As I never had a course which dealt with Hausdorff measures and every time I heard about Hausdorff measure I was only thinking using my intuition what that should be. So I decided to take a look at the definition, and try some typical examples. I am reading a chapter on Hausdorff measure of Real Analysis by Stein and Shakarchi.

I saw that they define the Hausdorff measure only for Borel measurable sets. My questions are:

  1. Can we define Hausdorff measure for Lebesgue measurable sets or not?
  2. If a set $E \subset \Bbb{R}^n$ has Lebesgue measure zero, then it's Hausdorff dimension is strictly smaller than $n$?

I am concerned about this since I sort of rushed out and 'proved' the following thing using Hausdorff measures:

If $E \subset \Bbb{R}^n$ has Lebesgue measure zero and $f :\Bbb{R}^n \to \Bbb{R}^n$ is Lipschitz continuous then $f(E)$ also has Lebesgue measure zero.

and I don't think my proof is right.


Hausdorff outer measure is defined for all sets, and then we use the definition of Caratheodory to restrict it to a subalgebra of "measurable" sets to get the Hausdorff measure. In $\mathbb R^n$, the $n$-dimensional Hausdorff outer measure is the same (up to a constant factor) as $n$-dimensional Lebesgue outer measure, so they have the same measurable sets as well.

Another counterexample to 2. In interval $[n,n+1]\;$ take a subset with Hausdorff dimension $1-1/n\;$. The union of these sets has Hausdorff dimension 1 but Lebesgue measure zero.

So to prove your fact about Lipschitz functions, you cannot do it by considering only sets with Hausdorff dimension ${}\lt 1$, you have to consider Lebesgue measure itself, and use its definition. Try it, it's not too hard!

  • $\begingroup$ How is the union of the sets you want to construct of Lebesgue measure zero if it is already of Hausdorff measure $1-1/n$ when restricted to $[n,n+1]$? Am I missing something here? $\endgroup$ – user20266 Apr 16 '12 at 17:08
  • $\begingroup$ (Did you intend to write 'of Hausdorff dimension $1-1/n$'?) $\endgroup$ – user20266 Apr 16 '12 at 17:25
  • $\begingroup$ Yes, corrected. $\endgroup$ – GEdgar Apr 16 '12 at 20:12
  • $\begingroup$ That's a much nicer counterexample. $\endgroup$ – Nate Eldredge Apr 16 '12 at 22:29
  • $\begingroup$ Why the union of that sets has Hausdorff dimension 1? $\endgroup$ – Lorban Mar 19 '13 at 17:37

Wikipedia has some useful information.

In $\mathbb{R}^d$, $d$-dimensional Hausdorff measure is equal to Lebesgue measure (up to scaling). So once you have defined it for Borel sets, you can extend it to Lebesgue-measurable sets in the same way: any Lebesgue-measurable set is of the form $A = B \cup C$ where $B$ is Borel and $C$ is Lebesgue measurable with $m(C) = 0$, so define the Hausdorff measure of $A$ to be the Hausdorff measure of $B$. Just as with Lebesgue measure, this gives you a countably additive complete measure.

Wiki also gives a counterexample to your question 2: the image of a 2-dimensional Brownian motion has Hausdorff dimension 2 but its 2-dimensional Hausdorff measure (i.e. its Lebesgue measure) is zero (almost surely).


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.