The second statement follows from the first since divisorial points are dense in $X$, see §2.1. To prove the first statement, pick an SNC model $\mathcal{X}$ of $X$ such that $x=x_{E}$ is associated to an irreducible component $E_{\omega}$ of the special fiber. By [BFJ11, Proposition 5.2] there exists $u\in\mathcal{D}(X)$ determined on $\mathcal{X}$ such that $-1\leq u\leq 0$ and $\omega+dd^{c}u$ is determined by an ample class in $N^{1}(\mathcal{X}/S)$. Then $\Capa_{\omega}\{x\}\geq\MA(u)\{x\}=b_{E}((\omega+dd^{c}u)|_{E})^{n}>0$, see §2.7. ∎

Given a Borel set $E\subset X$ we have

$$\int_{E}\MA(u)=\int_{E}(\omega+dd^{c}u)^{n}\leq\int_{E}(M\omega+dd^{c}u)^{n}=M^{n}\int_{E}(\omega+dd^{c}\frac{u}{M})^{n}\leq M^{n}\Capa_{\omega}(E).$$

Here the first inequality follows by writing $M\omega+dd^{c}u=(M-1)\omega+\omega+dd^{c}u$ and expanding the Monge-Ampère measure by multilinearity. ∎

After regularizing we may assume that all functions involved are model functions. Write, symbolically, $T=(\omega+dd^{c}\varphi_{2})\wedge\dots\wedge(\omega+dd^{c}\varphi_{n})$. Then

$$\int(\psi-\varphi)\MA(\varphi_{1},\dots,\varphi_{n})=\int(\psi-\varphi)\omega\wedge T+\int(\psi-\varphi)\,dd^{c}\varphi_{1}\wedge T.$$

Since $0\leq\int(\psi-\varphi)\omega\wedge T\leq M$, the first term in the right-hand side satisfies

$$\int(\psi-\varphi)\omega\wedge T\leq M^{\frac{1}{2}}\left(\int(\psi-\varphi)\omega\wedge T\right)^{\frac{1}{2}}.$$

By the Cauchy-Schwarz inequality (Corollary 3.3), the second term is bounded by

$$\left(\int(\psi-\varphi)\,dd^{c}(\varphi-\psi)\wedge T\right)^{\frac{1}{2}}\left(\int(-\varphi_{1})\,dd^{c}\varphi_{1}\wedge T\right)^{\frac{1}{2}}$$

By the assumption that $-M\leq u_{1}\leq 0$ and $\int\omega^{n}=1$ we have

$$0\leq\int(-\varphi_{1})\,dd^{c}\varphi_{1}\wedge T=\int\varphi_{1}\omega\wedge T-\int\varphi_{1}(\omega+dd^{c}\varphi_{1})\wedge T\leq M.$$

Similarly,

	$\displaystyle 0\leq\int(\psi-\varphi)\,dd^{c}(\varphi-\psi)\wedge T$	$\displaystyle=\int(\psi-\varphi)(\omega+dd^{c}\varphi)\wedge T-\int(\psi-\varphi)(\omega+dd^{c}\psi)\wedge T$
		$\displaystyle\leq\int(\psi-\varphi)(\omega+dd^{c}\varphi)\wedge T.$

Putting this together, and using the concavity of the square root, we get

$$\int(\psi-\varphi)\MA(\varphi_{1},\dots,\varphi_{n})\leq\\ M^{\frac{1}{2}}\left(\left(\int(\psi-\varphi)\omega\wedge T\right)^{\frac{1}{2}}+\left(\int(\psi-\varphi)(\omega+dd^{c}\varphi)\wedge T\right)^{\frac{1}{2}}\right)\\ \leq 2M^{\frac{1}{2}}\left(\int(\psi-\varphi)(\omega+dd^{c}\frac{\varphi}{2})\wedge T\right)^{\frac{1}{2}}.$$

The lemma follows (with the constant $4M/(2M)^{\frac{1}{2^{n}}}<4M$) by repeating this argument $n-1$ times, successively replacing $\varphi_{2},\dots,\varphi_{n}$ by $\varphi/2$. ∎

Let $(\varphi_{j})_{j}$ be a decreasing net of $\omega$-psh model functions converging to $\varphi$. We may assume that $-M\leq\varphi_{j}\leq 0$ for all $j$, where $M\geq 1$. For any $\omega$-psh function $\psi$ with $-1\leq\psi\leq 0$ it follows from Lemma 4.7 that

$$0\leq\int(\varphi_{j}-\varphi)\MA(\psi)\leq 4M\left(\int(\varphi_{j}-\varphi)\MA(\frac{\varphi}{2})\right)^{\frac{1}{2^{n}}}$$

and the right hand side tends to zero as $j\to\infty$ by Theorem 3.1. It therefore follows from the definition of the capacity and from Chebyshev’s inequality that for each integer $m\geq 1$ there exists $j_{m}$ such that the open set $G_{m}:=\{\varphi_{j_{m}}-\varphi>\frac{1}{m}\}$ has capacity $2^{-m}\varepsilon$. We can then set $G:=\bigcup_{m}G_{m}$ and $\varphi_{m}:=\varphi_{j_{m}}$. ∎

As above let $(\varphi_{j})_{j}$ be a decreasing net of $\omega$-psh model functions converging to $\varphi$. After adding a constant we may assume that $\varphi_{j}\leq 0$ for all $j$. For each integer $m\geq 1$, the net $(\max\{\varphi_{j},-m\})_{j}$ decreases to the bounded $\omega$-psh function $\max\{\varphi,-m\}$. We can therefore choose $j_{m}$ such that

(4.1)

$$0\leq\int\left(\max\{\varphi_{j_{m}},-m\}-\max\{\varphi,-m\}\right)\MA\left(\frac{\max\{\varphi,-m\}}{2}\right)\leq(2m)^{-2^{n+1}}.$$

We may further assume $j_{m+1}\geq j_{m}$ for all $m$. Set $\varphi_{m}:=\varphi_{j_{m}}$. We claim that the decreasing sequence $(\varphi_{m})_{m=1}^{\infty}$ converges to $\varphi$. By Theorem 2.10 it suffices to test this at any divisorial point $x\in X$. We have $0\geq\varphi(x)>-\infty$ and $\varphi_{m}(x)\geq\varphi(x)\geq-m$ for $m\geq-\varphi(x)\geq 0$. By (4.1), Lemma 4.7 and the definition of capacity we get

$$0\leq(\varphi_{m}(x)-\varphi(x))\Capa_{\omega}\{x\}\leq\frac{1}{m}$$

for $m\geq|\varphi(x)|$. Now $\Capa_{\omega}\{x\}>0$ by Lemma 4.2, thus $\varphi_{m}(x)$ converges to $\varphi(x)$, which concludes the proof. ∎