The Density Cost of a Convex Lipschitz Integrand

In the setting of Probability Measures on Euclidean Space and Random Vectors: Standing Notation, let $d$ be a natural number with $1\le d$, and let $\lambda_{d}$ be Lebesgue measure on $\mathcal{B}(\mathbb{R}^{d})$. Densities with respect to $\lambda_{d}$ are those of The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities, absolute continuity is that of that definition, and $\mathcal{P}_{2}(\mathbb{R}^{d})$ is the set of probability measures with finite second moment; its absolutely continuous members form the set $\mathcal{P}_{2}^{\mathrm{ac}}(\mathbb{R}^{d})$. Convex Lipschitz integrands are those of that definition, functions on the set $[0,\infty)$ of nonnegative reals.

(Density cost)¶ Let $L\in\mathbb{R}$ be nonnegative, let $\Phi$ be a convex Lipschitz integrand with constant $L$, and let $\mu\in\mathcal{P}_{2}^{\mathrm{ac}}(\mathbb{R}^{d})$. By The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §existence, $\mu$ has a density $\rho$ with respect to $\lambda_{d}$, and $\int_{\mathbb{R}^{d}}\rho\,d\lambda_{d}=\mu(\mathbb{R}^{d})=1$. The function $\tilde\Phi:\mathbb{R}\to\mathbb{R}$, $\tilde\Phi(s)=\Phi(\max\{s,0\})$, satisfies $|\tilde\Phi(s)-\tilde\Phi(s')|\le L|\max\{s,0\}-\max\{s',0\}|\le L|s-s'|$ by the Lipschitz bound of Convex Lipschitz Integrands §integrand (applied with $a,b$ the smaller and the larger of $\max\{s,0\}$ and $\max\{s',0\}$) and the elementary inequality $|\max\{s,0\}-\max\{s',0\}|\le|s-s'|$, so it is continuous, hence Borel by claim 3 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets; therefore $\Phi\circ\rho=\tilde\Phi\circ\rho$ is Borel, a composition of Borel maps being Borel as recorded in Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps. Taking $a=0$ and $b=\rho(x)$ in the Lipschitz bound of Convex Lipschitz Integrands §integrand, with $\Phi(0)=0$, gives $0\le\Phi(\rho(x))\le L\rho(x)$ for every $x\in\mathbb{R}^{d}$; by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, $\int_{\mathbb{R}^{d}}\Phi\circ\rho\,d\lambda_{d}\le L\int_{\mathbb{R}^{d}}\rho\,d\lambda_{d}=L$, so $\Phi\circ\rho$ is integrable with respect to $\lambda_{d}$. The density cost of $\mu$ with integrand $\Phi$ is the real number

and by claim 1 of Linearity and Monotonicity of the Lebesgue Integral it satisfies $0\le\mathcal{G}_{\Phi}(\mu)\le L$. It does not depend on the choice of $\rho$: two densities of $\mu$ with respect to $\lambda_{d}$ agree outside a set of $\lambda_{d}$-measure $0$ by The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities §uniqueness, and the integrals agree by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison.