If $\mathcal{H^s}$ is the Hausdorff measure, how to show:
$\bigcup_{i=1}^{\infty} E_{i}$ with $\mathcal{H^s}(E_i)<\infty$?
I tried to use the properties of the Hausdorff measure to show this.
I focused upon the fact that for all $E \subset \mathbb{R^n}$ with $\mathcal{H^s}(E)<\infty \Rightarrow \mathcal{H^r}(E)=0$ for all $r>s$ but I couldn't conclude this assertion.
By way of contradiction, if it were possible, then $$n=dim_H({\mathbb R}^n)=\sup_{i\in{\mathbb N}}\dim_H(E_i)\le s<n,$$ which is a contradiction.
Recall that ${\cal H}^s(E_i)<\infty$ implies that $dim_H(E_i)\le s$.