Internal faces cancel for glued boxes

For a box ${\prod }_i(a_i,b_i)$, its 2n faces are $x_i=a_i$ and $x_i=b_i$, with remaining coordinates in their closed intervals. Each is a compact subset of an affine regular hypersurface, with area element the ordinary product measure and outward normal $-e_i$ or $e_i$. Put every face boundary (at least two endpoint coordinates) into E. On each face it is a finite union of parameter-coordinate hyperplanes and is null: a bounded hyperplane strip has arbitrarily small volume by giving the fixed coordinate an arbitrarily short interval. Off E exactly one coordinate is an endpoint and the domain is locally on one side of that plane. Thus all three boxes satisfy the specified finite-face conditions.

The common face $S=\{0\}\times [0,1{]}^{n-1}$ has normal $e_1$ for $Q_{-}$ and $-e_1$ for $Q_{+}$. The same continuous F has the same trace on both sides, so the sum of its flux densities is $F\cdot e_1+F\cdot (-e_1)=0$ pointwise off E. F1 applies to both restrictions of F. Their volume integrals add to the integral on Q because the omitted plane S is volume-null by the strip argument in step 1.1. Their exposed face integrals add to the faces of Q, while the two integrals over S cancel. This proves the asserted identity and all required gluing data.

For $F(x)=x$, the divergence is n and each small box has volume one. On $Q_{-}$ the face $x_1=-1$ contributes 1 and $x_1=0$ contributes 0; on $Q_{+}$ the face $x_1=1$ contributes 1 and $x_1=0$ contributes 0. For each $i\ge 2$ the face $x_i=1$ contributes 1 on each box and $x_i=0$ contributes 0. Each box therefore has flux $1+(n-1)=n$, and Q has flux $2+2(n-1)=2n$, equal to its volume integral. For the further constant field $F=e_1$ the shared fluxes are explicitly 1 and -1, exhibiting nonzero cancellation.