PomeLemma 15.2.1.E9045CC3F2A3 Suppose $u,v\in \mathrm{PSH}(X,\theta )$, $v\preceq u$, and $\mathrm{MA}^{\mathrm{ext}}_\theta (u)(\{ u<v\} )=0$. If $0<\delta <1$ and $v_\delta =P_{\delta \theta }(v)\not\equiv -\infty $, then $u\geq v_\delta $.
Lemma 15.2.1E9045CC3F2A3
Proof. Choose $C$ with $v-C\leq u$. Since $v_\delta \leq v$, the obstacle $u-v_\delta $ is bounded below by $-C$ off the pole sets, and it is finite almost everywhere. Hence the envelope
\[ p=P_{(1-\delta )\theta }(u-v_\delta ) \]
is bounded by lemma 5.40.2, and by contact support (theorem 16.28.5) its measure is carried by $D=\{ p+v_\delta =u\} $. Moreover $p+v_\delta \leq u$ everywhere, and
\[ \theta +\mathrm{dd}^{\mathrm{c}}(p+v_\delta ) =((1-\delta )\theta +\mathrm{dd}^{\mathrm{c}}p)+(\delta \theta +\mathrm{dd}^{\mathrm{c}}v_\delta ) \]
is a sum of two positive currents. Positivity of mixed products and ordered contact therefore give
\[ \mathbf1_ D\mathrm{MA}_{(1-\delta )\theta }(p) \leq \mathbf1_ D\mathrm{MA}^{\mathrm{ext}}_\theta (u); \]
this inequality is first proved on common finite truncation levels and then extended by exhaustion. Off the pole sets, every point of $D\cap \{ p<0\} $ satisfies $u=p+v_\delta <v_\delta \leq v$. The hypothesis therefore gives $\mathrm{MA}_{(1-\delta )\theta }(p)(\{ p<0\} )=0$. Since $(1-\delta )\theta $ is semipositive and big, bounded domination (theorem 16.28.56, with proposition 16.28.55), applied to $p$ and the constant $0$ with coefficient zero, yields $p\geq 0$. Hence $u\geq p+v_\delta \geq v_\delta $. $\square$
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