Lemma 4.4.3621606E7390C

Proof. For any interior $a$, choose an entire superharmonic polar function infinite on the exceptional set and finite at $a$, adding a constant to make it nonnegative on $\overline D$. Add $\varepsilon $ times it to every upper competitor. This is a competitor for either set of boundary data; evaluation at $a$ and $\varepsilon \downarrow 0$ gives equality. Negate for lower solutions. For restriction, an upper competitor dominates $H_ f$ in $D$, so its boundary lower limits on interior boundary points of $W$ dominate $H_ f$ by harmonic continuity; on the remaining points they dominate $f$. Thus the restricted upper solution is at most $H_ f$, while the lower solution is at least $H_ f$. Bounded-domain comparison forces equality. $\square$