Theorem 13.15.74394319FDE01

Proof. Take $a=p/n$ in proposition 13.15.6. Since $F_{p/n}(u)$ takes values in $[0,n/p]$, proposition 13.15.2 with $A=n/p$ gives the total mass bound, after enlarging $C_\omega $. For the capacity bound put $B=\max (1,n/p)$ and $v=F_{p/n}(u)$. Then $v/B$ is a test function for the capacity, and

Consequently $\nu _{p,u}\leq \mathrm{MA}_\omega (v)\leq B^ n\mathrm{MA}_\omega (v/B)$, which gives the second bound. The measures $\nu _{p,u}$ do not charge pluripolar sets, because $\mu _ u$ does not. $\square$