9
$\begingroup$

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 $1$?

$\endgroup$
11
$\begingroup$

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 $1$.

$\endgroup$
  • $\begingroup$ 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 Feb 9 '16 at 3:16
  • $\begingroup$ Great answer, thanks. $\endgroup$ – John B Feb 9 '16 at 3:19
  • $\begingroup$ Uhm, varying the length of the intervals yields a "fat" Cantor set, with measure greater than zero. No? $\endgroup$ – bartgol Feb 9 '16 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 Feb 9 '16 at 3:29
  • 1
    $\begingroup$ 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 Feb 9 '16 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.