\otimes\rho)=\int\bar f\,d\rho=\sum_xp(x)\int f(x,r)\rho(dr)=\sum_xp(x)=1$ by (K1) and the definition of a probability mass function (whose sum over $\mathbb{R}^d$ equals its sum over $\mathsf{S}$, all other terms vanishing). Positivity: $g(\theta,r)>0$ if and only if some summand is positive, if and only if $f(x,r)>0$ for some $x$ (as $p>0$ on $\mathsf{S}$ and $\varphi_\eta>0$), if and only if $\bar f(r)>0$. For $g_j=g\circ\mathsf{T}_{a_j}$ the same statements follow from (P3) (measurability, invariance of the $\theta$-integral) and from $g_j(\theta,r)>0\iff g(\theta-a_j,r)>0\iff r\in \mathsf{R}_{+}$; for $\bar g$, a convex combination of $g,g_1,\dots,g_n$, they follow by linearity, and $\bar g>0$ exactly where $g>0$.

Step 2: claim 2. Fix $r\in\mathsf{R}$ and $i$. Differentiation under the sum. Fix $\theta\in\mathbb{R}^d$, let $I=(\theta_i-1,\theta_i+1)$, and for $t\in I$ let $\theta[t]$ be the point with $i$th coordinate $t$ and the other coordinates those of $\theta$. Define $F:I\times\mathsf{S}\to\mathbb{R}$ by $F(t,x)=p(x)f(x,r)\varphi_\eta(\theta[t]-x)$. We verify the hypotheses of Differentiation under the Integral Sign on $(\mathsf{S},2^{\mathsf{S}},\mathsf{c})$: (i) for each $t$, $|F(t,\cdot)|\le c_\eta p(\cdot)f(\cdot,r)$, which is $\mathsf{c}$-integrable with integral $c_\eta\bar f(r)<\infty$; (ii) for each $x$ and $t_0\in I$, the translate $\tau_x(\theta')=\varphi_\eta(\theta'-x)$ has, at $\theta[t_0]$, exactly the difference quotients of $\varphi_\eta$ at $\theta[t_0]-x$ in the $i$th variable, so by claim 2 of the weight lemma and Partial Derivative on a Euclidean Open Set its $i$th partial derivative at $\theta[t_0]$ exists and equals $\partial_i\varphi_\eta(\theta[t_0]-x)$; unwinding that definition (the point obtained from $\theta[t_0]$ by increasing the $i$th coordinate by $h$ is $\theta[t_0+h]$, which lies in the domain $\mathbb{R}^d$, and $t_0+h\in I$ for $|h|$ small) says verbatim that $t\mapsto\tau_x(\theta[t])$ is differentiable at $t_0$ with derivative $\partial_i\varphi_\eta(\theta[t_0]-x)$, whence $D_1F(t_0,x)=p(x)f(x,r)\,\partial_i\varphi_\eta(\theta[t_0]-x)$ by the constant-multiple rule (claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives); (iii) $|D_1F(t,x)|\le c_\eta(2\eta)^{-1/2}p(x)f(x,r)$ by claim 2 of the weight lemma, a $\mathsf{c}$-integrable bound. Hence $t\mapsto\int F(t,x)\,\mathsf{c}(dx)=g(\theta[t],r)$ is differentiable on $I$ with derivative $\sum_xp(x)f(x,r)\partial_i\varphi_\eta(\theta[t]-x)$, the summand being $\mathsf{c}$-integrable. This function of $t$ is the slice function of $g(\cdot,r)$ at $\theta$ in the $i$th variable with admissible radius $1$ (claim 2 of Slice Function and the Partial Derivative, with $I$ the interval so named there and $\mathbb{R}^d$ the open set), so $\partial_ig(\theta,r)$ exists and equals the displayed formula at $t=\theta_i$. The bound $|\partial_ig(\theta,r)|\le c_\eta(2\eta)^{-1/2}\bar f(r)$ follows from (iii) and monotonicity.

Continuity. Let $\theta^{k}\to\theta$ in $\mathbb{R}^d$. For each $x$, $\varphi_\eta(\theta^{k}-x)\to\varphi_\eta(\theta-x)$ and $\partial_i\varphi_\eta(\theta^{k}-x)\to\partial_i\varphi_\eta(\theta-x)$ by sequential continuity (claims 1 and 2 of the weight lemma), with the dominating functions $c_\eta p(x)f(x,r)$ and $c_\eta(2\eta)^{-1/2}p(x)f(x,r)$; DCT on $(\mathsf{S},\mathsf{c})$ gives $g(\theta^{k},r)\to g(\theta,r)$ and $\partial_ig(\theta^{k},r)\to\partial_ig(\theta,r)$. So $g(\cdot,r)$ and $\partial_ig(\cdot,r)$ are sequentially continuous, hence continuous at every point by (P5). For $r\in \mathsf{R}_{+}$, $g(\cdot,r)>0$ by claim 1, and clauses 1 and 3 of C^k Maps on a Euclidean Open Set make $g(\cdot,r)$ of class $C^1$ on $\mathbb{R}^d$.

Measurability and integrability of $\partial_ig$. By (P1) (integrable case), $\partial_ig$ is the pointwise limit of the finite sums $\sum_{x\in F_k}p(x)\partial_i\varphi_\eta(\theta-x)f(x,r)$, each measurable on the product as in Step 1, hence measurable. Next, with $e_i$ the $i$th unit vector, claim 2 of the weight lemma gives $|\partial_i\varphi_\eta(z)|=|Z_{e_i}(z)|\varphi_\eta(z)$ with $Z_{e_i}(z)=z_i/\eta$ and $\kappa_{e_i}=1/\eta$, so by claim 4 there, $\int Z_{e_i}^{2}\varphi_\eta\,d\lambda_d=1/\eta$, and (P4)-type Cauchy-Schwarz (claim 1 of the averaging lemma on $(\mathbb{R}^d,\lambda_d)$ with $\beta=\varphi_\eta$) gives $\int|Z_{e_i}|\varphi_\eta\,d\lambda_d\le(\int\varphi_\eta)^{1/2}(\int Z_{e_i}^{2}\varphi_\eta)^{1/2}=\eta^{-1/2}$; moreover $|z_k|\,|\partial_i\varphi_\eta(z)|=\eta\,|Z_{e_k}(z)Z_{e_i}(z)|\varphi_\eta(z)\le\tfrac{\eta}{2}(Z_{e_k}^{2}+Z_{e_i}^{2})\varphi_\eta(z)$, whose integral is at most $1$. Since $|(\theta-x)_k+x_k|\le|(\theta-x)_k|+|x_k|$ and $|x_k|\le\lVert x\rVert$, (P3) yields for every $x$

$$\int\Bigl(1+\sum_{k=1}^{d}|\theta_k|\Bigr)|\partial_i\varphi_\eta(\theta-x)|\,\lambda_d(d\theta)=\int\Bigl(1+\sum_{k=1}^{d}|z_k+x_k|\Bigr)|\partial_i\varphi_\eta(z)|\,\lambda_d(dz)\le\eta^{-1/2}+d+d\,\eta^{-1/2}\lVert x\rVert=:C_1+C_2\lVert x\rVert .$$

By monotonicity, $|\partial_ig(\theta,r)|\le\sum_xp(x)f(x,r)|\partial_i\varphi_\eta(\theta-x)|$, so by Tonelli, (P1) and (K1),

$$\int\Bigl(1+\sum_k|\theta_k|\Bigr)|\partial_ig|\,d(\lambda_d\otimes\rho)\le\sum_xp(x)\int f(x,r)\rho(dr)\,(C_1+C_2\lVert x\rVert)=C_1+C_2\sum_xp(x)\lVert x\rVert<\infty$$

by the first-moment condition. Finally, let $h(v)=1/v$ for $v>0$ and $h(0)=0$; as in the proof of claim 1 of the averaging lemma, $h\circ g$ is measurable (a pointwise limit of $g/(g^{2}+1/k)$), so $\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}(\partial_ig)^{2}/g=(\partial_ig)^{2}\,(h\circ g)$ is measurable, the two sides agreeing also off $\mathbb{R}^d\times \mathsf{R}_{+}$ by (P6) and claim 1. For $r\in \mathsf{R}_{+}$ and any $\theta$, apply (P4) with $\alpha(x)=p(x)f(x,r)\partial_i\varphi_\eta(\theta-x)$ and $\beta(x)=p(x)f(x,r)\varphi_\eta(\theta-x)$ (so $\sum_x\beta=g(\theta,r)>0$, $\alpha=0$ where $\beta=0$, and $\alpha^{2}/\beta=p(x)f(x,r)Z_{e_i}(\theta-x)^{2}\varphi_\eta(\theta-x)$ where $\beta>0$):

$$\frac{\partial_ig(\theta,r)^{2}}{g(\theta,r)}\le\sum_xp(x)f(x,r)\,Z_{e_i}(\theta-x)^{2}\varphi_\eta(\theta-x),$$

provided the right side is finite, and trivially otherwise. Integrating with Tonelli, (P1), (P3) and (K1), $\int\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}(\partial_ig)^{2}/g\,d(\lambda_d\otimes\rho)\le\sum_xp(x)\cdot1\cdot\int Z_{e_i}^{2}\varphi_\eta\,d\lambda_d=1/\eta$.

Step 3: claim 3. Decomposition. By linearity of the $\mathsf{c}$-integral, $\partial_ug(\theta,r)=\sum_xp(x)f(x,r)\sum_{i=1}^{d}u_i\partial_i\varphi_\eta(\theta-x)$, and since $u=\sum_jw_ja_j$, the definition of $R_{a_j}$ in claim 5 of the weight lemma gives $\sum_iu_i\partial_i\varphi_\eta(z)=\sum_jw_j\bigl(R_{a_j}(z)+\varphi_\eta(z)-\varphi_\eta(z-a_j)\bigr)$. Each $R_{a_j}$ is bounded (by $2c_\eta+c_\eta(2\eta)^{-1/2}\sum_i|(a_j)_i|$), so all the resulting summands are $\mathsf{c}$-integrable, and

$$\partial_ug=A_1+\sum_{j=1}^{n}w_jB_j,\qquad A_1(\theta,r)=\sum_{j=1}^{n}w_j\bigl(g(\theta,r)-g_j(\theta,r)\bigr),\qquad B_j(\theta,r)=\sum_xp(x)f(x,r)R_{a_j}(\theta-x),$$

because $\sum_xp(x)f(x,r)\varphi_\eta(\theta-x-a_j)=g(\theta-a_j,r)=g_j(\theta,r)$ by definition. $A_1$ is measurable by Step 1, and $B_j$ by (P1) ($R_{a_j}$ is sequentially continuous, hence measurable, by claim 5 of the weight lemma).

Reindexing $g_j$. Fix $(\theta,r)$ and $j$, and put $H(y)=\varphi_\eta(\theta-y)\,p_j(y)f_j(y,r)$ for $y\in\mathsf{S}$; $H$ vanishes off $\mathsf{S}_j^{+}$ (there $p_j(y)=p(y-a_j)=0$), and for $x\in\mathsf{S}$, $H(x+a_j)=\varphi_\eta(\theta-x-a_j)p(x)f(x,r)$ since $(x+a_j)-a_j=x\in\mathsf{S}$. By (R),

$$g_j(\theta,r)=\sum_{x\in\mathsf{S}}H(x+a_j)=\sum_{y\in\mathsf{S}}\varphi_\eta(\theta-y)\,p_j(y)f_j(y,r).$$

The main term. Fix $r\in \mathsf{R}_{+}$ and $\theta$. On $(\mathsf{S},\mathsf{c})$ put $\ell_0(y)=\varphi_\eta(\theta-y)p(y)f(y,r)$ and $\ell_j(y)=\varphi_\eta(\theta-y)p_j(y)f_j(y,r)$, nonnegative with finite integrals $\int\ell_0\,d\mathsf{c}=g(\theta,r)$ and $\int\ell_j\,d\mathsf{c}=g_j(\theta,r)$ (by the reindexing). Since $\bar g(\theta,r)=\mathsf{D}(g,g_1,\dots,g_n)(\theta,r)>0$ and $A_1=\mathsf{N}_w(g,g_1,\dots,g_n)$ in the notation of the averaging lemma, claim 3 there followed by claim 2(c) (homogeneity, with the factor $\varphi_\eta(\theta-y)>0$) gives

$$\frac{A_1(\theta,r)^{2}}{\bar g(\theta,r)}=\Phi_w\bigl(g,g_1,\dots,g_n\bigr)(\theta,r)\le\sum_{y\in\mathsf{S}}\varphi_\eta(\theta-y)\,\Phi_w\bigl(p(y)f(y,r),p_1(y)f_1(y,r),\dots,p_n(y)f_n(y,r)\bigr).$$

For each $y$, $r\mapsto\Phi_w(p(y)f(y,r),\dots)$ is $\mathcal{R}$-measurable by claim 3 of the averaging lemma (its arguments are measurable in $r$), so the right side is a measurable nonnegative function on the product by (P1). Integrating over $\theta$ first, with (P3) and $\int\varphi_\eta(\theta-y)\lambda_d(d\theta)=1$, then over $r$, using Tonelli and (P1),

$$\int\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}\frac{A_1^{2}}{\bar g}\,d(\lambda_d\otimes\rho)\le\sum_{y\in\mathsf{S}}\int_{\mathsf{R}}\Phi_w\bigl(p(y)f(y,r),p_1(y)f_1(y,r),\dots,p_n(y)f_n(y,r)\bigr)\rho(dr)=\mathsf{J}^{\mathrm{sym}} ,$$

where $\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}A_1^{2}/\bar g=A_1^{2}\,(h\circ\bar g)$ is measurable as in Step 2.

The remainder terms. Fix $j$, $r\in \mathsf{R}_{+}$ and $\theta$. Since $\bar g\ge\frac12g>0$, $B_j^{2}/\bar g\le2B_j^{2}/g$. Apply (P4) with $\alpha(x)=p(x)f(x,r)R_{a_j}(\theta-x)$ and $\beta(x)=p(x)f(x,r)\varphi_\eta(\theta-x)$: $B_j(\theta,r)^{2}/g(\theta,r)\le\sum_xp(x)f(x,r)R_{a_j}(\theta-x)^{2}/\varphi_\eta(\theta-x)$ (trivially if the right side is infinite). Integrating with Tonelli, (P1), (P3) and claim 5 of the weight lemma ($\int R_{a_j}^{2}/\varphi_\eta\,d\lambda_d=\exp(\kappa_j)-1-\kappa_j$) and (K1),

$$\int\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}\frac{B_j^{2}}{\bar g}\,d(\lambda_d\otimes\rho)\le2\sum_xp(x)\bigl(\exp(\kappa_j)-1-\kappa_j\bigr)=2\bigl(\exp(\kappa_j)-1-\kappa_j\bigr).$$

Conclusion. By claim 3 of Image Measures, Measures with Densities, and Change of Variables, $\nu(E)=\int\mathbf{1}_E\,\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}\,\bar g\,d(\lambda_d\otimes\rho)$ is a measure on the product $\sigma$-algebra with $\nu(\mathbb{R}^d\times\mathsf{R})=\int\bar g\,d(\lambda_d\otimes\rho)=1$ (as $\bar g$ vanishes off $\mathbb{R}^d\times \mathsf{R}_{+}$ by (P6)), so it makes the product a probability space, and $\int F^{2}\,d\nu=\int F^{2}\,\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}\,\bar g\,d(\lambda_d\otimes\rho)$ for measurable $F$. For a measurable real $A$ on the product put $F_A=A\,(h\circ\bar g)$, measurable, equal to $A/\bar g$ on $\mathbb{R}^d\times \mathsf{R}_{+}$ and to $0$ elsewhere; then $\int F_A^{2}\,d\nu=\int\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}A^{2}/\bar g\,d(\lambda_d\otimes\rho)$. Whenever this is finite, $F_A$ is square-integrable under $\nu$ with mean-square norm $\lVert F_A\rVert_2$ equal to its square root; $F_{A+A'}=F_A+F_{A'}$ and $F_{cA}=cF_A$, so the triangle inequality (claim 2 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm, extended to finitely many summands by induction) and the homogeneity $\lVert cF\rVert_2=|c|\lVert F\rVert_2$ (immediate from Square-Integrable Random Variables and the Mean-Square Inner Product) apply. If $\mathsf{J}^{\mathrm{sym}}=\infty$ the claimed inequality holds trivially. Otherwise $F_{A_1}$ and every $F_{B_j}$ are square-integrable by the two bounds above, hence so is $F_{\partial_ug}=F_{A_1}+\sum_jw_jF_{B_j}$, and

$$\Bigl(\int\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}\frac{(\partial_ug)^{2}}{\bar g}\,d(\lambda_d\otimes\rho)\Bigr)^{1/2}=\lVert F_{\partial_ug}\rVert_2\le\lVert F_{A_1}\rVert_2+\sum_{j=1}^{n}|w_j|\,\lVert F_{B_j}\rVert_2\le(\mathsf{J}^{\mathrm{sym}})^{1/2}+\sum_{j=1}^{n}|w_j|\bigl(2(\exp(\kappa_j)-1-\kappa_j)\bigr)^{1/2},$$

the square root being increasing on $[0,\infty)$ (Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field) and $(2t)^{1/2}=2^{1/2}t^{1/2}$ for $t\ge0$ (the right side is nonnegative with square $2t$). This is claim 3; the measurability of $\mathbf{1}_{\mathbb{R}^d\times \mathsf{R}_{+}}(\partial_ug)^{2}/\bar g=(\partial_ug)^{2}(h\circ\bar g)$ was noted above. $\blacksquare$