Since the metric is semipositive and toric, by Proposition 5.67 the function $\psi$ is concave. Since, moreover it is algebraic, by Theorem 5.49 it is defined by a toric model $({\mathcal{X}}_{\Pi},D_{\psi},e)$ of $(X_{\Sigma},D_{\Psi})$ in the equivalence class determined by $\psi$. As in Remark 4.66, the irreducible components of ${\mathcal{X}}_{\Pi,o}$ are in bijection with the vertices of $\Pi$. For each vertex $v\in\Pi^{0}$, let $\xi_{v}$ be the point of $X_{\Sigma}^{{\text{\rm an}}}$ corresponding to the generic point of $V(v)$ defined by equation (2.15). Then, by equation (2.29),

$$c_{1}(\overline{L})^{n}\land\delta_{X_{\Sigma}}=\frac{1}{e^{n}}\sum_{v\in\Pi^{0}}\nu_{v}\deg_{D_{\psi}}V(v)\delta_{\xi_{v}}.$$

Thus, by Corollary 5.40,

$$({\operatorname{val}}_{K})_{\ast}(c_{1}(\overline{L})^{n}\land\delta_{X_{\Sigma}})=\frac{1}{e^{n}}\sum_{v\in\Pi^{0}}\nu_{v}\deg_{D_{\psi}}V(v)\delta_{v}.$$

But, using Proposition 3.95 and Proposition 4.105, the Monge-Ampère measure is given by

	$\displaystyle\mathcal{M}_{M}(\psi)$	$\displaystyle=\frac{1}{e^{n}}\mathcal{M}_{M}(e\psi)$
		$\displaystyle=\frac{1}{e^{n}}\sum_{v\in\Pi^{0}}\operatorname{vol}_{M}(v^{\ast})\delta_{v}$
		$\displaystyle=\frac{1}{n!e^{n}}\sum_{v\in\Pi^{0}}\nu_{v}\deg_{D_{\psi}}V(v)\delta_{v}.$

Since $\mathcal{M}_{M}(\psi)$ is a finite sum of Dirac deltas, we obtain that

$${\overline{\mathcal{M}}}_{M}(\psi)=\frac{1}{n!e^{n}}\sum_{v\in\Pi^{0}}\nu_{v}\deg_{D_{\psi}}V(v)\delta_{v}.$$

Hence we have proved (5.71). To prove equation (5.72) we just observe that $x_{v}=(\theta_{0}\circ{\operatorname{\mathbf{e}}}_{K})(v)$. ∎