12
$\begingroup$

I would like to ask the following:

Are there "many" sets, say in the interval $[0,1]$, with zero Lebesgue measure but with Hausdorff dimension $1$?

The motivation for this question is the dichotomy between measure and category. There are certainly dense sets with zero Lebesgue measure. But a dense set need not have positive Hausdorff dimension (for example, the rationals are dense but have zero Hausdorff dimension).

Honestly, I would already be satisfied with an answer to the following question:

Is there any set in $[0,1]$ with zero Lebesgue measure but with Hausdorff dimension $1$?

$\endgroup$
2

2 Answers 2

Reset to default
18
$\begingroup$

For any $r<1$, you can construct a Cantor set with Hausdorff dimension $r$ by varying the lengths of the intervals in the usual Cantor set construction. In particular, you can let $C_n\subset[0,1]$ be a Cantor set of Hausdorff dimension $1-1/n$ for each $n$. The union $C=\bigcup C_n$ then has Lebesgue measure $0$ because each $C_n$ does, but Hausdorff dimension $1$.

$\endgroup$
6
  • $\begingroup$ Then to answer the first question, can't you add/subtract any countable set of points without disturbing the measure (and, I think the dimension) so you get continuum many sets? $\endgroup$
    – Ross Millikan
    Commented Feb 9, 2016 at 3:16
  • $\begingroup$ Great answer, thanks. $\endgroup$
    – John B
    Commented Feb 9, 2016 at 3:19
  • $\begingroup$ Uhm, varying the length of the intervals yields a "fat" Cantor set, with measure greater than zero. No? $\endgroup$
    – bartgol
    Commented Feb 9, 2016 at 3:22
  • 1
    $\begingroup$ @bartgol: It depends how you vary them. If you always remove a constant proportion of each of the intervals you have so far (as in the standard construction where you always remove $1/3$), you'll always get a set of measure zero, but the Hausdorff dimension will depend on the proportion. If you let the proportion you're removing get smaller and smaller fast enough, you get a Cantor set of positive measure. $\endgroup$
    – Eric Wofsey
    Commented Feb 9, 2016 at 3:29
  • 3
    $\begingroup$ If you remove the middle $r$ proportion of each interval, then at the $n$th stage of the construction you have a set of measure $(1-r)^n$, which goes to $0$ as $n\to\infty$. So as long as the proportion is always the same (even if it's smaller than $1/3$), you get a set of measure $0$. $\endgroup$
    – Eric Wofsey
    Commented Feb 9, 2016 at 3:49
4
$\begingroup$

For a "naturally occurring" example, let $b_1$ and $b_2$ be positive integers $\geq 2$ such that no positive integer power of $b_1$ equals a positive integer power of $b_2$ (i.e. $(b_1)^m = (b_2)^n$ has no solution where $m$ and $n$ are positive integers). Kenji Nagasaka proved in 1979 that the set of real numbers normal to base $b_1$ but not normal to base $b_2$ is a measure zero set with Hausdorff dimension $1.$ See my 5 July 2002 sci.math post Numbers normal to one base but not to another base. (Note: In that post I seem to have reversed the definitions of multiplicatively dependent and multiplicatively independent.)

Actually, Nagasaka only proved the Hausdorff dimension $1$ part. The measure zero part follows from the long-known fact that all real numbers except for a set of measure zero are normal to every base.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .