Let $\mathcal{H^s}$ be the s-dimensional outer Hausdorff measure.

For all $t \in (0,n)$ there is none countable cover $\lbrace F_i \rbrace^\infty_{i=1}$

of $\mathbb{R^n}=\bigcup_{i=1}^{\infty} F_{i}$ with $\mathcal{H^s}(F_i)<\infty$

I used the contradiction::

If it would be possible then

$n=$dim$_H(\mathbb{R^n})=$sup dim$_H(F_i)\leq s<n$

So there can't be such a cover.

Can it be proved like this or is there something to improve?


That looks fine, provided that you can prove $$\dim_H \left( \bigcup_i F_i \right) = \sup_i \dim_H(F_i).$$ It shouldn't be a difficult proof.

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