Variation of a finite vector Radon measure is Radon

If $m=0$, the vector measure and $V$ vanish and the zero field gives the asserted supremum. If $\mathrm{\Omega }=\mathrm{\varnothing }$, the conclusion is immediate. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, then $\mathrm{\Omega }$ is a singleton, $V(\mathrm{\Omega })=\mathrm{\mid }\mu (\mathrm{\Omega })\mathrm{\mid }$, and the constant field $\mu (\mathrm{\Omega })\mathrm{/}\mathrm{\mid }\mu (\mathrm{\Omega })\mathrm{\mid }$ attains the supremum when the vector is nonzero (use the zero field otherwise). Assume henceforth $n,m\ge 1$. For every Borel $A$ and every countable Borel partition $(A_j)$ of $A$, the Euclidean norm satisfies $\mathrm{\mid }\mu (A_j)\mathrm{\mid }\le {\sum }_{i=1}^m\mathrm{\mid }{\mu }_i(A_j)\mathrm{\mid }$. Summing over $j$ gives $V(A)\le \lambda (A)$ for $\lambda :={\sum }_{i=1}^m\mathrm{\mid }{\mu }_i\mathrm{\mid }$, so $V$ is finite and $V(\mathrm{\Omega })<\mathrm{\infty }$.

If $A={⨆}_{k\ge 1}{E}_k$ is a disjoint Borel union, then intersecting any partition of $A$ with each $E_k$ and using countable additivity of each coordinate shows $V(A)\le {\sum }_{k}V(E_k)$. Conversely, for fixed $N$ and $\epsilon >0$, choose for each of the finitely many $k\le N$ a partition of $E_k$ whose sum exceeds $V(E_k)-\epsilon \mathrm{/}N$. Their union, together with $A\setminus {\bigcup }_{k\le N}{E}_k$, is a partition of $A$, so $V(A)\ge {\sum }_{k\le N}V(E_k)-\epsilon$. Letting $\epsilon \downarrow 0$ and then $N\to \mathrm{\infty }$ proves countable additivity. Thus $V$ is a finite positive Borel measure.

For every $f\in C_c(\mathrm{\Omega };{\mathbb{R}}^m)$ with $\mathrm{\mid }f\mathrm{\mid }\le 1$, approximate $f$ uniformly by finite-valued Borel simple functions whose values have norm at most one. The partition definition bounds each simple integral by $V(\mathrm{\Omega })$, and finiteness of the coordinate measures lets the uniform errors tend to zero. Hence the displayed supremum is at most $V(\mathrm{\Omega })$.

Since $V\le \lambda$ and $\lambda ={\sum }_i\mathrm{\mid }{\mu }_i\mathrm{\mid }$ is a finite regular Borel measure by [F1], for every Borel $A$ and $\epsilon >0$ there is compact $K\subseteq A$ with $\lambda (A\setminus K)<\epsilon$, whence $V(A\setminus K)<\epsilon$. There is also an open $G\supseteq A$ with $\lambda (G\setminus A)<\epsilon$, whence $V(G\setminus A)<\epsilon$. Thus $V$ is a regular Borel measure in the sense of Regular Borel measure on an LCH space, and in particular a Radon measure.

If $V(\mathrm{\Omega })=0$, the zero field proves the reverse inequality. Otherwise fix $0<\epsilon <V(\mathrm{\Omega })\mathrm{/}4$ and choose a countable Borel partition $(A_j)$ with sum greater than $V(\mathrm{\Omega })-\epsilon$; truncate after $N\ge 1$ so the tail is below $\epsilon$. Omit indices with $\mu (A_j)=0$ or $V(A_j)\le \epsilon \mathrm{/}(4N)$; the sum of the omitted partition values is at most $\epsilon \mathrm{/}4$. For every retained index, compact inner regularity [F2] gives a nonempty compact $K_j\subset A_j$ with $V(A_j\setminus K_j)<\epsilon \mathrm{/}(8N)$. Since there are finitely many pairwise disjoint compact sets $K_j$ in the metric space $\mathrm{\Omega }$, choose pairwise disjoint open neighborhoods $N_j$ with compact closures in $\mathrm{\Omega }$. Outer regularity [F2] gives open $G_j\supset K_j$ with $V(G_j\setminus K_j)<\epsilon \mathrm{/}(8N)$. Put $U_j=G_j\cap N_j$. Then $K_j\subset U_j$, $V(U_j\setminus K_j)<\epsilon \mathrm{/}(8N)$, and the closures of the $U_j$ remain pairwise disjoint compact subsets of $\mathrm{\Omega }$. Define ${\psi }_j(x):=\frac{\mathrm{dist}(x,\mathrm{\Omega }\setminus U_j)}{\mathrm{dist}(x,\mathrm{\Omega }\setminus U_j)+\mathrm{dist}(x,K_j)}\mathrm{.}$ This is continuous and compactly supported, equals one on $K_j$, and vanishes outside $U_j$. Set $e_j=\mu (A_j)\mathrm{/}\mathrm{\mid }\mu (A_j)\mathrm{\mid }$ and $f={\sum }_j{e}_j{\psi }_j$. The disjoint closures of the $U_j$ imply disjoint supports and $\mathrm{\mid }f\mathrm{\mid }\le 1$. For bounded Borel vector fields, the partition definition and uniform simple-function approximation give $\mathrm{\mid }\int g\cdot d\mu \mathrm{\mid }\le \int \mathrm{\mid }g\mathrm{\mid }dV$. Thus replacing $e_j{\mathbf{1}}_{A_j}$ by $e_j{\psi }_j$ loses at most $V(A_j\setminus K_j)+V(U_j\setminus K_j)$ for each retained $j$. Summing these errors, the omitted terms, and the tail gives $\int f\cdot d\mu >V(\mathrm{\Omega })-3\epsilon$. Letting $\epsilon \downarrow 0$ proves the reverse bound.