# Problem with Hausdorff dimension

Let $\varepsilon>0,s\geq0$ and $C\subseteq \mathbb{R}^d$ be randomly given. Now define: $$\mathcal{H}^s_\varepsilon(C)= \inf\biggl\{\,\sum^\infty_{n=1}(\rho (A_n))^s\biggm| C\subseteq \bigcup^\infty_{n=1}A_n, \rho(A_n)<\varepsilon\,\biggl\}$$ where $\rho(C)=\sup_{x,y\in C}|x-y|$.

Then the Hausdorff measure is given by $\mathcal{H}^s(C)=\lim_{\varepsilon \to 0} \mathcal{H}^s_\varepsilon(C)$.

Show that for any $C$ there exists a $s_0$ such that $$\mathcal{H}^s(C)=\begin{cases}+\infty, & s<s_0\\ 0, & s >s_0 \end{cases}$$ This $s_0$ is called the Hausdorff dimension of $C$. Can anyone help me with this exercise? I am completely stumped. If someone could at least give me a hint that would be wonderful.

• If $\mathcal H^s(C)>0$, then $\mathcal H^r(C)=\infty$ for all $r<s$.
• If $\mathcal H^s(C)<\infty$, then $\mathcal H^r(C)=0$ for all $r>s$.
Assume now that the above claims have been proven. Show that $\mathcal H^{d+1}(\mathbb R^d)=0$ and thus $\mathcal H^{d+1}(C)=0$. (Sketch: Cover the cube $[0,1]^d$ with smaller cubes and use this cover to show that $\mathcal H^{d+1}([0,1]^d)=0$. Then use the fact that $\mathbb R^d$ is a union of countably many such cubes.) From the second claim above we get $\mathcal H^s(C)=0$ for all $s\geq d+1$. Therefore $s_0=\inf\{s\geq0;\mathcal H^s(C)=0\}$ is a number in $[0,\infty)$.
By the definition of $s_0$ we have $\mathcal H^s(C)=0$ for all $s>s_0$. Take then any $s<s_0$; we need to show that $\mathcal H^s(C)=\infty$. Take any number $r\in(s,s_0)$. It follows from the definition of $s_0$ that $\mathcal H^r(C)>0$. It then follows from the first claim above that $\mathcal H^s(C)=\infty$.
• I'm trying to use the fact that $\mathcal{H}^s(C)$ is a decreasing function in $s$ but I'm still running into some difficulty. Because if the sum converges a $s$ it's hard to say anything for an arbitrarily larger power. Could you maybe elaborate a bit more? – D. Vente Oct 3 '14 at 11:39