Cutoffs around surface-null edges

If E is empty use eta=0. Otherwise subdivide the compact chart preimages into finitely many smaller closed boxes contained in chart domains, with interiors covering them. On each box the derivative has a bound L, enlarged to at least one, and its density J has a positive lower bound c by regularity and compactness (F1). The parameter subset mapping into E is compact and null: $c{\lambda }_{n-1}(A)\le {\int }_{A}J=0$. The segment integral of DX in the convex box gives $\mathrm{\mid }X(y)-X(z)\mathrm{\mid }\le L\mathrm{\mid }y-z\mathrm{\mid }$.

Fix delta,A_0>0. F2 covers each null preimage by closed cubes with sum of side lengths to the power n-1 as small as desired. Make them open by enlarging the kth side by a positive amount with added volume below a prescribed geometric error $2^{-k}$ times the budget. Subdivide beforehand if needed so every resulting side is smaller than a prescribed positive bound. Intersect with the chart box. Compactness of each null preimage retains finitely many covering cubes; discard empty intersections with that preimage and choose one such point in each retained cube. The Lipschitz bound from step 1.1 places its image in a ball centered at the chosen image point of E with radius $r_j=2L\sqrt{n-1}{\mathrm{\ell }}_j$. Choosing the finitely many chart budgets and side bounds sufficiently small gives a finite open ball cover of E with $r_j<\delta$ and ${\sum }_j{r}_j^{n-1}<A_0$.

Use the fixed bump of F3 and put $b_j(x)=b((x-c_j)\mathrm{/}{r}_j)$ and $\eta =1-{\prod }_j(1-b_j)$. It lies between zero and one and is one on the union of the covering balls, hence near E. Its support lies in the union of the doubled balls, within distance $2\delta$ of E. The finite product rule and $0\le 1-b_j\le 1$ give $\mathrm{\mid }D\eta \mathrm{\mid }\le {\sum }_j\mathrm{\mid }D{b}_j\mathrm{\mid }$, so F3 yields $\int \mathrm{\mid }D\eta \mathrm{\mid }\le C_n{\sum }_j{r}_j^{n-1}<C_n{A}_0$.

The support volume is at most $2^n\mathrm{\mid }{B}_1\mathrm{\mid }{\sum }_j{r}_j^n\le 2^n\mathrm{\mid }{B}_1\mathrm{\mid }\delta A_0$ by volume dilation F4 and translation invariance. Take $\delta =\epsilon \mathrm{/}4$ and $A_0=\epsilon \mathrm{/}(2\max (1,C_n))$. Then the support is within epsilon of E, its gradient integral is less than epsilon, and its volume is bounded by a fixed dimensional constant times ${\epsilon }^2$, tending to zero.