# Hausdorff measure for Lebesgue measurable sets?

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!

• 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? – user20266 Apr 16 '12 at 17:08
• (Did you intend to write 'of Hausdorff dimension $1-1/n$'?) – user20266 Apr 16 '12 at 17:25
• Yes, corrected. – GEdgar Apr 16 '12 at 20:12
• That's a much nicer counterexample. – Nate Eldredge Apr 16 '12 at 22:29
• Why the union of that sets has Hausdorff dimension 1? – 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).