Question concerning fractal dimension Hausdorff Measures.

[blackfedora]

Homework Statement

I'm currently trying to work my way through Frank Morgan's Geometric Measure Theory and I had a quick question regarding his definition of Hausdorff measure. Which is:

Homework Equations

$$\mathscr{H}^m(A) = \lim_{\delta \rightarrow 0} \inf \left\{ \sum \alpha_m \left( \frac{diam(S_j)}{2}\right) ^m : A \subset \cup S_j \right\}$$

where $$\alpha_m$$ is the Lebesgue measure of a closed unit ball in R^m.

The Attempt at a Solution

My understanding was that we wanted to define Hausdorff measure because an m-dimensional Lebesgue measure is hard to assign to an m-dimensional subset of Rⁿ. Should $$\alpha_m$$ actually be $$\alpha_n$$, since I thought we were covering the subset of Rⁿ with n-dimensional closed balls? I'm asking this question because I'm attempting to show that the Hausdorff dimension of the Cantor set is ln(2)/ln(3) and I don't really know what a closed unit ball in R^ln(2)/ln(3) would be, let alone Lebesgue measure it would have.

 

[mighty2000]

Dear student,

Thank you for your question regarding Frank Morgan's definition of Hausdorff measure in his book Geometric Measure Theory. I can understand your confusion about the use of \alpha_m instead of \alpha_n in the definition. Let me explain this in more detail.

Firstly, the reason for defining Hausdorff measure is indeed to assign a measure to an m-dimensional subset of Rn. However, this subset may not necessarily be covered by n-dimensional balls. In fact, it may not even be possible to cover it with n-dimensional balls in a way that gives a meaningful measure. This is where the concept of Hausdorff measure comes in.

Hausdorff measure is defined using a covering of the subset A with m-dimensional balls, where the dimension m can be different from the dimension n of the ambient space Rn. This allows us to assign a measure to A even if it cannot be covered by n-dimensional balls.

Now, coming to the use of \alpha_m in the definition, it is indeed correct. The reason is that we are covering the subset A with m-dimensional balls, so we need to use the Lebesgue measure of an m-dimensional ball, denoted by \alpha_m. This is different from the Lebesgue measure of an n-dimensional ball, which is denoted by \alpha_n. So, in essence, the use of \alpha_m is to account for the dimension of the balls we are using to cover A.

I hope this clarifies your doubts regarding the definition of Hausdorff measure and the use of \alpha_m. As for your question about the Hausdorff dimension of the Cantor set, it can be shown that the closed unit ball in Rln(2)/ln(3) is equivalent to the Cantor set, so the Lebesgue measure of this ball would be equal to the Hausdorff measure of the Cantor set.

I hope this helps. Keep up the good work with your studies!