We already know by Theorem 5.2 that the equation holds on the set of vertices. Furthermore it is clear from the definition, that $c_{1}\left(\overline{\mathcal{O}}^{h\circ p_{\mathfrak{X}^{\prime}}}\right)^{n}$ is supported on the vertices of $\mathfrak{D}$. What remains to show is that this also holds for $\MA(h)$.

Let $U:=\tau\setminus\left\{u\in\tau\;\Big|\;u\text{ is a vertex of }\mathfrak{D}\right\}$. We want to show $\MA(h)(U)=0$. Let $\Delta_{1},...,\Delta_{r}$ be the open faces of $\mathfrak{D}$ of dimension at least one. For every $j\in\{1,...,r\}$ there is a $v_{j}\in\mathbb{R}^{n}\setminus\{0\}$ such that for all $y\in\Delta_{j}$ there exists $\epsilon\in\mathbb{R}_{+}$ such that $y\pm\epsilon v_{j}\in\Delta_{j}$. Furthermore $h_{j}:=h\Big|_{\Delta_{j}}=\boldsymbol{m}_{j}\boldsymbol{x}+v(\alpha_{j})$ for some $\boldsymbol{m}_{j}\in\mathbb{Z}^{n}$ and $\alpha_{j}\in K^{\times}$ and we define $h_{j}^{lin}:=\boldsymbol{m}_{j}\boldsymbol{x}$. Now let $y\in U$. Then there is an $i$ such that $y\in\Delta_{i}$. For $p\in\nabla h(y)$ and $\epsilon$ as above it follows

	$\displaystyle\epsilon\langle v_{i},p\rangle$	$\displaystyle=h_{i}(y)+\langle y+\epsilon v_{i}-y,p\rangle-h_{i}(y)$
		$\displaystyle\leq h_{i}(y+\epsilon v_{i})-h_{i}(y)$
		$\displaystyle=h_{i}^{lin}(\epsilon v_{i})$
		$\displaystyle=\epsilon h_{i}^{lin}(v_{i}),$

hence

$$\langle v_{i},p\rangle\leq h_{i}^{lin}(v_{i}).$$

A similar argument shows

$$-\epsilon\langle v_{i},p\rangle\leq-\epsilon h_{i}^{lin}(v_{i})$$

and hence

$$\langle v_{i},p\rangle\geq h_{i}^{lin}(v_{i}).$$

We conclude $\langle v_{i},p\rangle=h_{i}^{lin}(v_{i})$ and $p$ lies in a hypersurface which depends on $i$ but not on $y$. Hence $\bigcup_{y\in U}\nabla h(y)$ is contained in the union of $r$ hypersurfaces. Therefore

$$\MA(h)(U)=\lambda\left(\bigcup_{y\in U}\nabla h(y)\right)=0,$$

∎