Region charge equals boundary electric flux

For every basis label $(\mathbf{n},\mathbf{m})$, summing ${\gamma }_x(\mathbf{n},\mathbf{m})$ over $x\in R$ cancels each link with both endpoints in $R$, leaves $+\text{\hspace{0.17em}⁣}{n}_e$ on a link directed out of $R$ and $-\text{\hspace{0.17em}⁣}{n}_e$ on a link directed into $R$, and subtracts ${\sum }_{x\in R}(m_x-b_x)$. Therefore ${\sum }_{x\in R}{\gamma }_x(\mathbf{n},\mathbf{m})={\sum }_{e\in \mathrm{\partial }R}ϵ(e,R)n_e-{\sum }_{x\in R}(m_x-b_x)\mathrm{.}$

The equality of basis multipliers in step 1.1 is an equality of diagonal operators on vectors for which both sides are defined. Every $\psi \in {\mathcal{D}}_R$ is in all $D(G_x)$ by physicality and in each boundary $D(E_e)$ by definition, while $N_R$ is bounded; applying the multiplier identity gives ${\sum }_{x\in R}{G}_x\psi =({\sum }_{e\in \mathrm{\partial }R}ϵ(e,R)E_e-N_R)\psi$.

If $\psi \in {\mathcal{D}}_R\subseteq {\mathcal{H}}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$, then $G_x\psi =0$ for every $x\in R$. Substitution in step 2.1 gives the claimed regional charge/flux identity.

For $R=V$, every link has both endpoints in $R$, so $\mathrm{\partial }R=\mathrm{\varnothing }$ and the flux sum is zero. Step 3.1 then yields $N_V\psi =0$, which is exactly ${\sum }_x{n}_x\psi =({\sum }_x{b}_x)\psi$ on all of ${\mathcal{H}}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$.