Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 12G / Solution

Using the diameter normalization, define ${\mathcal{H}}_{\delta }^d(E)=\inf \{{\sum }_j(\mathrm{diam}{U}_j{)}^d:E\subseteq {\bigcup }_j{U}_j,\mathrm{diam}{U}_j\le \delta \}$ and the Hausdorff measure ${\mathcal{H}}^d(E)={\lim }_{\delta \downarrow 0}{\mathcal{H}}_{\delta }^d(E)$. The Hausdorff dimension is $\inf \{d:{\mathcal{H}}^d(E)=0\}$, equivalently $\sup \{d:{\mathcal{H}}^d(E)=\infty \}$.

The Cantor set is the intersection of the sets left by successively removing open middle thirds from $[0,1]$. At level $k$ it has $2^k$ Cantor cylinders of diameter $3^{-k}$. For $s=\log 2/\log 3$, their total $s$-cost is $2^k{3}^{-ks}=1$, so ${\mathcal{H}}^s(C)\le 1$.

The random series $\xi ={\sum }_{n\ge 1}{X}_n{3}^{-n}$ converges absolutely, and its ternary digits are all zero or two. Thus $\xi \in C$ and its law is the Cantor Bernoulli measure. Each level-$k$ cylinder has probability $2^{-k}$. Distinct such intervals have separation at least $3^{-k}$; a set $U$ of smaller diameter meets at most one. If $3^{-(k+1)}\le \mathrm{diam}U<3^{-k}$, thenSets of diameter zero have probability zero, since the measure has no measure atoms; diameters at least one give the bound trivially. For arbitrary nonmeasurable sets use outer probability. Any countable cover of $C$ therefore satisfies $1\le {\sum }_j{\mathbb{P}}^{\ast }(\xi \in U_j)\le 2{\sum }_j(\mathrm{diam}{U}_j{)}^s$. HenceThe last implication follows by multiplying the $s$-costs by powers of the cover diameter: measures above $s$ vanish and those below $s$ are infinite.