4.1.8 Continuous potentials and polar sets12EDE2E332B0

Proof. The sets $\{ |f|>N\} $ decrease to the $\mu $-null set $\{ |f|=\infty \} $ as $N\to \infty $, and $\mu $ is finite, so we can choose $N$ with $\mu (\{ |f|>N\} )<\varepsilon /3$. The truncation $f_ N=\max \{ -N,\min \{ f,N\} \} $ is bounded, hence belongs to $L^1(\mu )$. By proposition 2.3.5 there are continuous functions $g_ j$, $j\geq 1$, with $\int |g_ j-f_ N|\, \mathrm d\mu <\varepsilon 2^{-2j}/3$. By Chebyshev's inequality, the set $E_ j=\{ |g_ j-f_ N|>2^{-j}\} $ satisfies $\mu (E_ j)<\varepsilon 2^{-j}/3$, so $\mu (\bigcup _ jE_ j)<\varepsilon /3$. On the Borel set $F=\mathbb R^ d\setminus (\{ |f|>N\} \cup \bigcup _ jE_ j)$ we have $f=f_ N$ and $|g_ j-f|\leq 2^{-j}$ for every $j$, so $g_ j\to f$ uniformly on $F$. Since $\mu (\mathbb R^ d\setminus F)<2\varepsilon /3$, inner regularity gives a compact set $L\subset F$ with $\mu (F\setminus L)<\varepsilon /3$. Then $\mu (\mathbb R^ d\setminus L)<\varepsilon $, and $f|_ L$ is finite and continuous as a uniform limit of continuous functions. $\square$

Proof. (1) Since $p$ is upper semicontinuous (proposition 3.4.11), $\varlimsup _{z\to x}p(z)\leq p(x)$; in particular $p$ is continuous at $x$ if $p(x)=-\infty $. Assume that $p(x)$ is finite; we must show that $\varliminf _{z\to x}p(z)\geq p(x)$. For $z\in \mathbb R^ d$ choose a point $z'\in K$ closest to $z$. Then $|z'-x|\leq |z'-z|+|z-x|\leq 2|z-x|$, so $z'\to x$ as $z\to x$, and $|z'-y|\leq |z'-z|+|z-y|\leq 2|z-y|$ for every $y\in K$. Consequently, for $y\in K$,

Fix $\delta >0$ and split $p=p'_\delta +p''_\delta $, where $p'_\delta (w)=\int _{B(x,\delta )}E_ d(w-y)\, \mathrm d\mu (y)$ is the near part and $p''_\delta $ the far part. The far part is continuous at $x$, since $E_ d(w-y)\to E_ d(x-y)$ uniformly for $y\in K$ with $|y-x|\geq \delta $ as $w\to x$, by the uniform continuity of $E_ d$ on compact subsets of $\mathbb R^ d\setminus \{ 0\} $. The near part, restricted to $K$, is continuous at $x$, being the difference of $p|_ K$ and $p''_\delta |_ K$. Since $\mu $ is carried by $K$, we may integrate the inequalities above with respect to $\mu $ over $B(x,\delta )$; letting $z\to x$, we obtain for $d\geq 3$

and for $d=2$ the bound $\varliminf _{z\to x}p(z)\geq p(x)-(2\pi )^{-1}\mu (B(x,\delta ))\log 2$. Since $p(x)$ is finite, the function $E_ d(x-\cdot )$ is $\mu $-integrable and $\mu (\{ x\} )=0$. By dominated convergence and continuity of $\mu $ from above, $p'_\delta (x)\to 0$ and $\mu (B(x,\delta ))\to 0$ as $\delta \downarrow 0$, which proves $\varliminf _{z\to x}p(z)\geq p(x)$.

(2) Let $M\geq 0$ be an upper bound of $E_ d$ on the compact set $K-K$. If $\nu $ is the restriction of $\mu $ to a Borel set, then $p=E_ d*\nu +E_ d*(\mu -\nu )$ and $E_ d*(\mu -\nu )\leq M\mu (K)$ on $K$, so $E_ d*\nu \geq p-M\mu (K)$ on $K$; hence $E_ d*\nu $ is finite $\nu $-almost everywhere.

We construct inductively pairwise disjoint compact sets $K_1,K_2,\ldots \subset K$ such that the residual measures $\nu _ j=\mu |_{K\setminus (K_1\cup \cdots \cup K_ j)}$ satisfy $\nu _ j(\mathbb R^ d)<2^{-j}$, and such that each $E_ d*(\mu |_{K_ j})$ is finite and continuous on $\mathbb R^ d$. Put $\nu _0=\mu $, and suppose $K_1,\ldots ,K_{j-1}$ have been chosen. Apply lemma 4.1.9 to $\nu =\nu _{j-1}$ and $E_ d*\nu $: there is a compact set $L$ with $\nu (\mathbb R^ d\setminus L)<2^{-j-1}$ on which $E_ d*\nu $ is finite and continuous. Since $\nu $ is carried by $K$ and does not charge $K_1\cup \cdots \cup K_{j-1}$, inner regularity gives a compact set $K_ j\subset (L\cap K)\setminus (K_1\cup \cdots \cup K_{j-1})$ with $\nu (\mathbb R^ d\setminus K_ j)<2^{-j}$. Put $\mu _ j=\nu |_{K_ j}=\mu |_{K_ j}$; then $\nu _ j=\nu -\mu _ j$ has mass less than $2^{-j}$.

It remains to show that $E_ d*\mu _ j$ is finite and continuous. On $K_ j$ the two functions $E_ d*\mu _ j$ and $E_ d*(\nu -\mu _ j)$ are upper semicontinuous and less than $+\infty $, and their sum $E_ d*\nu $ is finite and continuous; hence both are finite on $K_ j$. Moreover, for $x\in K_ j$ and $K_ j\ni w\to x$,

so the restriction of $E_ d*\mu _ j$ to $K_ j$ is continuous. The support of $\mu _ j$ is contained in $K_ j$, so part (1), applied to $\mu _ j$ and the compact set $K_ j$, shows that $E_ d*\mu _ j$ is continuous at every point of $K_ j$. At a point $x\notin K_ j$, the function $E_ d*\mu _ j$ is finite and continuous, because $E_ d(w-y)\to E_ d(x-y)$ uniformly in $y\in K_ j$ as $w\to x$, as for the far part in (1). Thus $E_ d*\mu _ j$ is finite and continuous on $\mathbb R^ d$. Finally, $\mu -\sum _{i\leq j}\mu _ i=\nu _ j$ has mass less than $2^{-j}$, so $\mu =\sum _ j\mu _ j$. $\square$

Proof. Fix a bounded open set $W\supset \overline A$; we construct $\nu $ with support in $\overline W$. Let $a\in A$. By definition 3.4.26, after shrinking the neighborhood as explained above, there are an open ball $V\subset W$ centered at $a$ and a nontrivial function $\psi \in \mathrm{SH}(V)$ with $A\cap V\subset \{ \psi =-\infty \} $; by proposition 3.4.8, $\psi $ is locally integrable. Choose an open ball $B_ a$ centered at $a$ with $\overline{B_ a}\subset V$, and a cutoff $\eta \in C_ c^\infty (V)$ with $0\leq \eta \leq 1$ and $\eta =1$ near $\overline{B_ a}$. By the Riesz decomposition (theorem 3.4.14), $\psi =E_ d*(\eta \mu _\psi )+h$ on $B_ a$, where $\mu _\psi $ is the Riesz measure of $\psi $ and $h$ is harmonic, hence finite; therefore $E_ d*(\eta \mu _\psi )=-\infty $ on $A\cap B_ a$. The measure $\eta \mu _\psi $ is finite, positive and supported in $\overline W$. Since $\mathbb R^ d$ has a countable base, countably many of the balls $B_ a$ cover $A$. Let $\nu _ j$ be the corresponding measures.

Choose $c_ j>0$ with $\sum _ jc_ j\nu _ j(\mathbb R^ d)<\infty $, and put $\nu =\sum _ jc_ j\nu _ j$, a finite positive measure supported in $\overline W$. Let $S\subset \mathbb R^ d$ be compact and let $M$ be an upper bound of $E_ d$ on the compact set $S-\overline W$. For $x\in S$, Tonelli's theorem applied to the nonnegative functions $M-E_ d(x-\cdot )$ gives

Thus $E_ d*\nu (x)=-\infty $ whenever $E_ d*\nu _ j(x)=-\infty $ for some $j$; in particular $E_ d*\nu =-\infty $ on $A$. By proposition 3.4.11, $E_ d*\nu $ is subharmonic and locally integrable, so it is not identically $-\infty $.

Now let $A=\bigcup _ kA_ k$ with polar sets $A_ k$, let $x\in \mathbb R^ d$, and let $B'$ be an open ball centered at $x$. Apply the construction above to the bounded polar sets $A_ k\cap B'$, with one and the same bounded open set $W\supset \overline{B'}$. It gives measures $\nu ^{(k)}$ supported in $\overline W$ with $E_ d*\nu ^{(k)}=-\infty $ on $A_ k\cap B'$. Choosing $c_ k>0$ with $\sum _ kc_ k\nu ^{(k)}(\mathbb R^ d)<\infty $, the Tonelli argument above shows that $\nu =\sum _ kc_ k\nu ^{(k)}$ satisfies $E_ d*\nu =-\infty $ on $A\cap B'$. Since $E_ d*\nu $ is subharmonic on $B'$ and not identically $-\infty $ there, $A$ is polar. $\square$

Proof. The set. Put $a_ j=\mathrm e^{-4^ j}$ for $j\geq 1$, and let $K$ be the set of all sums $\sum _{j\geq 1}\varepsilon _ ja_ j$ with $\varepsilon _ j\in \{ 0,1\} $. Since $a_{j+1}/a_ j=\mathrm e^{-3\cdot 4^ j}\leq \mathrm e^{-12}<1/2$, the tails $R_ k=\sum _{j>k}a_ j$ satisfy

In particular $K\subset [0,R_0]\subset [0,2\mathrm e^{-4}]\subset [0,1/10]$. Let $\mathcal S=\prod _{j\geq 1}\{ 0,1\} $ carry the product topology, and let $c\colon \mathcal S\to \mathbb R$ be the coding map $c(\varepsilon )=\sum _ j\varepsilon _ ja_ j$. If $\varepsilon $ and $\varepsilon '$ agree in their first $k$ digits, then $|c(\varepsilon )-c(\varepsilon ')|\leq R_ k$; hence $c$ is continuous, and $K=c(\mathcal S)$ is compact by lemma 2.7.2. If $\varepsilon $ and $\varepsilon '$ first differ at the $k$-th digit, say $\varepsilon _ k=1$ and $\varepsilon '_ k=0$, then $c(\varepsilon )-c(\varepsilon ')\geq a_ k-R_ k>0$; hence $c$ is injective.

For a word $\sigma \in \{ 0,1\} ^ k$, let $K_\sigma $ be the image of the sequences beginning with $\sigma $. The $2^ k$ sets $K_\sigma $ are compact, pairwise disjoint by injectivity, cover $K$, and have diameter at most $R_ k$. Consequently each $K_\sigma $ is open and closed in $K$, and two distinct points of $K$ lie in different sets $K_\sigma $ once $k$ is large; thus $K$ is totally disconnected. Changing the $k$-th digit of $\varepsilon $ produces a point of $K$ different from $c(\varepsilon )$ at distance $a_ k$, so $K$ has no isolated points. Hence $K$ is a Cantor set: a nonempty compact metric space that is totally disconnected and has no isolated points.

The measure. For $t\in [0,1)$ let $\varepsilon _ j(t)$ be its binary digits, using the terminating expansion at dyadic rationals, and put $\Phi (t)=\sum _ j\varepsilon _ j(t)a_ j$. Each $\varepsilon _ j$ is a Borel step function, so $\Phi $ is Borel as a uniform limit of Borel functions. Let $\mu $ be the image of Lebesgue measure on $[0,1)$ under $\Phi $. It is a probability measure carried by the compact set $K$, and it is a Radon measure by proposition 2.2.2. For $\sigma \in \{ 0,1\} ^ k$, injectivity of $c$ shows that $\Phi (t)\in K_\sigma $ if and only if the first $k$ binary digits of $t$ form the word $\sigma $, and this happens exactly on a dyadic interval of length $2^{-k}$. Hence $\mu (K_\sigma )=2^{-k}$. Every $x\in K$ lies in one of the sets $K_\sigma $ with $\sigma $ of length $k$, so $\mu (\{ x\} )\leq 2^{-k}$ for every $k$; thus $\mu $ has no atoms.

The potential. Since $\log |z-a|^2=4\pi E_2(z-a)$, we have $p=4\pi E_2*\mu $, which is subharmonic and locally integrable on $\mathbb C$ by proposition 3.4.11. Let $z_0\notin K$ and $\delta =\frac12\operatorname {dist}(z_0,K)$. For $|z-z_0|<\delta $ and $a\in K$, the kernel and all its derivatives in $z$ are bounded uniformly, so differentiation under the integral sign shows that $p$ is smooth on $B(z_0,\delta )$ with $\Delta p(z)=\int \Delta _ z\log |z-a|^2\, \mathrm d\mu (a)=0$. Thus $p$ is finite and harmonic on $\mathbb C\setminus K$.

Polarity. Let $x\in K$ and $k\geq 1$, and let $K_\sigma $, with $\sigma $ of length $k$, be the piece containing $x$. Since $\operatorname {diam}K\leq R_0<1$, every term $\log |x-a|^2$ with $a\in K$ is nonpositive, and for $a\in K_\sigma $ we have $|x-a|\leq R_ k\leq 2a_{k+1}$. Consequently

The right side tends to $-\infty $ as $k\to \infty $, so $p(x)=-\infty $. Since $p$ is subharmonic on $\mathbb C$ and not identically $-\infty $, the set $K$ is polar. For $D>\operatorname {diam}K$, the logarithmic potential of the probability measure $\mu $ is $U^\mu =(2\pi )^{-1}\log D-p/(4\pi )$, which equals $+\infty $ on $K$. Since $\mu $ is carried by $K$, Tonelli's theorem gives $I_ D(\mu )=\int U^\mu \, \mathrm d\mu =+\infty $. $\square$