Lemma 5.47.21FA3C3BD45E57

Proof. Replacing $t$ by $\max \{ t,0\} $, which is plurisubharmonic (proposition 5.9.5) and has the same set $\{ t>0\} $, we may assume that $t\ge 0$. Then $t$ is locally bounded, because plurisubharmonic functions are bounded above on compact sets, and $u+\varepsilon t$ is a locally bounded plurisubharmonic function for every $\varepsilon >0$. For $\varepsilon >0$ put $w_\varepsilon =\max \{ u+\varepsilon t,v\} $. On $\{ t>0\} $ we have $u+\varepsilon t>u=v$, so $\{ t>0\} \subset \{ u+\varepsilon t>v\} $, and proposition 7.3.10, applied to $u+\varepsilon t$ and $v$, gives

Multiplying by $t$, which vanishes outside $\{ t>0\} $, we obtain $t(\mathrm{dd}^{\mathrm{c}}w_\varepsilon )^ p\wedge T=t(\mathrm{dd}^{\mathrm{c}}(u+\varepsilon t))^ p\wedge T$. As $\varepsilon $ decreases to zero along a sequence, $w_\varepsilon $ decreases to $\max \{ u,v\} $ and $u+\varepsilon t$ decreases to $u$, because $t\ge 0$. The weighted form of theorem 7.3.7, with the constant sequence of weights equal to $t$, therefore yields

Interchanging the roles of $u$ and $v$, we also get $t(\mathrm{dd}^{\mathrm{c}}\max \{ u,v\} )^ p\wedge T=t(\mathrm{dd}^{\mathrm{c}}v)^ p\wedge T$, and hence $t(\mathrm{dd}^{\mathrm{c}}u)^ p\wedge T=t(\mathrm{dd}^{\mathrm{c}}v)^ p\wedge T$. Finally, multiply the coefficient measures of both sides by $1/(t+\delta )$ and let $\delta \downarrow 0$. Since $t/(t+\delta )\to \mathbf1_{\{ t>0\} }$ pointwise and boundedly, dominated convergence with respect to the total variations of the coefficient measures gives the assertion. $\square$