Bounded-Lipschitz dual discrepancy for finite signed measures

Let $\omega ,\nu$ be finite signed Borel measures on ${\mathbb{R}}^d$ such that $\mathrm{\mid }\omega -\nu \mathrm{\mid }({\mathbb{R}}^d)<\mathrm{\infty }$. For each bounded Lipschitz test function, integration against $\omega -\nu$ is defined by Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|); boundedness and finite total variation make every test in the following set integrable. Set $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}:={\sup }_x\mathrm{\mid }\varphi (x)\mathrm{\mid }$ and $\mathrm{Lip}(\varphi ):={\sup }_{x\mathrm{\ne }y}\mathrm{\mid }\varphi (x)-\varphi (y)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }$, with $\mathrm{Lip}(\varphi )=0$ when $d=0$, using the Euclidean norm from The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$. Define the bounded-Lipschitz dual discrepancy $d_{BL}(\omega ,\nu ):=\sup \{\mid \int \varphi d(\omega -\nu )\mid :\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}\le 1,\mathrm{Lip}(\varphi )\le 1\}\mathrm{.}$ The integral bound by total variation shows this supremum is finite. This page uses only the displayed dual quantity and makes no separate metric-separation claim. No Jordan decomposition, positivity, or probability normalization is needed.