Theorem 4.4.26A856D11F5895

Proof. We give the kernel and measure details used in both directions. If $U\Subset B$ is open, compact exhaustion and the equilibrium measures give a measure $\mu _ U$ on $\partial U$ with mass $C(U)$ and $G_ B\mu _ U=1$ on $U$. Indeed their masses are bounded by $C(\overline U)$; weak compactness gives a limit supported on $\partial U$, and for each interior $x$ the eventual supports avoid a neighborhood of $x$, so the continuous kernel can be tested there. A cutoff equal to one around $\overline U$ gives the limiting mass. Also, if $\nu $ is supported on a closed $F$ and $G_ B\nu \ge b$ quasi-everywhere on $F$, then $\nu (F)\ge b C_*(F)$. To see this, integrate against each compact equilibrium measure $\mu _ K$, $K\subset F$, use reciprocity and $G_ B\mu _ K\le 1$, and then take the compact supremum.

Here is the uniform separated-shell estimate. For $\Phi (y-a)\in [\lambda ^ j,\lambda ^{j+1}]$ and $\Phi (z-a)\notin (\lambda ^{j-1},\lambda ^{j+2})$,

In $d\ge 3$, the radii of the two shells differ by at least the fixed factor $\lambda ^{1/(d-2)}$; the reverse triangle inequality bounds $|y-z|$ below by a fixed positive multiple of $|z-a|$. In the plane, write the radii as $e^{-t}$. In the outer case their ratio is at most $q=e^{-(\lambda -1)}<1$, hence $\log (1/|y-z|)\le \Phi (z-a)+\log (1/(1-q))$. In the inner case the same estimate with the variables reversed gives $\Phi (y-z)\le \lambda ^{-1}\Phi (z-a)+\log (1/(1-q))$. Since $z\in B$ gives $\Phi (z-a)\ge \log 2$, these are uniform ratio bounds. The same argument applies to shells of width three whose indices are separated by six. If $G_ B\nu (a)<\infty $ and $\nu $ is finite, then $\int \Phi (z-a)d\nu =G_ B\nu (a)+\Phi (1/2)\nu (B)<\infty $. Thus the potential contributed by all sufficiently separated shells is bounded by a constant independent of their index and target point.

Suppose first the series converges. Choose open $U_ j\Supset E_ j$ with compact closure in $B$, avoiding $\{ \Phi >\lambda ^{j+2}\} $, and $C(U_ j)\le C^*(E_ j)+\varepsilon _ j$, where $\sum \lambda ^ j\varepsilon _ j<\infty $. Finite initial indices outside $B$ may be discarded. Their open equilibrium potentials $p_ j$ satisfy $p_ j=1$ on $U_ j$ and $p_ j(a)\le \lambda ^{j+2}C(U_ j)$. Hence $v_ k=\sum _{j\ge k}p_ j$ is superharmonic, finite at $a$, with $v_ k(a)\to 0$, while $v_ k\ge 1$ on all sufficiently near points of $E\setminus \{ a\} $. For large $k$ this is the strict-gap criterion.

Conversely use an amplified thinness potential $v=G_ B\nu $, with $\nu $ finite and $v(a)<\infty $. To obtain it on this fixed ball, first extend the local amplified barrier exactly to an entire superharmonic function, then restrict its Riesz measure to $B$. That measure is finite because the function is defined on a neighborhood of $\overline B$. The discarded harmonic function is bounded near $a$ and has no effect on the infinite approach. Remove $a$ from $E$ and discard empty shells. Set $b_ j=\min (j,\inf _{E_ j}v)$, so $b_ j$ is finite and tends to infinity. The Borel sets

contain $E_ j$ and are capacitable. Choose compact $K_ j\subset U_ j$ with $C(K_ j)>C(U_ j)-\varepsilon _ j$, with summable weighted errors. Fix a residue class modulo six and put $K=\{ a\} \cup \bigcup K_{6j+p}$. This is compact. Its reduction $w=\widehat R_ Kv=G_ B\eta $ has $w(a)\le v(a)<\infty $, is supported on $K$, and $\eta (\{ a\} )=0$. The separated-shell estimate bounds the contribution of all the other $K_{6k+p}$ at points of $K_{6j+p}$ by a fixed $\beta $. Since $w=v>b_{6j+p}-1$ quasi-everywhere on that compact set, for all sufficiently large $j$ its own contribution is at least one quasi-everywhere. The preceding mass-capacity inequality gives $\eta (K_{6j+p})\ge C(K_{6j+p})$. Finally $\int \Phi (z-a)d\eta <\infty $, so

Apply this to all six classes, restore the summable errors, and use $C^*(E_ j)\le C(U_ j)$. This proves necessity. The last assertion is exactly the already proved equivalence between irregularity and thinness of the complement. $\square$