Lemma 5.29.16.F56E1BEB2997 Let $A\geq 0$ and let $0\leq k\leq n$ be an integer. There is a constant $C_ A$ such that all the positive mixed measures
\[ \omega _{v_1}\wedge \cdots \wedge \omega _{v_ k}\wedge \omega ^{n-k}, \qquad v_ i\in \mathrm{PSH}(X,\omega ), \quad \| v_ i\| _\infty \leq A, \]
are bounded above on Borel sets by $C_ A\operatorname {Cap}_\omega $. In particular, their total masses are uniformly bounded. If uniformly bounded functions tend to zero in capacity, then their integrals against these measures tend to zero uniformly in the $v_ i$.
Proof.
Translate the $v_ i$ so that $0\leq v_ i\leq 2A$, and put $q=(v_1+\cdots +v_ k)/B$, where $B\geq \max \{ 2k,2kA,1\} $. Then $0\leq q\leq 1$ and
\[ \omega _ q=B^{-1}\sum _{i=1}^ k\omega _{v_ i}+(1-k/B)\omega , \]
where all coefficients are positive. Expanding $\omega _ q^ n$ shows that the given mixed product is bounded by a constant multiple of $\omega _ q^ n$, and the definition of capacity proves the first claim. The case $k=0$ follows by testing with $v=0$. For a bounded function $z$, split its integral into the parts over $\{ |z|>t\} $ and over its complement; first use convergence in capacity for fixed $t>0$, and then let $t\downarrow 0$.
$\square$
Pome
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