The thin-collar flux error is uniform over all smooth comparison regions

Assume first that $S\mathrm{\ne }\mathrm{\varnothing }$. Give $\mathrm{\Omega }$ its induced smooth Riemannian metric. Cover this compact manifold with boundary by its boundary and interior charts and choose a subordinate smooth partition $({\chi }_i)$. Compactness and local finiteness imply that only finitely many supports meet $\mathrm{\Omega }$: finitely many neighborhoods, each meeting only finitely many supports, cover $\mathrm{\Omega }$. Each support is closed in compact $\mathrm{\Omega }$ and lies in its chart $U_i$, hence is compact in $U_i$. Orient each $U_i$ separately and apply [F3] to the compactly supported field ${\chi }_{i}X$, where $X={\mathrm{\nabla }}_h\rho$ is smooth up to the boundary by [F1]. The only manifold boundary of this relatively open chart is $U_i\cap \mathrm{\partial }\mathrm{\Omega }$; the field vanishes near its artificial chart edges. In a positive oriented chart the Riemannian volume form integrates as its density, and the same is true of the induced outward-normal-first boundary form. Thus each local divergence identity is an equality of density integrals, without a chosen global orientation.

For $w=\sqrt{\det G}$, [F5] gives ${\mathrm{div}}_h({\chi }_{i}X)=w^{-1}{\sum }_k{\mathrm{\partial }}_k(w{\chi }_i{X}^k)={\chi }_i{\mathrm{div}}_{h}X+X({\chi }_i)$. Summing the finitely many local identities from step 1.1, locality and linearity of density integration give ${\sum }_i{\mathrm{div}}_h({\chi }_{i}X)={\mathrm{div}}_{h}X$ and ${\sum }_{i}h({\chi }_{i}X,\nu )=h(X,\nu )$, since ${\sum }_i{\chi }_i=1$ and ${\sum }_{i}X({\chi }_i)=0$. Consequently ${\int }_{\mathrm{\Omega }}{\mathrm{\Delta }}_h\rho d{V}_h={\int }_{\mathrm{\partial }\mathrm{\Omega }}h({\mathrm{\nabla }}_h\rho ,\nu )dA$. At $S$, the Gaussian metric and inward-side condition imply $\nu =-{\mathrm{\partial }}_s=-{\mathrm{\nabla }}_h\rho$. Its flux is therefore $-\mathrm{\mid }S\mathrm{\mid }$, which yields the stated exact identity. The argument uses only local orientations and includes disconnected $D$.

On $D$, [F6] and $\mathrm{\mid }{\mathrm{\nabla }}_h\rho \mathrm{\mid }=\mathrm{\mid }\nu \mathrm{\mid }=1$ give $h({\mathrm{\nabla }}_h\rho ,\nu )\le 1$. For the other term, the smooth density $(C+{\mathrm{\Delta }}_h\rho )d{V}_h$ is nonnegative, so positivity and linearity of its integral give $-{\int }_{\mathrm{\Omega }}{\mathrm{\Delta }}_h\rho d{V}_h\le C{\mathrm{Vol}}_h(\mathrm{\Omega })\le CK\eta \mathrm{\mid }S\mathrm{\mid }$, using [F1]. All these integrals are finite on the compact smooth region. Inserting these two bounds into step 2.1 proves the displayed inequality in every dimension $n\ge 1$. When $n=3$, the region, ZFC, compact boundary, unit gradient, Laplacian bound and collar density supplied in the Statement and [F1] also meet every hypothesis of [F2], which independently gives this same comparison. No three-dimensional comparison is used to obtain the exact identity or to infer the result in other dimensions.

The coefficient $CK\mathrm{\mid }S\mathrm{\mid }$ depends only on the one fixed compact collar and boundary, not on $\mathrm{\Omega }$, any smoothing parameter, or the number of components of $D$. This proves the asserted uniform $O(\eta )$ error. If $S=\mathrm{\varnothing }$, [F1] identifies the distance sublevel with the empty collar, so $\mathrm{\Omega }=D=\mathrm{\varnothing }$ and the same conclusions hold with zero integrals.