Proof of Translation and Mollification Estimates in the Integrable Norm for a Density of Finite Fisher Information, and the Mollified Density Cost

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. The rules for adding inequalities, for multiplying them by nonnegative reals and for handling absolute values, from Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field and Properties of the Absolute Value in an Ordered Field, are used without further mention; so is the fact that for nonnegative reals $a,b$ one has $a\le b$ exactly when $a^{2}\le b^{2}$ (claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field). Integrals of nonnegative measurable functions are taken in $[0,\infty]$, and for a nonnegative integrable function this integral is the real integral, by the criterion recorded in Measure Spaces and the Lebesgue Integral: Standing Notation §integral.

Step 0 (Preliminaries).

(0.1) The density. By the meaning of a density in The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities, $\rho$ is measurable, $0\le\rho$, and $\mu$ is the measure with density $\rho$ with respect to $\lambda_{d}$ of claim 3 of that lemma. Hence, by that claim, $\int_{\mathbb{R}^{d}}f\,d\mu=\int_{\mathbb{R}^{d}}f(x)\rho(x)\,dx$ for every Borel $f:\mathbb{R}^{d}\to[0,\infty]$, and a Borel $f:\mathbb{R}^{d}\to\mathbb{R}$ is integrable with respect to $\mu$ exactly when $f\rho$ is integrable with respect to $\lambda_{d}$, with the same identity. By The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §uniqueness, applied with both densities equal to $\rho$, $\rho$ is integrable with respect to $\lambda_{d}$ and $\int_{\mathbb{R}^{d}}\rho(y)\,dy=\mu(\mathbb{R}^{d})=1$.

(0.2) Translations and reflections. For $a\in\mathbb{R}^{d}$, claim 2 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$ (with $n=d$) shows that for a Borel $f:\mathbb{R}^{d}\to[0,\infty]$ the functions $x\mapsto f(x+a)$ and $x\mapsto f(a-x)$ are measurable with $\int f(x+a)\,dx=\int f(a-x)\,dx=\int f(x)\,dx$, and claim 3 there gives the same for Borel integrable $f:\mathbb{R}^{d}\to\mathbb{R}$. Measurable real functions are closed under sums, real multiples, products and absolute values by claims 2, 3 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and continuous functions $\mathbb{R}^{d}\to\mathbb{R}$ are Borel (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps).

(0.3) Compact supports. Let $f:\mathbb{R}^{d}\to\mathbb{R}$. (a) Suppose there is a real $r\ge0$ with $f(x)=0$ whenever $r<\lVert x\rVert$. Then $\{x:f(x)\ne0\}$ is contained in the closed ball $\bar B(0_{\mathbb{R}^{d}},r)$ of $(\mathbb{R}^{d},d_{E})$, as $d_{E}(x,0_{\mathbb{R}^{d}})=\lVert x\rVert$ (claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$); that ball is closed and bounded by claims 3 and 2 of Elementary Properties of the Closed Ball in a Metric Space. By claims 3 and 2 of The Closure is the Smallest Closed Superset, the support $\operatorname{supp}f$ is a closed subset of that ball, hence closed and bounded, hence compact by Heine-Borel Theorem in $\mathbb{R}^n$: $f$ is compactly supported. (b) Conversely, if $f$ is compactly supported, $\operatorname{supp}f$ is bounded by Heine-Borel Theorem in $\mathbb{R}^n$, so by Bounded Subset of a Metric Space there are $x_{0}\in\mathbb{R}^{d}$ and a real $s>0$ with $\lVert w-x_{0}\rVert\le s$ for $w\in\operatorname{supp}f$; with $r=\lVert x_{0}\rVert+s$ the triangle inequality (claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$) gives $\lVert w\rVert\le r$ on $\operatorname{supp}f$, and since $\{f\ne0\}\subseteq\operatorname{supp}f$ (claim 1 of The Closure is the Smallest Closed Superset), $f(x)=0$ whenever $r<\lVert x\rVert$.

(0.4) The score. The space $L^{2}(\mu;\mathbb{R}^{d})$ is a real Hilbert space by Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation §fields, in particular a real inner product space, so The Cauchy-Schwarz Inequality in a Real Inner Product Space gives $|\langle\xi_{\mu},w\rangle_{\mu}|\le\lVert\xi_{\mu}\rVert_{\mu}\lVert w\rVert_{\mu}$ for $w\in L^{2}(\mu;\mathbb{R}^{d})$. As $0\le\lVert\xi_{\mu}\rVert_{\mu}$ and $\lVert\xi_{\mu}\rVert_{\mu}^{2}=\mathcal{I}(\mu)$ (Finite Fisher Information, the Score and the Fisher Information of a Probability Measure §information), $\lVert\xi_{\mu}\rVert_{\mu}=\sqrt{\mathcal{I}(\mu)}$.

Step 1 (Pairing with a test function). Claim. Let $z\in\mathbb{R}^{d}$ and let $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$ be a test function with $|\psi(x)|\le1$ for every $x$. Then the function $y\mapsto\psi(y)\bigl(\rho(y-z)-\rho(y)\bigr)$ is integrable with respect to $\lambda_{d}$ and

$$\int_{\mathbb{R}^{d}}\psi(y)\bigl(\rho(y-z)-\rho(y)\bigr)\,dy\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)} .$$

(1a) Translates of $\psi$. For $t\in\mathbb{R}$ let $\psi_{t}:\mathbb{R}^{d}\to\mathbb{R}$, $\psi_{t}(x)=\psi(x+tz)$. By Partial Derivatives, Continuity and $C^k$ Regularity under a Scaling Substitution, applied with its $n=d$, $m=1$, $U=\mathbb{R}^{d}$ (open by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous), $f=\psi$, $c=tz$ and both of its nonzero scalars equal to $1$, so that its set $V$ is $\mathbb{R}^{d}$ and its $g$ is $\psi_{t}$: $\psi_{t}$ is smooth (claim 5 there) and $\partial_{i}\psi_{t}(x)=\partial_{i}\psi(x+tz)$ for all $x$ and $i\in[d]$ (claim 2 there). By (0.3)(b) there is $r\ge0$ with $\psi(w)=0$ whenever $r<\lVert w\rVert$. If $r+|t|\lVert z\rVert<\lVert x\rVert$, then $\lVert x\rVert\le\lVert x+tz\rVert+|t|\lVert z\rVert$ (claims 5 and 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$) gives $r<\lVert x+tz\rVert$, so $\psi_{t}(x)=0$; by (0.3)(a), $\psi_{t}$ is compactly supported. Thus $\psi_{t}\in C_{c}^{\infty}(\mathbb{R}^{d})$, and $|\psi_{t}(x)|\le1$ for every $x$. Being continuous (claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous), $\psi_{t}$ is Borel, and being bounded it is integrable with respect to $\mu$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures).

(1b) The field. For $t\in\mathbb{R}$ let $\eta_{t}:\mathbb{R}^{d}\to\mathbb{R}^{d}$ be the map with components $\eta_{t,i}=z_{i}\psi_{t}$ ($i\in[d]$), that is $\eta_{t}(x)=\psi_{t}(x)\,z$. Each $\eta_{t,i}$ is a test function by The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure §linear, hence of class $C^{1}$ and compactly supported (Test Functions on Euclidean Space, Their Gradient Maps and Laplacians §gradient), and $\partial_{i}\eta_{t,i}(x)=z_{i}\,\partial_{i}\psi(x+tz)$ by claim 1 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set and (1a). So

$$\operatorname{div}\eta_{t}(x)=\sum_{i=1}^{d}\partial_{i}\psi(x+tz)\,z_{i}\qquad(x\in\mathbb{R}^{d}).\tag{1.1}$$

By The Score Integrates by Parts Against Every Compactly Supported Continuously Differentiable Vector Field §parts, applied to $\mu$ and $\eta_{t}$, the class of $\eta_{t}$ lies in $L^{2}(\mu;\mathbb{R}^{d})$, $\operatorname{div}\eta_{t}$ is integrable with respect to $\mu$, and $\langle\xi_{\mu},\eta_{t}\rangle_{\mu}=-\int_{\mathbb{R}^{d}}\operatorname{div}\eta_{t}\,d\mu$. For every $x$, $\lVert\eta_{t}(x)\rVert=|\psi_{t}(x)|\,\lVert z\rVert\le\lVert z\rVert$ (claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$), so, by the formula for the norm in Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation §l2mu, claim 1 of Linearity and Monotonicity of the Lebesgue Integral and claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space with $\mu(\mathbb{R}^{d})=1$, $\lVert\eta_{t}\rVert_{\mu}^{2}=\int\lVert\eta_{t}\rVert^{2}\,d\mu\le\lVert z\rVert^{2}$, that is $\lVert\eta_{t}\rVert_{\mu}\le\lVert z\rVert$. With (0.4),

$$\Bigl|\int_{\mathbb{R}^{d}}\operatorname{div}\eta_{t}\,d\mu\Bigr|=|\langle\xi_{\mu},\eta_{t}\rangle_{\mu}|\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)}\qquad(t\in\mathbb{R}).\tag{1.2}$$

(1c) Differentiation under the integral. Let $U$ be the open interval with endpoints $-1$ and $2$, and let $F:U\to\mathbb{R}$, $F(t)=\int_{\mathbb{R}^{d}}\psi(x+tz)\,\mu(dx)=\int_{\mathbb{R}^{d}}\psi_{t}\,d\mu$. We apply Differentiation under the Integral Sign on the measure space $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu)$ with $f(t,x)=\psi(x+tz)$. Condition (i) holds by (1a). For (ii), fix $x$: by claim 2 of Derivatives Along a Segment for C^1 Functions on a Euclidean Open Set, with $W=\mathbb{R}^{d}$, the $C^{1}$ function $\psi$ (Test Functions on Euclidean Space, Their Gradient Maps and Laplacians §gradient), its $x$ and $h$ equal to our $x$ and $z$, and $J=U$, an interval every point of which is interior by An Open Interval is an Interval All of Whose Points Are Interior, the function $s\mapsto\psi(x+sz)$ is differentiable at every $s\in U$ with derivative $D_{1}f(s,x)=\sum_{i=1}^{d}\partial_{i}\psi(x+sz)\,z_{i}$. For (iii), let $K\ge0$ be as in The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure §gradient, so $\lVert\nabla\psi(w)\rVert\le K$ for all $w$; the number $\sum_{i}\partial_{i}\psi(w)z_{i}$ is the dot product $z\cdot\nabla\psi(w)$, whence $|D_{1}f(s,x)|\le\lVert z\rVert K$ by Cauchy-Schwarz Inequality for the Euclidean Dot Product, and the constant $\lVert z\rVert K$ is integrable with respect to $\mu$ (claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space). Hence $F$ is differentiable at every $t\in U$ and, by (1.1) and (1.2),

$$F'(t)=\int_{\mathbb{R}^{d}}\sum_{i=1}^{d}\partial_{i}\psi(x+tz)\,z_{i}\,\mu(dx)=\int_{\mathbb{R}^{d}}\operatorname{div}\eta_{t}\,d\mu,\qquad|F'(t)|\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)} .$$

(1d) Mean value. By Differentiability at an Interior Point Implies Continuity There, $F$ is continuous relative to $U$ at every point of $U$, for the absolute-value metric; since the defining condition at a point $t$ only becomes weaker when the points $s$ it quantifies over are restricted to the smaller set $[0,1]\subseteq U$, the restriction $F_{0}$ of $F$ to the closed interval $[0,1]$ is continuous on $[0,1]$. At each $t\in(0,1)$ the difference quotients of $F_{0}$ and of $F$ agree for all increments small enough that the point stays in $(0,1)$, so $F_{0}$ is differentiable at $t$ with derivative $F'(t)$ (Derivative at an Interior Point). By Mean Value Theorem on a Closed Real Interval there is $c\in(0,1)$ with $F(1)-F(0)=F'(c)$, and so

$$F(1)-F(0)\le|F'(c)|\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)} .\tag{1.3}$$

(1e) Identification. The function $k(y)=\psi(y)\rho(y-z)$ is measurable by (0.2) and satisfies $|k(y)|\le\rho(y-z)$; the latter function is integrable by (0.2) (with $a=-z$) and (0.1), so $k$ is integrable (claim 1 of Linearity and Monotonicity of the Lebesgue Integral and the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §integral). By (0.2) with $a=z$, $\int k(y)\,dy=\int k(x+z)\,dx=\int\psi(x+z)\rho(x)\,dx$, and by (0.1), applied to the bounded Borel function $\psi_{1}$, this equals $\int\psi_{1}\,d\mu=F(1)$. Likewise $\psi\rho$ is integrable with $\int\psi(y)\rho(y)\,dy=\int\psi_{0}\,d\mu=F(0)$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral the function $y\mapsto\psi(y)(\rho(y-z)-\rho(y))$ is integrable with integral $F(1)-F(0)$, and (1.3) proves the claim.

Step 2 (Clause 1). Fix $z\in\mathbb{R}^{d}$ and put $f_{z}(y)=\rho(y-z)-\rho(y)$. By (0.1) and (0.2), $f_{z}$ is Borel and, as a difference of integrable functions, integrable with respect to $\lambda_{d}$ (claim 2 of Linearity and Monotonicity of the Lebesgue Integral); this is the first assertion of clause 1. Let $\delta\in\mathbb{R}$ be positive.

(2a) A Lipschitz sign. The function $|f_{z}|$ is Borel and nonnegative, so claim 3 of Image Measures, Measures with Densities, and Change of Variables provides the measure $\kappa$ with density $|f_{z}|$ with respect to $\lambda_{d}$, $\kappa(A)=\int\mathbf{1}_{A}|f_{z}|\,dy$ for $A\in\mathcal{B}(\mathbb{R}^{d})$; $\kappa(\mathbb{R}^{d})=\int|f_{z}|\,dy<\infty$. As $\mathcal{B}(\mathbb{R}^{d})$ is the Borel $\sigma$-algebra of the metric space $(\mathbb{R}^{d},d_{E})$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §spaces), $\kappa$ is a Borel measure on $(\mathbb{R}^{d},d_{E})$. The set $P=\{y:f_{z}(y)>0\}$ is Borel (Measure Spaces and the Lebesgue Integral: Standing Notation §measurable). By Inner and Outer Regularity of a Finite Borel Measure on a Metric Space, and Lipschitz Approximation of Indicators §lipschitz, applied to $X=\mathbb{R}^{d}$, $\kappa$, $B=P$ and $\delta$, there are a map $h:\mathbb{R}^{d}\to\mathbb{R}$, Lipschitz with some constant $\Lambda\ge0$ and with $0\le h\le1$, and $N\in\mathcal{B}(\mathbb{R}^{d})$ with $\kappa(N)\le\delta$, such that $h=\mathbf{1}_{P}$ on $\mathbb{R}^{d}\setminus N$. Put $s=2h-1$. Then $|s(x)|\le1$, and $|s(x)-s(x')|=2|h(x)-h(x')|\le2\Lambda\,d_{E}(x,x')$, so $s$ is Lipschitz and hence continuous by A Lipschitz Map is Uniformly Continuous.

(2b) The pointwise inequality. Let $y\in\mathbb{R}^{d}$. If $y\notin N$ and $f_{z}(y)>0$, then $y\in P$, $s(y)=1$ and $s(y)f_{z}(y)=|f_{z}(y)|$; if $y\notin N$ and $f_{z}(y)\le0$, then $y\notin P$, $s(y)=-1$ and $s(y)f_{z}(y)=|f_{z}(y)|$; if $y\in N$, then $s(y)f_{z}(y)\ge-|f_{z}(y)|$. In all cases

$$s(y)f_{z}(y)\ge|f_{z}(y)|-2\,\mathbf{1}_{N}(y)\,|f_{z}(y)| .\tag{2.1}$$

Both sides are integrable (each is dominated in absolute value by $|f_{z}|$ or $2|f_{z}|$), and integrating (2.1) with claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives $\int|f_{z}|\,dy-2\kappa(N)\le\int s f_{z}\,dy$, hence

$$\int_{\mathbb{R}^{d}}|f_{z}|\,dy\le\int_{\mathbb{R}^{d}}s f_{z}\,dy+2\delta .\tag{2.2}$$

(2c) Approximation by test functions. By Test Functions Approximate Bounded Continuous Functions Pointwise, Determine a Finite Borel Measure, and Detect a Vanishing Density §approximation, applied to $s$ with $M=1$, there are $\psi_{n}\in C_{c}^{\infty}(\mathbb{R}^{d})$ ($n\in\mathbb{N}$) with $|\psi_{n}|\le1$ and $\psi_{n}(y)\to s(y)$ for every $y$. By Step 1, $\int\psi_{n}f_{z}\,dy\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)}$ for every $n$. The functions $\psi_{n}f_{z}$ are measurable, converge pointwise to $sf_{z}$, and satisfy $|\psi_{n}f_{z}|\le|f_{z}|$; by the dominated convergence theorem on $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\lambda_{d})$, $\int\psi_{n}f_{z}\,dy\to\int sf_{z}\,dy$. Given a positive real $\varepsilon'$, choose $n$ with $\int sf_{z}\,dy<\int\psi_{n}f_{z}\,dy+\varepsilon'\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)}+\varepsilon'$; by Comparison of Real Numbers with Arbitrary Positive Slack §slack-above, $\int sf_{z}\,dy\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)}$. With (2.2), $\int|f_{z}|\,dy\le\lVert z\rVert\sqrt{\mathcal{I}(\mu)}+2\delta$ for every positive $\delta$, and Comparison of Real Numbers with Arbitrary Positive Slack §slack-above gives clause 1.

Step 3 (Clause 2). Let $\eta$, $\varepsilon$, $\eta_{\varepsilon}$ and $\eta_{\varepsilon}*\mu$ be as in clause 2.

(3a) The kernel. By Rescaling a Mollifier Kernel, with $n=d$, radius $1$ and the kernel $\eta$, $\eta_{\varepsilon}$ is a mollifier kernel of radius $\varepsilon$: it is smooth, $0\le\eta_{\varepsilon}$, $\eta_{\varepsilon}(y)=0$ whenever $\varepsilon<\lVert y\rVert$, and $\eta_{\varepsilon}$ is integrable with $\int\eta_{\varepsilon}(z)\,dz=1$. It is continuous by claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous, compactly supported by (0.3)(a) with $r=\varepsilon$, and therefore bounded by claim 1 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable; fix a real $C_{\eta}\ge0$ with $|\eta_{\varepsilon}(y)|\le C_{\eta}$ for all $y$.

(3b) Regularity of $\eta_{\varepsilon}*\mu$. By The Convolution of a Bounded Continuous Function with a Probability Measure on Euclidean Space: Continuity, Boundedness and Differentiation Under the Integral §continuous, with $q=d$, the measure $\mu$, $H=\eta_{\varepsilon}$ and $C=C_{\eta}$, for every $y$ the function $x\mapsto\eta_{\varepsilon}(y-x)$ is Borel and integrable with respect to $\mu$, and $\eta_{\varepsilon}*\mu$ is continuous with $|(\eta_{\varepsilon}*\mu)(y)|\le C_{\eta}$. It is nonnegative by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, the integrand being nonnegative, and Borel by (0.2). This is the first assertion of clause 2.

(3c) The integral formula. Fix $y$ and let $k_{y}(x)=\eta_{\varepsilon}(y-x)\rho(x)$, a nonnegative Borel function by (3b), (0.1) and (0.2). By (0.1), $(\eta_{\varepsilon}*\mu)(y)=\int k_{y}(x)\,dx$, and by (0.2) with $a=y$, $\int k_{y}(x)\,dx=\int k_{y}(y-z)\,dz=\int\eta_{\varepsilon}(z)\rho(y-z)\,dz$. So the nonnegative measurable function $z\mapsto\eta_{\varepsilon}(z)\rho(y-z)$ has the finite integral $(\eta_{\varepsilon}*\mu)(y)$; it is therefore integrable, and the displayed formula of clause 2 holds.

(3d) A pointwise bound. Fix $y$ and put $D(y)=(\eta_{\varepsilon}*\mu)(y)-\rho(y)$. As $\int\eta_{\varepsilon}(z)\,dz=1$, claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives $\rho(y)=\int\eta_{\varepsilon}(z)\rho(y)\,dz$, and then, with (3c),

$$D(y)=\int_{\mathbb{R}^{d}}\eta_{\varepsilon}(z)\bigl(\rho(y-z)-\rho(y)\bigr)\,dz,\qquad|D(y)|\le\int_{\mathbb{R}^{d}}Q(y,z)\,dz,\quad Q(y,z)=\eta_{\varepsilon}(z)\,\bigl|\rho(y-z)-\rho(y)\bigr|,\tag{3.1}$$

the inequality by the bound $|\int f|\le\int|f|$ of claim 2 of Linearity and Monotonicity of the Lebesgue Integral and $0\le\eta_{\varepsilon}$.

(3e) Joint measurability. Let $\iota=\iota^{d,d}$, $\mathrm{pr}_{1}$ and $\mathrm{pr}_{2}$ be the concatenation map and coordinate projections of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs. By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §product, $\iota$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{d})\otimes\mathcal{B}(\mathbb{R}^{d})$ and $\mathcal{B}(\mathbb{R}^{d+d})$, and $\mathrm{pr}_{1},\mathrm{pr}_{2}$ are Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections; so, by claim 4 of Borel Measurability and Bounded Integration on a Metric Space, the maps $(y,z)\mapsto\mathrm{pr}_{1}(\iota(y,z))=y$ and $(y,z)\mapsto\mathrm{pr}_{2}(\iota(y,z))=z$ are measurable from $(\mathbb{R}^{d}\times\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d})\otimes\mathcal{B}(\mathbb{R}^{d}))$ to $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}))$. By claims 2 and 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets their components are measurable, the components $y_{k}-z_{k}$ of $(y,z)\mapsto y-z$ are measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and hence $(y,z)\mapsto y-z$ is measurable into $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}))$. Composing (claim 4 of Borel Measurability and Bounded Integration on a Metric Space), for every Borel $u:\mathbb{R}^{d}\to\mathbb{R}$ the functions $(y,z)\mapsto u(y-z)$, $u(y)$, $u(z)$ are $\mathcal{B}(\mathbb{R}^{d})\otimes\mathcal{B}(\mathbb{R}^{d})$-measurable, and so are their sums, products and absolute values (Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions). In particular $Q$ is a nonnegative measurable function on the product.

(3f) Tonelli. $\lambda_{d}$ is $\sigma$-finite by Lebesgue Measure on Euclidean Space is Sigma-Finite §sigma-finite, so Tonelli's theorem applies to $Q$ on $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\lambda_{d})\otimes(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\lambda_{d})$:

$$\int_{\mathbb{R}^{d}}\Bigl(\int_{\mathbb{R}^{d}}Q(y,z)\,dz\Bigr)dy=\int_{\mathbb{R}^{d}}\Bigl(\int_{\mathbb{R}^{d}}Q(y,z)\,dy\Bigr)dz .$$

Fix $z$. By claim 1 of Linearity and Monotonicity of the Lebesgue Integral and clause 1, $\int Q(y,z)\,dy=\eta_{\varepsilon}(z)\int|\rho(y-z)-\rho(y)|\,dy\le\eta_{\varepsilon}(z)\lVert z\rVert\sqrt{\mathcal{I}(\mu)}$. If $\lVert z\rVert\le\varepsilon$ this is at most $\varepsilon\sqrt{\mathcal{I}(\mu)}\,\eta_{\varepsilon}(z)$; if $\varepsilon<\lVert z\rVert$ then $\eta_{\varepsilon}(z)=0$ and both sides vanish. Integrating in $z$ with claim 1 of Linearity and Monotonicity of the Lebesgue Integral and $\int\eta_{\varepsilon}=1$, the right side of the Tonelli identity is at most $\varepsilon\sqrt{\mathcal{I}(\mu)}$. By (3.1) and monotonicity, $\int|D(y)|\,dy\le\varepsilon\sqrt{\mathcal{I}(\mu)}$. As $D$ is Borel by (3b) and (0.1), and $\int|D|<\infty$, $D$ is integrable (Measure Spaces and the Lebesgue Integral: Standing Notation §integral). This proves clause 2.

Step 4 (Clause 3). Let $L$ and $\Phi$ be as in clause 3.

(4a) The two costs. Since $\mathcal{P}_{2}^{\mathcal{I}}(\mathbb{R}^{d})\subseteq\mathcal{P}_{2}(\mathbb{R}^{d})$ (Finite Fisher Information, the Score and the Fisher Information of a Probability Measure §finite) and $\mu$ is absolutely continuous, $\mu\in\mathcal{P}_{2}^{\mathrm{ac}}(\mathbb{R}^{d})$, and by The Density Cost of a Convex Lipschitz Integrand §cost, computed with the density $\rho$, the function $\Phi\circ\rho$ is Borel and integrable and $\mathcal{G}_{\Phi}(\mu)=\int\Phi(\rho(y))\,dy$, a nonnegative real. The Lipschitz bound of Convex Lipschitz Integrands §integrand, applied with $a,b$ the smaller and the larger of two nonnegative reals, gives $|\Phi(a)-\Phi(b)|\le L|a-b|$ for all $a,b\ge0$; with $a=0$ and $\Phi(0)=0$ it gives $0\le\Phi(b)\le Lb$. By Mollifying a Probability Measure on Euclidean Space: Kernel Bounds, the Translation Estimate, the Mollified Density and Its Cost §cost, applied with the present $d$, $\eta$, $\varepsilon$, $\mu$, $L$ and $\Phi$, the function $\Phi\circ(\eta_{\varepsilon}*\mu)$ is Borel and integrable with respect to $\lambda_{d}$, and by The Mollified Density Cost of a Probability Measure with Finite Second Moment §cost, $\mathcal{G}_{\Phi,\varepsilon}(\mu)=\int\Phi((\eta_{\varepsilon}*\mu)(y))\,dy$.

(4b) Upper bound. For every $y$, $\Phi(\rho(y))-\Phi((\eta_{\varepsilon}*\mu)(y))\le L|D(y)|$ by (4a). Integrating with claim 2 of Linearity and Monotonicity of the Lebesgue Integral and using clause 2,

$$\mathcal{G}_{\Phi}(\mu)-\mathcal{G}_{\Phi,\varepsilon}(\mu)=\int_{\mathbb{R}^{d}}\bigl(\Phi(\rho(y))-\Phi((\eta_{\varepsilon}*\mu)(y))\bigr)\,dy\le L\int_{\mathbb{R}^{d}}|D(y)|\,dy\le L\,\varepsilon\sqrt{\mathcal{I}(\mu)} .$$

(4c) Lower bound. By claim 3 of Image Measures, Measures with Densities, and Change of Variables let $\pi_{\varepsilon}$ be the measure with density $\eta_{\varepsilon}$ with respect to $\lambda_{d}$; $\pi_{\varepsilon}(\mathbb{R}^{d})=\int\eta_{\varepsilon}=1$, so $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\pi_{\varepsilon})$ is a probability space. Fix $y$ and let $h_{y}(z)=\rho(y-z)$, a nonnegative Borel function (0.2). By claim 3 of Image Measures, Measures with Densities, and Change of Variables and (3c), $\int h_{y}\,d\pi_{\varepsilon}=\int\eta_{\varepsilon}(z)\rho(y-z)\,dz=(\eta_{\varepsilon}*\mu)(y)<\infty$, so $h_{y}$ is integrable with respect to $\pi_{\varepsilon}$. Jensen's inequality for convex Lipschitz integrands, applied on this probability space to $\Phi$ and $h_{y}$, and claim 3 of Image Measures, Measures with Densities, and Change of Variables for the nonnegative Borel function $\Phi\circ h_{y}$, give

$$\Phi\bigl((\eta_{\varepsilon}*\mu)(y)\bigr)=\Phi\Bigl(\int h_{y}\,d\pi_{\varepsilon}\Bigr)\le\int\Phi\circ h_{y}\,d\pi_{\varepsilon}=\int_{\mathbb{R}^{d}}Q'(y,z)\,dz,\qquad Q'(y,z)=\eta_{\varepsilon}(z)\,\Phi(\rho(y-z)).$$

By (3e), applied to the Borel functions $\eta_{\varepsilon}$ and $\Phi\circ\rho$, $Q'$ is a nonnegative measurable function on the product. Integrating in $y$ (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) and applying Tonelli's theorem as in (3f),

$$\mathcal{G}_{\Phi,\varepsilon}(\mu)\le\int_{\mathbb{R}^{d}}\Bigl(\int_{\mathbb{R}^{d}}Q'(y,z)\,dz\Bigr)dy=\int_{\mathbb{R}^{d}}\Bigl(\int_{\mathbb{R}^{d}}Q'(y,z)\,dy\Bigr)dz .$$

For fixed $z$, claim 1 of Linearity and Monotonicity of the Lebesgue Integral and (0.2) with $a=-z$, applied to the nonnegative Borel function $\Phi\circ\rho$, give $\int Q'(y,z)\,dy=\eta_{\varepsilon}(z)\int\Phi(\rho(y-z))\,dy=\eta_{\varepsilon}(z)\,\mathcal{G}_{\Phi}(\mu)$; integrating in $z$ with $\int\eta_{\varepsilon}=1$ yields $\mathcal{G}_{\Phi}(\mu)$. Hence $\mathcal{G}_{\Phi,\varepsilon}(\mu)\le\mathcal{G}_{\Phi}(\mu)$, that is $0\le\mathcal{G}_{\Phi}(\mu)-\mathcal{G}_{\Phi,\varepsilon}(\mu)$. Together with (4b) this proves clause 3.