Distance Laplacian and normal-collar volume density are uniformly bounded

First take $b>0$ such that $F(S\times [0,b])$ lies inside a larger collar where the Gaussian metric and distance identities of [F1] hold, and fix $0<a_0<b$. Every path starting at $S$ that leaves the width-$b$ collar must first cross its outer face. On its initial portion, the Gaussian metric gives $\mathrm{\mid }\dot{\gamma }{\mathrm{\mid }}_h\ge \mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }$, hence length at least $b$. Thus every point outside that collar has boundary distance at least $b$: take the infimum of these lengths; a component with no boundary has distance $+\mathrm{\infty }$. Inside the collar $\rho =s$, so $\{0\le \rho \le \eta \}=F(S\times [0,\eta ])$ for every $0\le \eta \le a_0$. If $S=\mathrm{\varnothing }$, all these sets are empty, their volumes vanish, and the conclusions are vacuous with any positive $a_0,C,K$; henceforth take $S\mathrm{\ne }\mathrm{\varnothing }$.

In a boundary chart the Gaussian metric determinant is $\det h_s$ and $\mathrm{\nabla }\rho ={\mathrm{\partial }}_s$. The fixed sign convention and the coordinate divergence formula give ${\mathrm{\Delta }}_h\rho =(\det h_s{)}^{-1\mathrm{/}2}{\mathrm{\partial }}_s\sqrt{\det h_s}=\frac{1}{2}\mathrm{tr}(h_s^{-1}{\mathrm{\partial }}_s{h}_s)\mathrm{.}$ The last equality follows by differentiating the determinant column by column, giving ${\mathrm{\partial }}_s\det h_s=(\det h_s)\mathrm{tr}(h_s^{-1}{\mathrm{\partial }}_s{h}_s)$. The resulting scalar is smooth up to $S$ and on the compact collar $F(S\times [0,a_0])$, so its absolute value has a finite upper bound $C$.

In boundary coordinates set $j(p,s)=\sqrt{\det h_s\mathrm{/}\det h_0}$. A change of boundary coordinates multiplies both determinants by the same squared Jacobian, so the ratio is intrinsic. It is smooth and positive and equals one at $s=0$; compactness gives constants $0<m\le j\le K<\mathrm{\infty }$ on $S\times [0,a_0]$. In the product coordinates the pulled-back volume density has coefficient $\sqrt{\det h_s}=j\sqrt{\det h_0}$, proving the stated smooth-density formula by the density pullback law. The pullback is first taken on the larger relatively open collar, where $F$ is a local diffeomorphism, and then restricted to this closed truncation.

Let $E\subset S\times [0,a_0]$ be Borel. Choose a finite boundary-chart cover $U_i$ of compact $S$, and partition $S$ into the disjoint Borel sets $P_1=U_1$, $P_i=U_i\setminus {\bigcup }_{k<i}{U}_k$. In the collar chart $(x_i,s)$ over $U_i$, [F5] computes the volume of $F(E\cap (P_i\times [0,a_0]))$ by the Euclidean integral of ${\mathbf{1}}_E\sqrt{\det h_s}$. The maps in question are homeomorphisms onto their images, so these sets are Borel. When $\dim S\ge 1$, [F6] splits this Euclidean measure as the product of the boundary-coordinate Lebesgue measure and $ds$. Tonelli then makes that integral the iterated integral of ${\mathbf{1}}_{E}j\sqrt{\det h_0}$. The boundary chart-integration formula identifies the inner integral with the $d{A}_{h_0}$ integral over $P_i$; summing the finite disjoint partition and applying Tonelli to $d{A}_{h_0}\times ds$ gives ${\mathrm{Vol}}_h(F(E))={\int }_{E}jd(d{A}_{h_0}\times ds)\mathrm{.}$ The measures are sigma-finite by [F4], and all integrands are nonnegative Borel and hence measurable on the completed products by [F6]–[F7]; [F8] supplies countable choice. If $\dim S=0$, compact $S$ is finite and its induced measure is counting measure by [F3]–[F4]; the same equality follows by summing the one-dimensional interval chart integrals, without applying the positive-dimensional Euclidean product theorem to a zero-dimensional factor.

The lower bound $j\ge m>0$ and upper bound $j\le K$ show from step 3.1 that the pulled-back volume and the product measure have exactly the same Borel null sets. Consequently the homeomorphism $F$ identifies their completion domains as well: write a completed measurable set as a Borel set together with a subset of a Borel null set, transport this representation by $F$, and use the same zero measure on the null subset. Applying this to both directions extends the measure identity of step 3.1 to the completions. This justifies the measure interpretation of the displayed pullback formula for every measurable region asserted in the Statement.

Compact-finiteness gives $\mathrm{\mid }S\mathrm{\mid }<\mathrm{\infty }$. For $0\le \eta \le a_0$ and a completed-volume-measurable $\mathrm{\Omega }\subset \{0\le \rho \le \eta \}$, put $E=F^{-1}(\mathrm{\Omega })\subset S\times [0,\eta ]$. Steps 3.1 and 4.1 and nonnegativity yield ${\mathrm{Vol}}_h(\mathrm{\Omega })={\int }_{E}jd\overline{(d{A}_{h_0}\times ds)}\le K(d{A}_{h_0}\times ds)(S\times [0,\eta ])=K\eta \mathrm{\mid }S\mathrm{\mid }\mathrm{.}$ The rectangle value follows also by Tonelli on its indicator. It is zero when $\eta =0$. The constants $C,K,a_0$ were fixed before choosing $\eta$ and $\mathrm{\Omega }$, proving the claimed uniformity.