Proposition 4.1.12EBEA4A14BCD2

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 $. $\square$