We introduce the sets

$$D_{j}=\{z\in{\mathbb{C}}^{n}:\,u(z)<m_{j}\}\;,\;\;K_{j}=\{z\in{\mathbb{C}}^{n}:\,u(z)\leq m_{j}\}.$$

Since $u$ is a continuous psh exhaustion function, $K_{j}$ is a compact set. Let

$$\rho_{j}=\gamma_{j}\max\{u-m_{j-1},0\}-j,\;j\geq 0.$$

Then $\rho_{j}$ is psh on ${\mathbb{C}}^{n}$ and (1) implies that

(2)

$$\rho_{j}(z)=\rho_{j-1}(z)\;{\rm if}\;u(z)=m_{j},\;j\geq 1.$$

We claim that

(3)

$$\rho_{j}(z)\geq u(z)\;{\rm if}\;z\in{\mathbb{C}}^{n}\setminus D_{j},\;j\geq 0.$$

Indeed, since $\gamma_{j}\geq 1$ and using (1) we obtain

	$\displaystyle\rho_{j}(z)-u(z)$	$\displaystyle=$	$\displaystyle(\gamma_{j}-1)u(z)-\gamma_{j}m_{j-1}-j\geq(\gamma_{j}-1)m_{j}-\gamma_{j}m_{j-1}-j$
		$\displaystyle=$	$\displaystyle(\gamma_{j-1}-1)m_{j}-\gamma_{j-1}m_{j-2}-j+1$
		$\displaystyle\geq$	$\displaystyle(\gamma_{j-1}-1)m_{j-1}-\gamma_{j-1}m_{j-2}-(j-1).$

So $x_{j}:=(\gamma_{j}-1)m_{j}-\gamma_{j}m_{j-1}-j\geq x_{0}=0$, and (3) is proved.

Let $\varphi_{j}=\max\{\varphi,-j\}$. We construct by induction on $j\geq 1$ a sequence of continuous psh functions $\psi_{j}$ on ${\mathbb{C}}^{n}$ with the following properties:

(4)		$\displaystyle\psi_{j}(z)>\varphi_{j}(z)\;{\rm for}\;z\in X\;,\;\;\int_{X\cap K_{j-1}}(\psi_{j}-\varphi_{j})<2^{-j}.$
(5)		$\displaystyle\psi_{j}(z)\geq\rho_{j}(z)\;{\rm for}\;z\in D_{j}\;,\;\;\psi_{j}(z)=\rho_{j}(z)\;{\rm for}\;z\in{\mathbb{C}}^{n}\setminus D_{j}.$
(6)		$\displaystyle\psi_{j}(z)<\psi_{j-1}(z)\;{\rm for}\;z\in K_{j-1},\;{\rm where}\;\psi_{0}=\rho_{0}=\max\{u,0\}.$

Here the integral in (4) is with respect to the area measure on each irreducible component, i.e.

$$\int_{X\cap K}f:=\sum\int_{Y\cap K}f\,\beta^{\dim Y},$$

where the sum is over all irreducible components $Y$ of $X$ which intersect $K$ and $\beta$ is the standard Kähler form on ${\mathbb{C}}^{n}$. (Note that this is a finite sum.)

Assume that the function $\psi_{j-1}$ is constructed with the desired properties. We construct $\psi_{j}$ by applying Proposition 1.2 with $\chi=\varphi_{j}$ and $v=\psi_{j-1}$. (If $j=1$, $\psi_{1}$ is constructed in the same way by applying Proposition 1.2 with $\chi=\varphi_{1}$ and $v=\psi_{0}$.) By (4), $\varphi_{j}\leq\varphi_{j-1}<\psi_{j-1}$ on $X$ (and for $j=1$, clearly $\varphi_{1}<\psi_{0}$ on $X$). Therefore Proposition 1.2 yields a psh function $\widetilde{\varphi}_{j}$ on ${\mathbb{C}}^{n}$ so that $\widetilde{\varphi}_{j}\,|_{{}_{X}}=\varphi_{j}$ and $\widetilde{\varphi}_{j}<\psi_{j-1}$ on $K_{j}$. Using the standard regularization of $\widetilde{\varphi}_{j}$ and the dominated convergence theorem (as $\varphi_{j}\geq-j$) we obtain a continuous psh function $\widetilde{\psi}_{j}$ on ${\mathbb{C}}^{n}$ which verifies

$$\widetilde{\psi}_{j}(z)>\varphi_{j}(z)\;{\rm for}\;z\in X\;,\;\;\int_{X\cap K_{j}}(\widetilde{\psi}_{j}-\varphi_{j})<2^{-j}.$$

Moreover, since $\psi_{j-1}$ is continuous, we can ensure by the Hartogs lemma that we also have $\widetilde{\psi}_{j}(z)<\psi_{j-1}(z)$ for $z\in K_{j}$.

We now define

$$\psi_{j}(z)=\left\{\begin{array}[]{ll}\max\{\widetilde{\psi}_{j}(z),\rho_{j}(z)\},\;{\rm if}\;z\in D_{j},\\ \rho_{j}(z),\;{\rm if}\;z\in{\mathbb{C}}^{n}\setminus D_{j}.\end{array}\right.$$

By (5) and (2) we have $\widetilde{\psi}_{j}<\psi_{j-1}=\rho_{j-1}=\rho_{j}$ on $\partial D_{j}$ (for $j=1$, recall that $\psi_{0}=\rho_{0}$ by definition). So $\psi_{j}$ is a continuous psh function on ${\mathbb{C}}^{n}$ which verifies (5). On $X\setminus D_{j}$ we have by (3) that $\psi_{j}=\rho_{j}\geq u>\varphi_{j}$, while on $X\cap D_{j}$, $\psi_{j}\geq\widetilde{\psi}_{j}>\varphi_{j}$. Since $\rho_{j}=-j\leq\varphi_{j}<\widetilde{\psi}_{j}$ on $X\cap K_{j-1}$, we see that $\psi_{j}=\widetilde{\psi}_{j}$ on $X\cap K_{j-1}$ so

$$\int_{X\cap K_{j-1}}(\psi_{j}-\varphi_{j})\leq\int_{X\cap K_{j}}(\widetilde{\psi}_{j}-\varphi_{j})<2^{-j}.$$

Hence $\psi_{j}$ verifies (4). Finally, we have by (5), $\rho_{j}=-j<\rho_{j-1}\leq\psi_{j-1}$ on $K_{j-1}$ (and for $j=1$, $\rho_{1}=-1<\psi_{0}=0$ on $K_{0}$). Since $\widetilde{\psi}_{j}<\psi_{j-1}$ on $K_{j}$ we conclude that $\psi_{j}<\psi_{j-1}$ on $K_{j-1}$, so (6) is verified.

So we have constructed a sequence of continuous psh functions $\psi_{j}$ on ${\mathbb{C}}^{n}$ verifying properties (4)-(6). Since $\bigcup_{j\geq 1}D_{j}={\mathbb{C}}^{n}$, we have by (6) that the function

$$\psi(z)=\lim_{j\to\infty}\psi_{j}(z)$$

is well defined and psh on ${\mathbb{C}}^{n}$. As $\ldots<\psi_{j+2}<\psi_{j+1}<\psi_{j}$ on $K_{j}$, it follows that $\psi<\psi_{j}$ on $K_{j}$.

Suppose now that $z\in K_{j}\setminus D_{j-1}$, for some $j\geq 2$, so $m_{j-1}\leq u(z)\leq m_{j}$. By the above construction and property (5), we have

$$\widetilde{\psi}_{j}(z)<\psi_{j-1}(z)=\rho_{j-1}(z)\Longrightarrow\psi(z)<\psi_{j}(z)\leq\max\{\rho_{j-1}(z),\rho_{j}(z)\}\leq\gamma_{j}u(z).$$

Similarly, for $z\in K_{1}$ we have

$$\psi(z)<\psi_{1}(z)\leq\max\{\rho_{0}(z),\rho_{1}(z)\}\leq\gamma_{1}\max\{u(z),0\}.$$

Hence $\psi$ satisfies the desired global upper estimates on ${\mathbb{C}}^{n}$.

Property (4) implies that $\psi(z)\geq\varphi(z)$ for every $z\in X$. Let $K$ be a compact in ${\mathbb{C}}^{n}$ and $Y$ be an irreducible component of $X$ so that $\varphi\,|_{{}_{Y}}\not\equiv-\infty$. By (4) we have that for all $j$ sufficiently large

$$0\leq\int_{Y\cap K}(\psi_{j}-\varphi)=\int_{Y\cap K}(\psi_{j}-\varphi_{j})+\int_{Y\cap K}(\varphi_{j}-\varphi)\leq 2^{-j}+\int_{Y\cap K}(\varphi_{j}-\varphi).$$

Hence by dominated convergence, $\int_{Y\cap K}(\psi-\varphi)=0$, which shows that $\psi=\varphi$ on $Y$.

Assume now that $Y$ is an irreducible component of $X$ so that $\varphi\,|_{{}_{Y}}\equiv-\infty$. Then using (4) and the monotone convergence theorem we conclude that

$$\int_{Y\cap K}\psi=\lim_{j\to\infty}\int_{Y\cap K}\psi_{j}=\lim_{j\to\infty}\left(\int_{Y\cap K}(\psi_{j}-\varphi_{j})+\int_{Y\cap K}\varphi_{j}\right)=-\infty,$$

so $\psi\,|_{{}_{Y}}\equiv-\infty$. Therefore $\psi=\varphi$ on $X$, and the proof is finished. ∎