Let ${\mathcal{X}}$, ${\mathcal{X}}^{\prime}$ be proper dlt models of $X$, with ${\mathcal{X}}^{\prime}$ dominating ${\mathcal{X}}$ via a proper birational morphism $\rho\colon{\mathcal{X}}^{\prime}\to{\mathcal{X}}$. Let ${\mathcal{L}}^{\#}=({\mathcal{L}},\psi_{0})$ be a residually metrized model of $K_{X}$ consisting of a model ${\mathcal{L}}$ of $K_{X}$ determined on ${\mathcal{X}}$ and a continuous metric $\psi_{0}$ on ${\mathcal{L}}_{0}$. Set ${\mathcal{L}}^{\prime}=\rho^{*}{\mathcal{L}}$, $\psi^{\prime}_{0}=\rho^{*}\psi_{0}$ and ${\mathcal{L}}^{\prime\#}=({\mathcal{L}}^{\prime},\psi^{\prime}_{0})$. We must prove that $\mu_{{\mathcal{L}}^{\prime\#}}=\mu_{{\mathcal{L}}^{\#}}$.

Let $\sigma^{\prime}$ be a top-dimensional face of $\Delta({\mathcal{L}}^{\prime})$, $Y^{\prime}$ the associated stratum of ${\mathcal{X}}^{\prime}_{0}$, $Y$ the minimal stratum of ${\mathcal{X}}_{0}$ containing $\rho(Y^{\prime})$ and $\sigma=\sigma_{Y}$ the associated simplex of $\Delta({\mathcal{X}})$. Then $\sigma$ and $\sigma^{\prime}$ have the same dimension, and if we (somewhat abusively) identify $\sigma$ and $\sigma^{\prime}$ with their images in $\operatorname{Sk}(\phi_{\mathcal{L}})\subset X^{\mathrm{an}}$, then $\sigma^{\prime}$ is a rational subsimplex of $\sigma$. It suffices to prove that $\mu_{{\mathcal{L}}^{\prime\#}}(\sigma^{\prime})=\mu_{{\mathcal{L}}^{\#}}(\sigma^{\prime})$.

Now $\rho$ restricts to a birational morphism of $Y^{\prime}\to Y$, so since $\lambda_{\sigma}|_{\sigma^{\prime}}=\lambda_{\sigma^{\prime}}$ and $b_{\sigma}=b_{\sigma^{\prime}}$, it suffices to prove that $\operatorname{Res}_{Y^{\prime}}({\mathcal{L}}^{\prime\#})=\rho^{*}\operatorname{Res}_{Y}({\mathcal{L}}^{\#})$. But this is formal. Indeed, we have $(\rho|_{Y})_{*}(B_{Y^{\prime}}^{{\mathcal{L}}^{\prime}})=B_{Y}^{\mathcal{L}}$ and we can identify $(\rho|_{Y})^{*}K_{(Y,B_{Y}^{\mathcal{L}})}$ with $K_{(Y^{\prime},B_{Y^{\prime}}^{{\mathcal{L}}^{\prime}})}$ in such a way that the restriction of $\psi^{\prime}_{0}$ to ${\mathcal{L}}^{\prime}|_{Y^{\prime}}=K_{(Y^{\prime},B_{Y^{\prime}}^{{\mathcal{L}}^{\prime}})}$ coincides with the pullback under $\rho|_{Y^{\prime}}$ of the restriction of $\psi_{0}$ to ${\mathcal{L}}|_{Y}=K_{(Y,B_{Y}^{\mathcal{L}})}$. ∎

Pick a representative ${\mathcal{L}}^{\#}=({\mathcal{L}},\psi_{0})$ of $\psi^{\#}$ such that ${\mathcal{L}}$ is defined on a proper snc model ${\mathcal{X}}$. Let ${\mathcal{X}}^{\prime}$ be the normalized base change by ${t}={t}^{\prime m}$.

Let $\sigma$ be a $d$-dimensional face of $\Delta({\mathcal{L}})$. By Lemma 5.13, $p^{-1}(\sigma)$ is the union of $g_{\sigma}$ distinct isomorphic faces $\sigma^{\prime}_{\alpha}$ of $\Delta({\mathcal{X}})$ such that

$$b_{\sigma^{\prime}_{\alpha}}=b_{\sigma}/\gcd(m,b_{\sigma})$$

(6.2)

$$\operatorname{Vol}(\sigma^{\prime}_{\alpha})=m^{d}\operatorname{Vol}(\sigma).$$

(6.3)

Further, the induced map $Y^{\prime}_{\sigma^{\prime}_{\alpha}}\to Y$ is generically finite, of degree $f_{\sigma}$ independent of $\alpha$, and we have $f_{\sigma}g_{\sigma}=\gcd(m,b_{\sigma})$. Pick a toroidal modification ${\mathcal{X}}^{\prime\prime}\to{\mathcal{X}}^{\prime}$ with ${\mathcal{X}}^{\prime\prime}$ snc, denote by $\rho\colon{\mathcal{X}}^{\prime\prime}\to{\mathcal{X}}$ the composition, and set ${\mathcal{L}}^{\prime\prime}:=\rho^{*}{\mathcal{L}}$.

Each face $\sigma^{\prime}_{\alpha}$ above is subdivided into simplices $\sigma^{\prime\prime}_{\alpha\beta}$ of $\Delta({\mathcal{L}}^{\prime\prime})$ of dimension $d$, each corresponding to a stratum $Y^{\prime\prime}_{\alpha\beta}$ of ${\mathcal{X}}^{\prime\prime}_{0}$, and $\rho|_{Y^{\prime\prime}_{\alpha\beta}}\colon Y^{\prime\prime}_{\alpha\beta}\to Y$ is generically finite, of degree $f_{\sigma}$. Further, (6.2) implies that

$$b_{\sigma^{\prime\prime}_{\alpha\beta}}=b_{\sigma^{\prime}_{\alpha}}=b_{\sigma}/\gcd(m,b_{\sigma})\quad\text{for all $\alpha,\beta$}.$$

(6.4)

We shall need the following result:

Grant this result for the moment. Lemma 6.8 implies

$$\int_{Y^{\prime\prime}_{\alpha\beta}}\operatorname{Res}_{Y^{\prime\prime}_{\alpha\beta}}({\mathcal{L}}^{\prime\prime\#})=f_{\sigma}\gcd(m,b_{\sigma})^{-2}\int_{Y}\operatorname{Res}_{Y}({\mathcal{L}}^{\#}),$$

and hence

$$(p_{*}\mu^{\prime})(\sigma)=\sum_{\alpha,\beta}\mu^{\prime}(\sigma^{\prime\prime}_{\alpha\beta})=\sum_{\alpha,\beta}\left(\int_{Y^{\prime\prime}_{\alpha\beta}}\operatorname{Res}_{Y^{\prime\prime}_{\alpha\beta}}({\mathcal{L}}^{\prime\prime\#})\right)b_{\sigma^{\prime\prime}_{\alpha\beta}}^{-1}\operatorname{Vol}(\sigma^{\prime\prime}_{\alpha\beta})\\ =f_{\sigma}\gcd(m,b_{\sigma})^{-2}\left(\int_{Y}\operatorname{Res}_{Y}({\mathcal{L}}^{\#})\right)b_{\sigma}^{-1}\gcd(m,b_{\sigma})\sum_{\alpha}\operatorname{Vol}(\sigma^{\prime}_{\alpha})\\ =m^{d}\left(\int_{Y}\operatorname{Res}_{Y}({\mathcal{L}}^{\#})\right)b_{\sigma}^{-1}\operatorname{Vol}(\sigma)=m^{d}\mu(\sigma),$$

thanks to (6.3) and (6.4). ∎

Pick a closed point $\xi^{\prime\prime}\in\mathring{Y^{\prime\prime}}$ and set $\xi=\rho(\xi^{\prime\prime})\in\mathring{Y}$. We use the notation at the end of §6.2 with $p=d$. Namely, pick local coordinates $(z_{i})_{0\leq i\leq n}$ at $\xi$ and $(z^{\prime\prime}_{j})_{0\leq j\leq n}$ at $\xi^{\prime\prime}$ such that $E_{i}=\{z_{i}=0\}$ for $0\leq i\leq d$ and $E^{\prime\prime}_{j}=\{z_{j}^{\prime\prime}=0\}$ for $0\leq j\leq d$. We have $\rho^{*}z_{i}=u_{i}\prod_{j=0}^{d}(z^{\prime\prime}_{j})^{c_{ij}}$ for $0\leq i\leq d$, where $c_{ij}\in{\mathbb{Z}}_{\geq 0}$ and $u_{i}\in{\mathcal{O}}_{{\mathcal{X}}^{\prime\prime},\xi^{\prime\prime}}$ is a unit. Further, by Lemma 5.13, the matrix $(c_{ij})$ has determinant $\pm e_{\sigma}$, where $e_{\sigma}=m/\gcd(m,b_{\sigma})$.

Set

$$\Omega_{1}:=\frac{dz_{0}}{z_{0}}\wedge\dots\wedge\frac{dz_{d}}{z_{d}}{\quad\text{and}\quad}\Omega_{2}:=dz_{d+1}\wedge\dots\wedge dz_{n},$$

and define $\Omega_{1}^{\prime\prime}$, $\Omega^{\prime\prime}_{2}$ similarly. Then $\Omega:=\Omega_{1}\wedge\Omega_{2}$ and $\Omega^{\prime\prime}:=\Omega^{\prime\prime}_{1}\wedge\Omega_{2}^{\prime\prime}$ are local ${\mathbb{Q}}$-generators of $K^{\mathrm{log}}_{\mathcal{X}}$ and $K^{\mathrm{log}}_{{\mathcal{X}}^{\prime\prime}}$ at $\xi$ and $\xi^{\prime\prime}$, respectively. Further,

$$\operatorname{Res}_{Y}(\Omega)=\Omega_{2}|_{Y}{\quad\text{and}\quad}\operatorname{Res}_{Y^{\prime\prime}}(\Omega^{\prime\prime})=\Omega^{\prime\prime}_{2}|_{Y^{\prime\prime}}.$$

Now

$$\rho^{*}\Omega_{1}=\pm e_{\sigma}\Omega^{\prime\prime}_{1}+\frac{1}{z_{0}^{\prime\prime}\dots z^{\prime\prime}_{d}}\tilde{\Omega}^{\prime\prime}_{1},$$

where $\tilde{\Omega}^{\prime\prime}_{1}$ is a regular $(d+1)$-form vanishing at $\xi^{\prime\prime}$, and

$$\rho^{*}\Omega_{2}=q\Omega^{\prime\prime}_{2}+\tilde{\Omega}^{\prime\prime}_{2},$$

where $q\in{\mathcal{O}}_{{\mathcal{X}},\xi^{\prime\prime}}$ and $\tilde{\Omega}^{\prime\prime}_{2}$ is a regular $(n-d)$-form at $\xi^{\prime\prime}$ satisfying $\Omega^{\prime\prime}_{1}\wedge\tilde{\Omega}^{\prime\prime}_{2}=0$. On the one hand, this leads to

$$(\rho|_{Y^{\prime\prime}})^{*}\operatorname{Res}_{Y}(\Omega)=(\rho|_{Y^{\prime\prime}})^{*}(\Omega_{2}|_{Y})=q\Omega^{\prime\prime}_{2}|_{Y^{\prime\prime}}=q\operatorname{Res}_{Y^{\prime\prime}}(\Omega^{\prime\prime}).$$

On the other hand, we also get

$$\rho^{*}\Omega=\pm qe_{\sigma}(1+h)\Omega^{\prime\prime},$$

with $q$ as above and $h\in{\mathcal{O}}_{{\mathcal{X}}^{\prime\prime},\xi^{\prime\prime}}$ vanishing along $Y^{\prime\prime}$.

Define $\Omega^{\mathrm{rel}}$ and $\Omega^{{}^{\prime\prime}\mathrm{rel}}$ by $\frac{d{t}}{{t}}\otimes\Omega^{\mathrm{rel}}=\Omega$ and $\frac{d{t}^{\prime}}{{t}^{\prime}}\otimes\Omega^{{}^{\prime\prime}\mathrm{rel}}=\Omega^{\prime\prime}$, respectively. Then

$$m\frac{d{t}^{\prime}}{{t}^{\prime}}\otimes\rho^{*}\Omega^{\mathrm{rel}}=\rho^{*}(\frac{d{t}}{{t}})\otimes\rho^{*}\Omega^{\mathrm{rel}}=\rho^{*}\Omega=\pm qe_{\sigma}(1+h)\Omega^{\prime\prime}=\pm qe_{\sigma}(1+h)\frac{d{t}^{\prime}}{{t}^{\prime}}\otimes\Omega^{{}^{\prime\prime}\mathrm{rel}},$$

so that

$$\rho^{*}\Omega^{\mathrm{rel}}=\pm\frac{e_{\sigma}}{m}q(1+h)\Omega^{{}^{\prime\prime}\mathrm{rel}}.$$

As a consequence,

$$\rho^{*}|{t}^{\kappa_{\min}}\Omega^{\mathrm{rel}}|_{\psi_{0}}=\frac{e_{\sigma}}{m}|q||(1+h)||({t}^{\prime})^{\kappa^{\prime}_{\min}}\Omega^{{}^{\prime\prime}\mathrm{rel}}|_{\psi^{\prime}_{0}}.$$

Since $h$ vanishes along $Y^{\prime\prime}$, this finally leads to

$$(\rho|_{Y^{\prime\prime}})^{*}\operatorname{Res}_{Y}({\mathcal{L}}^{\#})=\frac{(\rho|_{Y^{\prime\prime}})^{*}|\operatorname{Res}_{Y}(\Omega)|^{2}}{(\rho|_{Y^{\prime\prime}})^{*}|{t}^{\kappa_{\min}}\Omega^{\mathrm{rel}}|^{2}_{\psi_{0}}}=\frac{|(\rho|_{Y^{\prime\prime}})^{*}(\Omega_{2}|_{Y})|^{2}}{\frac{e^{2}_{\sigma}}{m^{2}}|q|^{2}|({t}^{\prime})^{\kappa^{\prime}_{\min}}\Omega^{{}^{\prime\prime}\mathrm{rel}}|_{\psi^{\prime}_{0}}^{2}}\\ =\left(\frac{m}{e_{\sigma}}\right)^{2}\frac{|\operatorname{Res}_{Y^{\prime\prime}}(\Omega^{\prime\prime})|^{2}}{|({t}^{\prime})^{\kappa^{\prime}_{\min}}\Omega^{{}^{\prime\prime}\mathrm{rel}}|_{\psi^{\prime}_{0}}^{2}}=\left(\frac{m}{e_{\sigma}}\right)^{2}\operatorname{Res}_{Y^{\prime\prime}}({\mathcal{L}}^{{}^{\prime\prime}\#}),$$

which completes the proof since $e_{\sigma}=\frac{m}{\gcd(b_{\sigma},m)}$. ∎