Lemma 7.11.19546063653198

Proof. Since $\varphi $ tends to zero uniformly at $\partial B$, there is an open neighborhood $V$ of $\partial B$ such that $\varphi >-T$ on $V\cap B$; there $\varphi _ T=\varphi $. Thus $\varphi _ T$ is bounded, nonpositive and plurisubharmonic, it has zero boundary values, and it coincides with $\varphi $ outside the compact set $\overline B\setminus V\subset B$. By lemma 7.11.18, the measures $(\mathrm{dd}^{\mathrm{c}}\varphi _ T)^ n$ and $(\mathrm{dd}^{\mathrm{c}}\varphi )^ n$ have the same total mass, which is finite; in particular $\varphi _ T\in \mathcal E_0(B)$.

Put $F_ T=\{ \varphi \leq -T\} $. By maximum locality (proposition 7.3.10), the two measures agree on the complement $\{ \varphi >-T\} $. Since their total masses are equal and finite, they also assign the same mass $m_ T$ to $F_ T$, and Chebyshev's inequality gives

As $\varphi _ T=-T$ on $F_ T$, we get

The two measures differ only on $F_ T$, where each has mass $m_ T$; this proves (7.11.20). Finally, $(\varphi _ T+T)/T$ is plurisubharmonic with values in $[0,1]$, so it is a competitor for $\operatorname {Cap}_ B$ (definition 7.4.2), and therefore $\int _ E(\mathrm{dd}^{\mathrm{c}}\varphi _ T)^ n\leq T^ n\operatorname {Cap}_ B(E)$. Since the two measures agree on $E\setminus F_ T$, we conclude