[Math] Relationship between the Hausdorff dimension and the Box-counting dimension

analysisdimension-theory-analysisfractalsgeometric-measure-theorygeometry

In Fractal Geometry by Falconer the author writes:

If $1<\mathcal H^s(F)=\lim_{\delta\to0}\mathcal H_\delta^s(F)$
then $\log N_\delta(F)+s\log\delta>0$ if $\delta$ is sufficiently small.
Thus $s\leqslant\underline{\lim}_{\delta\to0}\log N_\delta(F)/-\log\delta$ so
$$\dim_HF\leqslant\underline\dim_BF\leqslant\overline\dim_BF\tag{3.17}$$
for every $F\subset\mathbb R^n$.
We do not in general get equality here.
Although Hausdorff and box dimensions are equal for many ‘reasonably regular’
sets, there are plenty of examples where this inequality is strict.

What happens in the case $\mathcal{H}^s(F) \leq 1$?

Best Answer

The result holds true as long as $\mathcal{H}^s(F) > 0$, though a slightly different argument is required. The reason is that for $r>0$ we have $$\mathcal{H}^s(rF) = r^s\mathcal{H}(F),$$ where $rF=\{rx:x\in F\}$. Thus, for sufficiently large $r$ we have $$\dim_H(rF)\leq\underline\dim_B(rF)\leq\overline\dim_B(rF).$$ Furthermore, $\dim(rF)=\dim(F)$ for all these definitions of dimension.