Two Cantor sets with different logarithmic capacities

By [F3] and [F8] the Cantor measure ${\mu }_c$ is a Borel probability with ${\mu }_c(C)=1$ and no atoms, so ${\mu }_c\otimes {\mu }_c(\{(x,x):x\in C\})=\int {\mu }_c(\{x\})d{\mu }_c(x)=0$; by [F2] the set $C$ is compact and uncountable with ${\lambda }_1(C)=0$.

For every $m\ge 0$ and every word $b\in \{0,1{\}}^m$ the level-$m$ basic interval $I_b$ has ${\mu }_c(I_b)=2^{-m}$, and by [F2] the intervals $I_b$, $b\in \{0,1{\}}^m$, are the digit cylinders and are pairwise disjoint.

If $x=\mathrm{\Phi }(a)$ and $y=\mathrm{\Phi }(a^{\mathrm{\prime }})$ have different first $m$ digits, let $k<m$ be the first index with $a_k\mathrm{\ne }{a}_k^{\mathrm{\prime }}$; then $\mathrm{\mid }x-y\mathrm{\mid }\ge 2\cdot 3^{-k-1}-{\sum }_{j>k}2\cdot 3^{-j-1}=3^{-k-1}\ge 3^{-m}$, so $\{\mathrm{\mid }x-y\mathrm{\mid }<3^{-m}\}$ is contained in the set where the first $m$ digits agree, that is, in ${\bigcup }_b{I}_b\times I_b$ with the $I_b$ of step 1.2. Hence $({\mu }_c\otimes {\mu }_c)(\{\mathrm{\mid }x-y\mathrm{\mid }<3^{-m}\})\le {\sum }_{b\in \{0,1{\}}^m}{\mu }_c(I_b{)}^2=2^m\cdot 2^{-2m}=2^{-m}$.

The construction of the cells is licensed by [F7]. For $n=1$, $2{\mathrm{\ell }}_1=2e^{-4}<1={\mathrm{\ell }}_0$; for $n\ge 2$, ${\mathrm{\ell }}_n={\mathrm{\ell }}_{n-1}^4$ and ${\mathrm{\ell }}_{n-1}\le e^{-4}$, so $2{\mathrm{\ell }}_n\mathrm{/}{\mathrm{\ell }}_{n-1}=2{\mathrm{\ell }}_{n-1}^3\le 2e^{-12}<1$. Thus the two children of each level-$(n-1)$ cell are disjoint closed intervals of length ${\mathrm{\ell }}_n$ contained in it, and $(K_n)$ is a nested sequence of nonempty compact sets with $2^n$ cells of length ${\mathrm{\ell }}_n$ at level $n$. Therefore $K={\bigcap }_n{K}_n$ is compact and nonempty, and ${\lambda }_1(K)=0$ because the level-$n$ cells cover $K$ and their total length $2^n{\mathrm{\ell }}_n=2^n{e}^{-4^n}\to 0$.

Since ${\mu }_c$ has total mass $1$ and support in $[0,1]$, [F1] gives $I({\mu }_c)=\iint \log \frac{1}{\mathrm{\mid }x-y\mathrm{\mid }}d{\mu }_c(x)d{\mu }_c(y)$; applying [F4] on the product measure of [F5] and splitting the integral at $t=m\log 3$, $I({\mu }_c)={\int }_0^{\mathrm{\infty }}({\mu }_c\otimes {\mu }_c)(\{\mathrm{\mid }x-y\mathrm{\mid }<e^{-t}\})dt\le \log 3+{\sum }_{m\ge 0}(\log 3)({\mu }_c\otimes {\mu }_c)(\{\mathrm{\mid }x-y\mathrm{\mid }<3^{-m}\})\le \log 3+(\log 3){\sum }_{m\ge 0}{2}^{-m}=3\log 3<+\mathrm{\infty }$, where step 2.1 bounds each dyadic piece. Therefore $V_C\le I({\mu }_c)<+\mathrm{\infty }$ and $\mathrm{cap}(C)=e^{-V_C}\ge e^{-3\log 3}=\frac{1}{27}>0$.

For a Borel probability $\nu$ on $K$ put $m_{n,k}:=\nu (I_{n,k})$ for the $2^n$ level-$n$ cells $I_{n,k}$ of step 2.2; since $\nu$ is carried by $K\subseteq {\bigcup }_k{I}_{n,k}$ and the cells are pairwise disjoint, ${\sum }_k{m}_{n,k}=1$, so by [F6] with the constant list $1$ one has $1=({\sum }_k{m}_{n,k}{)}^2\le 2^n{\sum }_k{m}_{n,k}^2$, that is, ${\sum }_k{m}_{n,k}^2\ge 2^{-n}$.

With $\nu$, the cells $I_{n,k}$ and the masses $m_{n,k}$ of step 3.2, [F1] gives $I(\nu )=\iint \log \frac{1}{\mathrm{\mid }x-y\mathrm{\mid }}d\nu (x)d\nu (y)$ because $\nu$ is a probability on $[0,1]$; two points of one level-$n$ cell satisfy $\mathrm{\mid }x-y\mathrm{\mid }\le {\mathrm{\ell }}_n$, so for $t<4^n$ the event of lying in the same level-$n$ cell is contained in $\{\mathrm{\mid }x-y\mathrm{\mid }<e^{-t}\}$ and hence $(\nu \otimes \nu )(\{\mathrm{\mid }x-y\mathrm{\mid }<e^{-t}\})\ge {\sum }_k{m}_{n,k}^2\ge 2^{-n}$; by [F4] and [F5], $I(\nu )\ge {\sum }_{n\ge 1}{\int }_{4^{n-1}}^{4^n}{2}^{-n}dt={\sum }_{n\ge 1}(4^n-4^{n-1})2^{-n}=\frac{3}{4}{\sum }_{n\ge 1}{2}^n=+\mathrm{\infty }$.

Every $\nu \in P(K)$ has $I(\nu )=+\mathrm{\infty }$ by step 4.1, so the infimum $V_K$ is $+\mathrm{\infty }$ and $\mathrm{cap}(K)=0$ by [F1].

For each $b\in \{0,1{\}}^{\mathbb{N}}$ the cells $I_{n,b\upharpoonright n}$ form a nested family of closed intervals with lengths ${\mathrm{\ell }}_n\to 0$, so by [F7] their intersection contains exactly one point $\phi (b)\in K$; distinct infinite words differ at some level $n$, where their cells are disjoint, so $\phi$ is injective. If $K$ were countable then its subset $\phi (\{0,1{\}}^{\mathbb{N}})$ would be countable by [F8], and since $\{0,1{\}}^{\mathbb{N}}$ is in bijection with $C$ by [F2] and $C$ is uncountable, that is impossible; hence $K$ is uncountable. Thus $C$ has positive capacity and $K$ has zero capacity although both are uncountable compact Lebesgue-null sets.