In the chapter of Hausdorff measures in Wolff's notes on harmonic analysis, I'm trying to understand a piece of remark.
Fix $\alpha>0$, and let $E\subset\mathbb{R}^n$. For $\epsilon>0$, one defines
$$
H_\alpha^\epsilon(E)=\inf \sum_{j=1}^\infty r_j^\alpha,
$$
where the infimum is taken over all countable coverings of $E$ by discs $D(x_j,r_j)$ with $r_j<\epsilon$. It is clear that $H_\alpha^\epsilon(E)$ increases1 as $\epsilon$ decreases, and we define
$$
H_\alpha(E)=\lim_{\epsilon\to 0} H_\alpha^\epsilon(E).
$$
1. Note: I believe this should be understood as "not decrease" since if $H_\alpha^1(E)=0$, then $H_\alpha(E)=0$.
Remark. It is clear that $H_\alpha(E)=0$ for all $E$ if $\alpha>n$, since one can then cover $\mathbb{R}^n$ by discs $D(x_j,r_j)$ with $\sum_jr_j^\alpha$ arbitrarily small.
I don't understand the remark. Could anyone elaborate it?
By applying the inequality
$$
\sum_j r_j^\alpha \le \delta^{\alpha-n}\sum_jr_j^n,\quad r_j<\delta,
$$
one can show that if $H_n(E)<\infty$, then $H_\alpha(E)=0$ for $\alpha>n$. But the remark says something stronger.
Best Answer
For simplicity first assume that the set $E$ is inside the ball $B_{n}(R)$ of radius $R$. See that: $$ H_\alpha^\epsilon(E)\leq H_\alpha^\epsilon(B_n(R)). $$ It can be seen that $B_{n}$ can be covered approximately with $(R/\epsilon)^n$ balls of radius $\epsilon$ which means that: $$ H_\alpha^\epsilon(B_n(R))\leq (R/\epsilon)^n\epsilon^{\alpha}. $$ If $\alpha>n$, the right hand side goes to zero az $\epsilon\to 0$.
If $E$ is not a bounded set, consider $E\cap B_n(R)$ for which $H_\alpha(E\cap B_n(R))=0$ for $\alpha>n$ and let $R\to\infty$.