The single-direction flux identity on a simple solid region

Let $j\in {\mathrm{\Sigma }}^0$. By [F5] the flux integrand of $R{e}_k$ through $(D_j,{\phi }_j)$ is $R({\phi }_j)({\phi }_{j,u}\times {\phi }_{j,v}{)}_k$, which is continuous on $D_j$ because ${\phi }_j$ is $C^1$ there by [F3] and $R$ is continuous. By [F2] its second factor vanishes on $D_j^{\circ }$, and $D_j$ is the closure of $D_j^{\circ }$ by [F3], so a continuous function vanishing on $D_j^{\circ }$ vanishes on $D_j$. Hence that patch's flux is ${\int }_{D_j}0=0$.

By [F4] the set $E$ is compact and Jordan measurable and ${\sigma }_k[E]=K:=\{(w,t):w\in D,{\gamma }_1(w)\le t\le {\gamma }_2(w)\}$. The function $H(w,t):=({\mathrm{\partial }}_{k}R)({\sigma }_k^{-1}(w,t))$ is continuous on $K$, since ${\mathrm{\partial }}_{k}R$ is continuous on $O$ by [F5] and ${\sigma }_k^{-1}$ is linear, and $H\circ {\sigma }_k={\mathrm{\partial }}_{k}R$ on $E$. So [L4] applied with $E^{\mathrm{\prime }}=E$ gives ${\int }_E{\mathrm{\partial }}_{k}R={\int }_{K}H$, both integrals existing by [L3] and [L6].

Fix $w\in D$ and put $G(t):=R({\sigma }_k^{-1}(w,t))$, defined and differentiable for every $t$ with ${\sigma }_k^{-1}(w,t)\in O$, with $G^{\mathrm{\prime }}(t)=({\mathrm{\partial }}_{k}R)({\sigma }_k^{-1}(w,t))=H(w,t)$ because varying $t$ moves only the $k$th coordinate. If ${\gamma }_1(w)<{\gamma }_2(w)$ then $G$ is differentiable on $[{\gamma }_1(w),{\gamma }_2(w)]$, whose points lie in $E\subseteq O$ by [F4], and $G^{\mathrm{\prime }}$ is continuous there hence integrable, so [L5] gives ${\int }_{{\gamma }_1(w)}^{{\gamma }_2(w)}H(w,t)dt=G({\gamma }_2(w))-G({\gamma }_1(w))=R^{{\gamma }_2}(w)-R^{{\gamma }_1}(w)$. If instead ${\gamma }_1(w)={\gamma }_2(w)$ then the interval is degenerate, so the integral is $0$, and the increment $R^{{\gamma }_2}(w)-R^{{\gamma }_1}(w)$ is also $0$; the identity holds in that case too.

The functions $R^{{\gamma }_1}$ and $R^{{\gamma }_2}$ are continuous on the compact Jordan base $D$, hence bounded and integrable over $D$ by [L6] and [F6]. By [L1] each $V_j$ with $j\in {\mathrm{\Sigma }}^{+}$ is a bounded Jordan subset of $D$ over which $R^{{\gamma }_2}$ is integrable, and by [F2] those sets are pairwise disjoint — so their pairwise intersections are empty and have content zero — and their union omits from $D$ only a set of content zero. So [L2] gives ${\sum }_{j\in {\mathrm{\Sigma }}^{+}}{\int }_{V_j}{R}^{{\gamma }_2}={\int }_D{R}^{{\gamma }_2}$, and by [L1] the left side is the sum of the upper faces' fluxes.

The same argument applied to the lower sublist gives ${\sum }_{j\in {\mathrm{\Sigma }}^{-}}{\int }_{V_j}{R}^{{\gamma }_1}={\int }_D{R}^{{\gamma }_1}$, and by [L1] each lower face's flux is $-{\int }_{V_j}{R}^{{\gamma }_1}$, so the lower faces' fluxes sum to $-{\int }_D{R}^{{\gamma }_1}$.

By step 1.3 the inner integral in [L3] is $R^{{\gamma }_2}(w)-R^{{\gamma }_1}(w)$ for every $w\in D$, a continuous function of $w$; so [L3] applied to $H$ on $K$ and step 1.2 give ${\int }_E{\mathrm{\partial }}_{k}R={\int }_{K}H={\int }_D(R^{{\gamma }_2}-R^{{\gamma }_1})={\int }_D{R}^{{\gamma }_2}-{\int }_D{R}^{{\gamma }_1}$, the last step by linearity of the integral over $D$.

By [F1] the flux over the presentation is the sum of the $P$ patch fluxes, which splits along the three supplied sublists. Step 1.1 makes the lateral sum zero, step 1.4 makes the upper sum ${\int }_D{R}^{{\gamma }_2}$ and step 1.5 makes the lower sum $-{\int }_D{R}^{{\gamma }_1}$, so the total is ${\int }_D{R}^{{\gamma }_2}-{\int }_D{R}^{{\gamma }_1}$.

Steps 2.1 and 2.2 give the same number for the two sides of the asserted identity, so it holds.