For simplicity we write $\omega_{\varphi}=\omega+dd^{c}\varphi$.

First we briefly indicate how to extend to $\omega$-psh of finite energy the calculus that we developed in §3. Let $\varphi,\psi\in\mathcal{E}^{1}(X,\omega)$. Since $E_{\omega}$ is convex, $(1-t)\varphi+t\psi\in\mathcal{E}^{1}(X,\omega)$ for any $t\in[0,1]$. For any $0\leq i\leq n$, define $\omega_{\varphi}^{i}\wedge\omega_{\psi}^{n-i}$ to be the unique probability measure such that

(8.1)

$$\sum_{k=0}^{n-1}\binom{n}{k}(1-t)^{k}t^{n-k}\omega_{\varphi}^{k}\wedge\omega_{\psi}^{n-k}=\MA((1-t)\varphi+t\psi)-(1-t)^{n}\,\MA(\varphi)-t^{n}\,\MA(\psi)$$

for any $t=j/n$ with $1\leq j\leq n-1$. By Proposition 6.9, we get $\omega_{\varphi_{j}}^{i}\wedge\omega_{\psi_{j}}^{n-i}\to\omega_{\varphi}^{i}\wedge\omega_{\psi}^{n-i}$ for any decreasing sequence of $\omega$-psh functions $\varphi_{j}\to\varphi$ and $\psi_{j}\to\psi$. In particular, $\omega_{\varphi}^{i}\wedge\omega_{\psi}^{n-i}$ is a probability measure. Replacing $\MA(\cdot)=(\omega+dd^{c}\cdot)^{n}$ by $(\omega+dd^{c}\cdot)^{i+j}\wedge\omega^{n-(i+j)}$ in  (8.1), we can further define probability measures of the same mass $\omega^{i}_{\varphi}\wedge\omega^{j}_{\psi}\wedge\omega^{n-(i+j)}$ as soon as $i,j\geq 0$ and $i+j\leq n$.

Observe that by definition and Lemma 6.10, these measures integrate $\omega$-psh functions of finite energy. By continuity, it also follows that the Cauchy-Schwarz inequality holds

$$\int-hdd^{c}g\wedge T\leq\left(\int hdd^{c}h\wedge T\right)^{1/2}\,\left(\int gdd^{c}g\wedge T\right)^{1/2}$$

for any $h,g$ lying in the vector space generated by $\mathcal{E}^{1}(X,\omega)$ and for any $T$ a positive linear combination of measures of the type $\omega^{i}_{\varphi}\wedge\omega^{j}_{\psi}\wedge\omega^{n-(i+j)}$ with $i+j\leq n$ and $\varphi,\psi\in\mathcal{E}^{1}(X,\omega)$.

Now pick $\varphi,\psi\in\mathcal{E}^{1}(X,\omega)$. We claim that

(8.2)

$$\int(\varphi-\psi)dd^{c}(\varphi-\psi)\wedge\omega^{n-1}\leq C\,\left(\int(\psi-\varphi)(\MA(\varphi)-\MA(\psi))\right)^{2^{1-n}}$$

for some constant $C$ depending on $\varphi$ and $\psi$.

Grant this claim, and suppose $\MA(\varphi)=\MA(\psi)$. We conclude the proof as in [YZ10]. We may assume $\sup\varphi=\sup\psi$. By Cauchy-Schwarz inequality, for any model function $h$ we get

$$\int(\varphi-\psi)dd^{c}h\wedge\omega^{n-1}\leq D^{1/2}\left|\int(\varphi-\psi)dd^{c}(\varphi-\psi)\wedge\omega^{n-1}\right|^{1/2}=0~,$$

with $0\leq D:=\int-hdd^{c}h\wedge\omega^{n-1}<+\infty$. Let $\mathcal{L}$ be any $\mathbf{R}$-line bundle in a model $\mathcal{X}$ whose numerical class is equal to $\omega$. The above equality applied to the model function determined in $\mathcal{X}$ by $\sum_{E}b_{E}(\varphi-\psi)(\ord_{E})E$ with $\mathcal{X}_{0}=\sum b_{E}E$ yields

$$\left(\sum_{E}b_{E}(\varphi-\psi)(\ord_{E})E\right)^{2}\cdot\mathcal{L}^{n-1}=0$$

which in turn implies $\sum_{E}b_{E}(\varphi-\psi)(\ord_{E})E$ to be proportional to $\mathcal{X}_{0}$ by [YZ10, Theorem 2.1.1(b)]. Since we normalized $\varphi,\psi$ by $\sup\varphi=\sup\psi$, we conclude that $\varphi=\psi$ on the vertices of $\Delta_{\mathcal{X}}$.

Now consider any (sufficiently) high model $\pi:\mathcal{X}^{\prime}\to\mathcal{X}$. By [BFJ11, Proposition 5.2] there exists a model function $h$ such that $\omega^{\prime}=\omega+dd^{c}h$ is induced by a ample divisor in $\mathcal{X}^{\prime}$. Then the functions $\varphi-h$ and $\psi-h$ are both $\omega^{\prime}$-psh, normalized by $\sup(\varphi-h)=\sup(\psi-h)$, and satisfy $(\omega^{\prime}+dd^{c}(\varphi-h))^{n}=(\omega^{\prime}+dd^{c}(\psi-h))^{n}$. By what precedes we get $\varphi=\psi$ on the vertices of $\mathcal{X}^{\prime}$. This implies $\varphi=\psi$ on $X^{\mathrm{div}}$, hence on $X^{\mathrm{qm}}$ by Proposition 2.8, hence on $X$ since $\varphi=\sup_{\mathcal{X}}\varphi\circ p_{\mathcal{X}}$ for any $\omega$-psh function by [BFJ11, Proposition 7.6].

We now prove the claim. For this we reproduce the argument of [Bło03]. By $C$ we will denote possibly different constants depending on $\omega,\varphi,\psi$. Set $\rho=\varphi-\psi$. For $k=0,1,...,n-1$ we will prove inductively that

$$0\leq\int-\rho dd^{c}\rho\wedge\omega_{\varphi}^{i}\wedge\omega_{\psi}^{j}\wedge\omega^{k}\leq Ca^{2^{-k}}$$

where

$$a=\int(\psi-\varphi)(\MA(\varphi)-\MA(\psi))=\int-\rho dd^{c}\rho\wedge T\geq 0$$

with $T=\sum_{l=0}^{n-1}\omega_{\varphi}^{l}\wedge\omega_{\psi}^{n-1-l}$, and $i,j$ are such that $i+j+k=n-1$. For $k=n-1$ we will then obtain the desired estimate.

If $k=0$, then

$$\int\rho dd^{c}\rho\wedge\omega_{\varphi}^{i}\wedge\omega_{\psi}^{j}\leq\int\rho dd^{c}\rho\wedge T=a$$

Assume that (3.1) holds for $0,1,...,k-1$. We have

$$\omega_{\varphi}^{i}\wedge\omega_{\psi}^{j}\wedge\omega^{k}=\omega_{\varphi}^{i+k}\wedge\omega_{\psi}^{j}-dd^{c}\varphi\wedge\alpha$$

where

$$\alpha=\omega_{\varphi}^{i}\wedge\omega_{\psi}^{j}\wedge\sum_{l=0}^{k-1}\omega_{\varphi}^{l}\wedge\omega^{k-1-l}$$

Therefore

	$\displaystyle 0\leq\int-\rho dd^{c}\rho\wedge\omega_{\varphi}^{i}\wedge\omega_{\psi}^{j}\wedge\omega^{k}$	$\displaystyle\leq\int-\rho dd^{c}\rho\wedge(T-dd^{c}\varphi\wedge\alpha)$
		$\displaystyle=-\int\rho dd^{c}\rho\wedge T-\int\rho dd^{c}\varphi\wedge\alpha\wedge dd^{c}\rho$

This means that

$$0\leq\int-\rho dd^{c}\rho\wedge\omega_{\varphi}^{i}\wedge\omega_{\psi}^{j}\wedge\omega^{k}\leq a-\int\rho dd^{c}\varphi\wedge\alpha\wedge dd^{c}\rho.$$

We have

$$-\int\rho dd^{c}\varphi\wedge\alpha\wedge dd^{c}\rho\leq\left|\int\rho dd^{c}\varphi\wedge\alpha\wedge\omega_{\varphi}\right|+\left|\int\rho dd^{c}\varphi\wedge\alpha\wedge\omega_{\psi}\right|.$$

If $\eta$ is equal to $\varphi$ or $\psi$, the Cauchy-Schwarz inequality gives

$$\left|\int\rho dd^{c}\varphi\wedge\alpha\wedge\omega_{\eta}\right|\leq\left(\int\rho dd^{c}\rho\wedge\alpha\wedge\omega_{\eta}\right)^{1/2}\,\left(\int\varphi dd^{c}\varphi\wedge\alpha\wedge\omega_{\eta}\right)^{1/2}$$

By the inductive assumption, we have $\int-\rho dd^{c}\rho\wedge\alpha\wedge\omega_{\eta}\leq Ca^{2^{-(k-1)}}$, and since $\varphi\in\mathcal{E}^{1}(X,\omega)$ we get $\int-\varphi dd^{c}\varphi\wedge\alpha\wedge\omega_{\eta}<+\infty$. The proof is complete. ∎