Corollary 5.27.15.F9517F96057F Let $u,v$ be locally quasi-subharmonic functions on a Riemann surface. If $u\leq v+C$ on an open set for some constant $C$, then $\mu _ u\geq \mu _ v$ there. Moreover,
6FC7AD4CB464
5.27.16
\begin{equation} \mu _{\max (u,v)}=\mu _ u\wedge \mu _ v, \end{equation}
where the right-hand side is the greatest measure dominated by both $\mu _ u$ and $\mu _ v$.
Proof.
Apply lemma 5.27.14 to $u-v-C$ in local coordinates. Since smooth functions have zero polar part, linearity of the polar part gives the first assertion. Applying it to $u-\max (u,v)$ and to $v-\max (u,v)$ gives the inequality $\leq $ in (5.27.16).
For the reverse inequality, add a common smooth correction to make $u$ and $v$ subharmonic on a coordinate disk. On a smaller disk, let $g$ be the Green potential of the restriction of $\lambda =\mu _ u\wedge \mu _ v$ to that disk, so that $\mathrm{dd}^{\mathrm{c}}g=\lambda $. The distributions $u-g$ and $v-g$ have positive Hessians and hence subharmonic representatives $a$ and $b$. Since $\max (u,v)=g+\max (a,b)$ almost everywhere, the Hessian of $\max (u,v)$ dominates $\lambda $. Restricting this inequality to a polar Borel set carrying $\lambda $ proves the reverse inequality on the smaller disk, and these local statements give it on the whole surface.
$\square$
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