ScalingStacks

Proof. Pretending everything is smooth, a standard integration by part gives

$$\begin{split}&\int_{X_{t}}(\psi_{t}-\phi_{CY,t})(\omega_{CY,t}^{n}-\omega_{\psi,t}^{n})\\ =&\int_{X_{t}}(\psi_{t}-\phi_{CY,t})dd^{c}(-\psi_{t}+\phi_{CY,t})\wedge(\omega_{CY,t}^{n-1}+\ldots+\omega_{\psi,t}^{n-1})\\ =&\int_{X_{t}}d(\psi_{t}-\phi_{CY,t})\wedge d^{c}(\psi_{t}-\phi_{CY,t})\wedge(\omega_{CY,t}^{n-1}+\ldots+\omega_{\psi,t}^{n-1})\\ \geq&\int_{X_{t}}d(\psi_{t}-\phi_{CY,t})\wedge d^{c}(\psi_{t}-\phi_{CY,t})\wedge\omega_{\psi,t}^{n-1}.\end{split}$$

The same calculations work for continuous $\omega_{FS,t}$-psh functions by standard pluripotential theory.

Combine $\left\lVert\phi_{CY,t}\right\rVert_{L^{\infty}}\leq C$ with the total variation bound in Lemma 4.2,

$$\int_{X_{t}}|\frac{|\log|t||^{n}\omega_{\psi,t}^{n}}{(L^{n})}-d\mu_{t}|<\delta,$$

we get

$$\int_{X_{t}}d(\psi_{t}-\phi_{CY,t})\wedge d^{c}(\psi_{t}-\phi_{CY,t})\wedge\omega_{\psi,t}^{n-1}\leq C\int_{X_{t}}|\omega_{CY,t}^{n}-\omega_{\psi,t}^{n}|\leq\frac{C\delta}{|\log|t||^{n}}.$$

Again by Lemma 4.2, the metric $\omega_{\psi,t}$ is uniformly controlled a.e. on $\text{Log}_{\mathcal{X}}^{-1}(W_{\delta})$, so

$$\int_{\text{Log}_{\mathcal{X}}^{-1}(W_{\delta})}d(\psi_{t}-\phi_{CY,t})\wedge d^{c}(\psi_{t}-\phi_{CY,t})\wedge(dd^{c}\phi_{0}\circ\text{Log}_{\mathcal{X}})^{n-1}\leq\frac{C\delta}{|\log|t||^{n}}.$$

∎