I am reading Hausdorff Dimension, Its Properties, and Its Surprises by Dierk Schleicher. Among the elementary properties of the Hausdorff dimension, the last one is:
If $X\subset \Bbb R^n$ has finite positive $d$-dimensional Lebesgue measure, then $\dim_H X = d$.
My work. It will be enough to show that $\mathcal H^s(X) = 0$ for all $s > d$, and $\mathcal H^s(X) = \infty$ for all $s < d$. As usual, $$H^s(X) = \lim_{\delta\to 0} H^s_\delta(X)$$
where $$\mathcal H^s_\delta(X) = \inf\left\{\sum_{i=1}^\infty |U_i|^s: \{U_i\} \text{ is a }\delta\text{-cover of }X \right\}$$
I'm unable to relate the Lebesgue measure with coverings of $X$, which would help me find a connection with $\mathcal H^s_\delta(X)$ for given $\delta > 0$.
Thanks a lot!
Best Answer
Edit: This answer only addresses the case $n = d$!
Note that an equivalent definition of the Hausdorff measure is obtained by using convex sets of diameter at most $\delta$ instead of arbitrary $\delta$-covers since any set of diameter $\delta$ is contained in a convex set of diameter $\delta$.
Let $|E|$ denote the diameter of a set $E$, let $\epsilon > 0$, and let $\{U_i\}$ be a cover of $X$ by convex sets such that $$ \sum_i |U_i|^d \le \mathcal H^d(X)+\epsilon. $$ You can show for any convex set $U\subset\mathbb R^d$, its Lebesgue measure satisfies $\mathcal L^d(U)\le c_d |U|^d$. (This is a form of the isodiametric inequality in $\mathbb R^d$.) Then, $$ \mathcal L^d(X)\le \sum_i\mathcal L^d(U_i) \le c_d\sum_i|U_i|^d\le c_d(\mathcal H^d(X)+\epsilon). $$ Letting $\epsilon\to 0$, this shows the inequality $\mathrm{dim}_HX\ge d$. For each $X\subset \mathbb R^d$, $\dim_HX\le d$, so this proves the claim in this special case.