Home > Geometry, Measure Theory, Real Analysis > Thoughts on Hausdorff Measure and Hausdorff Dimension

Thoughts on Hausdorff Measure and Hausdorff Dimension

The Hausdorff measure is a good way to study the fine properties of sets since it is more ‘sensible’ than the Lebesgue measure. For example in \Bbb{R}^3 the surfaces and paths have zero Lebesgue measure, but the Hausdorff dimension captures what Lebesgue measure misses for zero-measure sets, and it confirms our intuition that a path is in fact  one dimensional and a surface is two dimensional. The Hausdorff dimension of a set is the exact value of \alpha such that the \beta-dimensional Hausdorff measure is zero for \beta > \alpha and infinity for \beta <\alpha. The problem of finding the Hausdorff dimension of a set is not very simple, since the dimension doesn’t need to be an integer. There exist sets with fractional and other bizare dimensions. I will present a few problems below which can be approached using Hausdorff measure and Hausdorff dimension. A good reference for starters can be found in: Real Analysis.

1. Prove that if g:[0,1]\to [0,1]^2 is \alpha-Holder continuous surjective mapping then \alpha \leq 1/2.

Proof: The proof of this problem relies on a simple lemma, which says that Holder continuous functions and Hausdorff measures are strongly connected.

Lemma: Suppose that f defined on a compact set E of some \Bbb{R}^d is \gamma-Holder continuous. Then

  • m_\beta(f(E)) \leq M^\beta m_\alpha(E) if \beta=\alpha/\gamma.
  • \text{dim}f(E)\leq \frac{1}{\gamma} \text{dim}(E).

Proof of the lemma: Suppose \{F_k\} is a countable family of sets that covers E and $m_\alpha(E)$ is finite. Then \{f(E \cap F_k)\} covers f(E) and moreover f(E\cap F_k) has diameter less than M(\text{diam}\ F_k)^\gamma. Therefore

\displaystyle \sum_k (\text{diam}\ f(E\cap F_k))^{\alpha/\gamma}\leq M^{\alpha/\gamma}\sum_k (\text{diam} F_k)^\alpha,

and (i) now follows. The second part is merely a consequence of the first part.

Now the solution of the problem is a direct consequence of point (ii) in the lemma. Indeed, since g is \alpha-Holder continuous, we have 2=\text{dim} g([0,1])\leq \frac{1}{\alpha} \text{dim} [0,1]=\frac{1}{\alpha}.

2. Prove that there is no mapping f:\Bbb{R}^3 \to \Bbb{R}^2 such that |f(x)-f(y)| \geq |x-y| for all x,y \in \Bbb{R}^3.

Proof: Suppose such a mapping does exist. Then note that it is injective, i.e. one to one. Denote by E the image of f in \Bbb{R}^2. Then there exists the inverse mapping g: E \to \Bbb{R}^3 which is onto and satisfies |g(x)-g(y)|\leq |x-y|. Apply again the second part of the lemma for E\cap K_n, where K_n is a sequence of compacts which cover \Bbb{R}^2. Then we get 3=\text{dim}\ g(E) \leq\sup\limits_n \text{dim}\ g(E\cap K_n)\leq \sup\limits_n \text{dim}\ E\cap K_n \leq \text{dim}\ \Bbb{R}^2 =2, where we have used the fact that the Hausdorff dimension of a countable union of sets is less or equal than the supremum of the Hausdorff dimension taken on the considered sets. We have reached a contradiction, and therefore there isn’t such a function.

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