A unit vector density has scalar variation measure

The coordinate measures are well-defined by [F2]. For every countable Borel partition $(A_j)$ of a Borel set $A$, the triangle inequality for vector integrals gives ${\sum }_j\mathrm{\mid }\mu (A_j)\mathrm{\mid }\le {\sum }_j{\int }_{A_j}\mathrm{\mid }\nu \mathrm{\mid }d\lambda =\lambda (A)$. Taking the supremum in the definition [F1] gives $V(A)\le \lambda (A)$.

Fix a Borel $A$ and $0<\epsilon <1$. Partition the unit sphere in ${\mathbb{R}}^m$ into finitely many Borel pieces $C_j$ of diameter below $\epsilon$, by intersecting it with a sufficiently fine finite grid of half-open boxes in $[-1,1{]}^m$ (including the outer faces). Choose a unit vector $e_j$ in each nonempty piece, requiring only finite choice. The sets $A_j=A\cap {\nu }^{-1}(C_j)$ are a finite Borel partition of $A$, since $\nu$ is continuous. For $x\in A_j$, $e_j\cdot \nu (x)=1-\frac{1}{2}\mathrm{\mid }{e}_j-\nu (x){\mathrm{\mid }}^2\ge 1-{\epsilon }^2\mathrm{/}2$. Thus $\mathrm{\mid }\mu (A_j)\mathrm{\mid }\ge e_j\cdot \mu (A_j)\ge (1-{\epsilon }^2\mathrm{/}2)\lambda (A_j)$. Adding empty parts gives an admissible countable partition in [F1], so $V(A)\ge (1-{\epsilon }^2\mathrm{/}2)\lambda (A)$. Letting $\epsilon \downarrow 0$ and using step 1.1 proves $V(A)=\lambda (A)$ for every Borel $A$, including zero-mass sets.

Consequently the partition set function is the positive Radon measure $\lambda$ itself. This proves $\mathrm{\mid }\mu \mathrm{\mid }=\lambda$ on the stated space without any Euclidean-domain assumption or restriction of $A$ to an open set.