Measure 0 sets on the line with Hausdorff dimension 1

Question

I use $\dim_H(E)$ to denote the Hausdorff dimension of a set $E \subseteq \mathbb{R}$ and $|E|$ to denote its Lebesgue measure. It is easy to see from the definition of Hausdorff dimension that if $\dim_H(E) < 1$, then $|E| = 0$. The converse is not true, and there are many cases where $\dim_H(E) = 1$ yet $|E| = 0$. So the question:

What was the first (or most elementary) example of this phenomenon?

After some looking around, I was able to prove that a central Cantor set $C$ with ratio of dissection $r_k = 1/(2+\frac{1}{k})$ satisfies the condition I want. It is easy to see that $|C| = 0$ since at step $n$ of the process to construct this Cantor set, it has measure $2^n(r_1 \cdots r_n)$ which in this case limits to 0, but for the Hausdorff dimension I required a non-trivial result from the paper Sums of Cantor sets (Cabrelli, Hare, Molter) that gave the formula

$\dim_H(C) = \liminf_n \frac{n \ln 2}{\ln r_1 \cdots r_n}$.

This result is fairly recent and sophisticated, and I feel that there should be older and simpler examples.

Answer 1

Try a countable union of sets (such as Cantor sets) whose Hausdorff dimension tends to 1.

Answer 2

Perhaps Hausdorff's original paper? He uses gauge functions other than powers of x. And constructs Cantor sets corresponding to them. For example if you take $x |\log x|$ then you get a set of Hausdorff dimension $1$ but measure $0$.

Answer 3

It is a very common phenomenon in ergodic theory when the set of points which do not satisfy the Birkhoff ergodic theorem (i.e., a set of zero measure) has full Hausdorff dimension.

See, for instance, http://www.math.psu.edu/pesin/papers_www/birk.pdf

Answer 4

Perhaps the following charts "Universal measure zero sets with full Hausdorff dimension" http://mat.fsv.cvut.cz/Zindulka/papers/opava.pdf are of interest for you.