For short, denote $\mu=({\operatorname{val}}_{K})_{\ast}(c_{1}(\overline{L})^{n}\land\delta_{X_{\Sigma}})$. Let $\|\cdot\|_{l}$ be a sequence of semipositive smooth (respectively algebraic) metrics converging to $\|\cdot\|$. By Proposition 2.33, the measures $c_{1}(L,\|\cdot\|_{l})^{n}\land\delta_{X_{\Sigma}}$ converge to $c_{1}({\overline{L}})^{n}\land\delta_{X_{\Sigma}}$. Therefore, the measures $({\operatorname{val}}_{K})_{\ast}(c_{1}(L,\|\cdot\|_{l})^{n}\land\delta_{X_{\Sigma}})$ converge to the measure $\mu$ on $N_{\Sigma}$. Proposition 2.37 implies that the measure of $X_{\Sigma}^{{\text{\rm an}}}\setminus X_{0}^{{\text{\rm an}}}$ with respect to $c_{1}({\overline{L}})^{n}\land\delta_{X_{\Sigma}}$ is zero. Therefore $N_{\Sigma}\setminus N_{\mathbb{R}}$ has $\mu$-measure zero. Denote $\psi_{l}=\psi_{(\|\cdot\|_{l})_{\mathbb{S}}}$. By Proposition 3.108, the measures $\mathcal{M}_{M}(\psi_{l})$ converge to the measure $\mathcal{M}_{M}(\psi)$. Thus $\mu|_{N_{\mathbb{R}}}=n!\mathcal{M}_{M}(\psi)$. If we add to this that the measure of $N_{\Sigma}\setminus N_{\mathbb{R}}$ is zero, we deduce equation (5.82). The last statement of the theorem is clear from Theorem 5.33 and Theorem 5.70. ∎