I have the following exercise: Show that $H^s(A)=0 \iff H^s_\infty(A)=0$ for $A\subset \mathbb R^d$. Here $H^s(A)$ is he Hausdorff measure of the set $A$, so $H^s(A):=\lim_{\delta\to 0^+} H^s_\delta(A)$.
Now, it is clear that, since $H^s(A)\geq H^s_\delta(A)$ for any $\delta$, if $H^s(A)=0$ then $H^s_\infty(A)=0$.
How to do the converse? Do you have some hints to give me?
Here is the idea: if $H^s_\infty(A) = 0$ and $\epsilon > 0$, there exists a cover of $A$ by sets $E_k$ with $$\sum_k (\mathrm{diam}\ E_k)^s < \epsilon.$$ In particular, $(\mathrm{diam}\ E_k)^s < \epsilon$ for each $k$, so you in fact have $$H^s_{\epsilon^{1/s}} (A)< \epsilon.$$
This is true for every $\epsilon > 0$. What happens as $\epsilon \to 0^+$?