Theorem 4.5.3.2C4B36B5F3CD If $\mu \in \mathcal E^+$, $w\ge 0$ is superharmonic, and $G\mu \le w$ $\mu $-almost everywhere, then $G\mu \le w$ everywhere.
Theorem 4.5.32C4B36B5F3CD
Proof. The minimum $q=\min (w,G\mu )$ is a nonnegative superharmonic minorant of a potential, hence is a potential $G\nu $ with $\nu \in \mathcal E^+$. The hypothesis gives $\int (G\mu -G\nu )d\mu =0$, whereas $\int (G\mu -G\nu )d\nu \ge 0$. Their difference is $I(\mu -\nu )\le 0$. Strict definiteness makes $\mu =\nu $, proving the assertion. The nonnegativity of $w$ is necessary, including when $\mu =0$. $\square$
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