I have been having trouble on this question for a while now.
We define the Hausdorff on $B_\mathbb{R^n}$ measure this way :
$H^\beta(F) = lim_{\delta\to 0} \inf{\{\sum diam(U_i)^\beta : \delta \ge diam(U_i)}$ and $U_i$ covers F$\}, \forall F\in B_\mathbb{R^n}$
Is the measure $H^\beta$ dependant on the choice of norm on $\mathbb{R}^n$ ?
What I did so far, among other things :
- Tried using the equivalencies on the norms on $\mathbb{R}^n$ but that didn't lead anywhere
- Tried to maybe take two different norms and show that they are equal on a clan
- Let $N_1$ a norm and $N_2 = \lambda N_1$ a norm also (for $\lambda > 0$), we then have : $H^\beta(F)_1 = \lambda^\beta H^\beta(F)_2$ (where $H^\beta(F)_i$ is the hausdorff measure for the norm $N_i$. All we would need is to find a set $F$ such that $\infty >H^\beta(F)_1 > 0$ and we would be able to conclude, but I somehow can't find any set and norm for which I'm able to get a nonzero finite value when evaluating it in the Hausdorff measure
I thank you all in advance for your help
Edit :
Let $N_1$ be the euclidian norm, and let $N_2 = \lambda N_1$ for $\lambda>0$.
We take $\beta = n$, thus the hausdorff measure becomes proportional to the lebesgue measure.
We write $N_1 = \alpha_d \nu_n$. Thus for $F = B(0,1)$, we get $H^n(F)_1 = l \in \mathbb{R}_{>0}$ (we can find l explicitely).
We then have $H^\beta(F)_2 = \lambda^\beta H^\beta(F)_1 = \lambda^nl \not= l$, for $\lambda \not= 1$.
We can thus construct a counter-example very easily by taking a specific norm and a specific n.