I assume the Hausdorff Dimension of $\mathrm{R}^d$ is $d$. To prove this I guess one has to prove these two statements:
Unfortunately I am unable to prove anything here. I guess the trick is to use a clever partition of $\mathrm{R}^d$, but I don't come up with one.
How can I prove this statement?
Start with the unit hypercube $Q$. One can readily partition it into $\approx \frac 1{(\delta\sqrt d)^d}$ cubes of diameter $\le\delta$. This shows that $$H_\delta^\alpha(Q)\le \frac {\delta^\alpha}{(\delta\sqrt d)^d}\sim\delta^{\alpha-d}.$$ Especially, $H^\alpha(Q)=\lim_{\delta\to0}H_\delta^\alpha(Q)=0$ if $\alpha>d$.
On the other hand, let $U$ be any nonempty subset of $Q$ of diameter $\le \delta$ and pick $u\in U$. Then $U$ contains at most $\approx \left(\frac{2\operatorname{diam} U}\epsilon\right)^d$ points of $u+\epsilon\mathbb Z^d$, and likewise at most $\approx \left(\frac{2\operatorname{diam} U}\epsilon\right)^d$ points of $\epsilon\mathbb Z^d$. But $Q$ contains $\approx \frac{1}{\epsilon^d}$ such points. As all these points are covered, we conclude that $$ \left(\frac{2}\epsilon\right)^d\sum(\operatorname{diam} U)^d\ge\frac1{\epsilon^d}$$ i.e. $$ \sum(\operatorname{diam} U)^d\ge\frac1{2^d}$$ holds for any cover (strictly speaking, only approximately; but all uses of $\approx$ above can be made explicit by the use of the floor function and introducing a manageable error term). Now if $\alpha<d$ and all $\operatorname{diam}U$ are $\le\delta$, we have $$(\operatorname{diam} U)^\alpha\ge(\operatorname{diam} U)^d\frac1{\delta^{d-\alpha}},$$ hence $$ \sum(\operatorname{diam} U)^\alpha\ge\frac1{2^d\delta^{d-\alpha}}$$ As $\delta\to 0$, the right hand side goes $\to\infty$, hence $H^\alpha(Q)=\infty$ for $\alpha<d$.
Now from $Q\subseteq \mathbb R^d$ we already get for free that $H^\alpha(\mathbb R^d)=\infty $ for $\alpha<d$. To show that $H^\alpha(\mathbb R^d)=0 $ for $\alpha>d$, you can cover $\mathbb R^d$ by countably many copies of $Q$ and use finer and finer covers for these copies (e.g. the $n$th cube gets covered so that it contributes $\le 2^{-n}\epsilon$ to the sum).
