4.4.19 Thinness and fine limits78ACC7C20E8C

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. $\square$

Proof. Remove $a$ from $E$, which does not alter regularized reduction. Outside $E$ raw and regularized reductions agree. A deficit at $a$ therefore gives a defining superharmonic majorant $v$ with $v(a)<u(a)$. Lower semicontinuity of $u$ and $v\ge u$ on $E$ give the strict-gap criterion. Conversely a thinness barrier can be chosen bounded near $a$ by truncation above its value there; local exact extension followed by compact reduction gives a bounded nonnegative superharmonic $v$ on $\Omega $, with $v(a)<\alpha <\liminf _ Ev$. On a neighborhood $V$ of $a$ we have $v>\alpha $ along $E$. The peaking assumption gives $\gamma =u(a)-\sup _{\Omega \setminus V}u>0$. For sufficiently small $\lambda >0$, the function $u(a)+\lambda (v-\alpha )$ is nonnegative, majorizes $u$ on all of $E$, and is strictly below $u(a)$ at $a$. This proves the deficit.

Locally work on a ball of radius less than $1/2$. The fundamental potential $w$ of its Lebesgue measure is finite and continuous. For each $a$ in a smaller ball split off the measure of a small ball centered at $a$. Its potential $w_1$ peaks finitely at $a$: inside that ball it is a constant minus $c_ d|y-a|^2/(2d)$, and outside it is strictly radially decreasing by the spherical mean formula. The remaining potential $w_2$ is nonnegative on the original small ball. At a self-thin point of $E$,

The polar discrepancy theorem puts all these points in a polar set. Countably many smaller balls cover the domain. Polarity implies thinness by the prescribed-finite-point pole construction (remove the point itself first); the converse follows from this self-thin point assertion. $\square$

Proof. On $B(a,r)$ the function $u(y)=r-|y-a|$ peaks finitely at $a$. Nonthinness and the preceding theorem give $\widehat R_ Fu(a)=u(a)$. On the complement $w=u-\widehat R_ Fu$ is nonnegative superharmonic, tends to zero at $a$ by lower semicontinuity of the reduction, and is strictly positive: otherwise the strong minimum principle would make it zero on a component, contradicting $\Delta u=-(d-1)/|y-a|<0$ there.

Conversely, if $F$ were thin, its strict gap would give a subharmonic $v$ on a smaller ball with $v(a)=1$ and $v\le -1$ on $F\setminus \{ a\} $. On a still smaller concentric sphere, $v<0$ near $F$, and away from that neighborhood the positive lsc barrier $w$ has a positive lower bound. Choose $A$ so $v-Aw\le 0$ on that sphere. Subtracting a small positive fundamental pole at $a$ makes the boundary upper limit nonpositive at $a$ as well. The bounded-domain maximum principle gives $v\le Aw$ after the pole coefficient tends to zero. The upper limit of $v$ along the complement is at most zero and along $F$ at most $-1$, contradicting the canonical subharmonic value $v(a)=1$. The regularity equivalence now follows from the weak-barrier theorem on $\Omega \cap B(a,r)$ and the locality of finite boundary regularity. $\square$

Proof. Apply $\arctan $ to handle extended limits, and choose intervals $J_ j$ shrinking to the finite transformed limit. There are fine neighborhoods $V_ j$ on which $g\in J_ j$ along $E\setminus \{ a\} $. If infinitely many $V_ j$ are Euclidean neighborhoods the ordinary limit already exists. Otherwise discard finitely many and use amplified positive barriers $v_ j$ for their thin complements on a common ball. Choose $\alpha _ j>0$ with $\sum \alpha _ jv_ j(a)<\infty $. Put $F_ j=(B\setminus V_ j)\cap \{ \alpha _ jv_ j>j\} $ and $F'=\bigcup F_ j$. The sum $v=\sum \alpha _ jv_ j$ tends to infinity along $F'$ at $a$: terms with $j\ge p$ are already above $p$, while each of the finitely many remaining terms is above $p$ sufficiently close to $a$ on its complement. Thus $F'$ is thin. Let $V$ be the fine interior of its complement. Choose an actual Euclidean neighborhood $O_ j$ of $a$ with $O_ j\setminus V_ j\subset \{ \alpha _ jv_ j>j\} $. Then $O_ j\cap V\subset V_ j$, proving the ordinary limit on $E\cap V$.

For the second assertion choose an amplified $u\ge 0$, finite at $a$, infinite along $F$ approaching $a$. If $\liminf _{y\to a,\, y\notin F}(v+u)(y)=\infty $, fine continuity of $u$ implies the fine limit of $v$ is infinity. Otherwise call this liminf $\lambda $ and take $k>\lambda $. Near $a$, $u>k$ on $F\setminus \{ a\} $. Paste $\min (v+u,k)$ off $F$ to the value $k$ on $F\setminus \{ a\} $. The boundary lower limits justify superharmonic pasting. It is bounded below and extends across the polar singleton, with canonical value $\lambda $ at $a$. Its fine neighborhood $\{ w<k\} $ contains $a$; on its complement-of-$F$ portion $w=v+u$. Fine continuity of $w,u$ therefore gives the finite fine limit $\lambda -u(a)$ for $v$. $\square$