Digit-position density determines Hausdorff dimension

For an allowed prefix $b_1,\dots ,b_n$, with positions starting at one, put $p={\sum }_{k\le n}{b}_k{2}^{-k}$. The closed intervals $[p,p+2^{-n}]$, one for each prefix, form a $2^{a_S(n)}$-member cover $K_n$ of $A_S$, and $K_{n+1}\subseteq K_n\subseteq [0,1]$. Conversely, if $x\in {\bigcap }_n{K}_n$, among the finitely branching allowed prefixes whose intervals contain $x$ there is a branch: at each step take the first child with extensions of arbitrarily large depth, which exists because there are finitely many children. Its prefix sums tend to $x$ since the tail bound is $2^{-n}$. Thus $A_S={\bigcap }_n{K}_n$ is closed and bounded, hence compact. This retains both expansions at endpoints.

Write $d=\mathrm{lim\hspace{0.17em}inf}{a}_S(n)\mathrm{/}n$. For any $t>d$, choose $u$ with $d<u<t$. Infinitely many $n$ have $a_S(n)<un$; the corresponding covers have $t$-cost at most $2^{-n(t-u)}\to 0$ and diameters $2^{-n}\to 0$. At each fixed scale these arbitrarily cheap covers prove ${\mathcal{H}}^t(A_S)=0$. Thus ${\dim }_H{A}_S\le d$. If $S$ is finite then $A_S$ is finite, its singleton covers cost zero for positive exponents, and $d=0$; this includes the empty position set.

For infinite $S$ list its elements increasingly as $k_1<k_2<\cdots$. On $[0,1)$ put $b_j(u)=\lfloor 2^{j}u\rfloor -2\lfloor 2^{j-1}u\rfloor$ and $T(u)={\sum }_{j\ge 1}{b}_j(u)2^{-k_j}$. Each partial sum is a finite Borel step function, and convergence follows from the geometric tail. Hence $T$ is Borel measurable. Set $\nu (B)={\lambda }_1(\{u\in [0,1):T(u)\in B\})$ for Borel $B$. Borel preimages preserve disjoint unions, so countable additivity follows directly from that of Lebesgue measure. This is a probability with $\nu (A_S)=1$.

The level-$n$ cover has total length $2^{a_S(n)-n}$. An infinite complement means $n-a_S(n)\to \mathrm{\infty }$, so ${\lambda }_1(A_S)=0$ by those covers; compactness supplies measurability. The line equality gives ${\mathcal{H}}^1(A_S)=0$.

Each prescribed first $j$ binary digits of $u$ describes one half-open dyadic interval of length $2^{-j}$, hence mass $2^{-j}$. A real number has at most two binary expansions: at the first differing digit, equality of the sums requires the full possible tail ${\sum }_{k>n}{2}^{-k}=2^{-n}$, forcing the two opposite constant tails. Therefore a fibre $T^{-1}(\{x\})$ is contained, for every $n$, in at most two prefix events of mass $2^{-a_S(n)}$. Since $a_S(n)\to \mathrm{\infty }$, every singleton has $\nu$-mass zero. Away from endpoints, any level-$n$ dyadic cell can receive only its own allowed prefix. Its mass, with either closed or half-open endpoints, is consequently at most $2^{-a_S(n)}$.

If $0<t<d$, then for all sufficiently large $n$, $a_S(n)\ge tn$. A closed interval of length $r$ with $2^{-n}\le r<2^{1-n}$ meets at most three closed dyadic cells of length $2^{-n}$; the strict upper bound includes boundary contacts. Its mass is at most $32^{-a_S(n)}\le 3r^t$. Any nonempty bounded set lies in a closed interval of the same diameter, and diameter-zero sets have zero mass by the preceding step. Apply mass distribution with $\nu (A_S)=1$ to get ${\dim }_H{A}_S\ge t$. Let $t$ increase to $d$. When $d=0$, nonnegativity gives the lower bound directly. This proves the dimension formula also at $d=1$.

If $S$ and its complement are both infinite, insert arbitrary infinite bits successively at the positions $k_j$. Two different bit sequences first differ at some $k_j$; the maximal possible allowed tail is strictly less than $2^{-k_j}$ because some later position is forbidden. Their sums are distinct. Infinite bit sequences are uncountable by the diagonal argument (a purported enumeration is defeated by changing its $j$th bit at position $j$). Thus this injection proves uncountability of $A_S$.