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.