# Hausdorff dimension

Could you please give me some hints on (Exercise 1.7.21) of "A Course in Metric Geometry" by Burago, Burago, Ivanov.

We have a compact space $X$, which can be written as $X=\bigcup_{i=1}^n X_i$ (disjoint union), where $X_i=F_i(X)$ and $F_i$ are dilations with Lipschitz constant $c_i$. Then show that the Hausdorff dimension $d$ of $X$ satisfies $\sum_{i=1}^n c_i^d = 1$.

Let $\mu_d$ denote the $d$-dimensional Hausdorff measure, then thats how far I got:

Thanks to the corrections below I am able to show it in the case where $0<\mu_d(X)<\infty$ but I still do not know how to approach the two boundary cases. Any hints are appreciated! Thanks!

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This is obviously wrong. Consider $X=[0,1]$, $X_1=X$, $F_1(x)=x^2$. Then the minimal Lipschitz constant is $c_1=2$. Since $d=1$ for the one-dimensional line, the claim of the problem statement is $2^1=1$. – Hagen von Eitzen Nov 2 '12 at 22:50
I just read the linked original problem. It speaks of dilations, not of Lipschitz continuous functions. – Hagen von Eitzen Nov 2 '12 at 22:51
Ah, sorry. I didn't find this notion in the sections before so I just thought it meant Lipschitz continuous. I think it makes more sense now! Thank you! – zerhan Nov 2 '12 at 22:58
This is still wrong, unless you at least assume that the $X_i$ are disjoint. Otherwise you could just take all $F_i$ to be the identity. Also, what is $\mu_d$, and how do you know that $\mu_d(X)<\infty$? – Lukas Geyer Nov 2 '12 at 23:59

It helps to consider the Hausdorff content $\mathcal H^t_\infty$, which is defined like Hausdorff measure but without the requirement of covering by small sets. The content is subadditive and scales by $c^t$. It is equal to zero if and only if the $\mathcal H^t$ measure is equal to zero. And most helpfully for us, $\mathcal H^t_\infty$ is finite for every bounded set.
Therefore, $\mathcal H^t(X)\le \sum_{i}c_i^t \mathcal H^t(X)$. If $\sum_i c_i^t<1$, this implies that the content is zero, hence the measure is zero. We conclude that $\operatorname{dim}X\le d$ where $d$ is defined by $\sum_i c_i^d=1$.
For the opposite direction we need an additional assumption: $X$ is nonempty. :) Pick a point $a\in X$ and place a unit Dirac measure there: $\mu_0=\delta_a$. Then define $\mu_n$ inductively by pushing $\mu_{n-1}$ by the dilations and scaling it by $c_i^d$. Let $\mu$ be a weak* limit of these probability measures; since we are on a compact set, $\mu$ is also a probability measure. There is a constant $C$ such that $\mu(B(x,r))\le Cr^d$ for any ball $B(x,r)$ in $X$. (I am happy to leave the verification to you). It follows that for any cover of $X$ by balls $B(x_j,r_j$ we have $\sum_j r_j^d \ge C^{-1}\sum \mu(B(x_j,r_j))\ge C^{-1}$. This shows $\mathcal H^d(X)>0$.