Lemma 4.4.20201EE1A61AF0

Proof. Shrink a ball so the strict-gap criterion is represented by $G_ B\nu +h$ there, with $\nu $ finite on a larger compact neighborhood. The harmonic part has no effect on the gap. Finiteness at $a$ implies no atom there. For restrictions $\nu _ j=\nu |_{B(a,\varepsilon _ j)}$, their potentials $p_ j$ have $p_ j(a)\downarrow 0$. The difference $G_ B\nu -p_ j$ is harmonic near $a$, so every $p_ j$ retains the same strict limiting gap $\delta >0$ along $E$. Choose a subsequence with $\sum p_ j(a)<\infty $. Its sum is superharmonic, finite at $a$, and its first $m$ terms have limiting lower bound at least $m\delta $. This proves amplification. On every sufficiently small sphere, $E$ is contained in the Borel set $\{ w\ge M\} $. The supermean inequality bounds its relative outer area by $w(a)/M$; let $M\to \infty $.

Represent an amplified barrier locally as a fundamental potential of a finite compactly supported measure plus a harmonic function. The latter can be discarded. Push the measure forward by the Lipschitz map $T$. The decreasing fundamental kernel gives $U(T_\# \nu )(Ty)\ge U\nu (y)$, whereas equality holds at $Ta$ by the radial-distance hypothesis. The transformed potential is finite there. Any sequence in $T(E)$ tending to $Ta$ has preimages tending to $a$, again by radial distance. Thus its values tend to infinity, proving thinness. In dimension two the compact supports make all negative parts bounded; add a local constant as needed. $\square$