# Outer Hausdorff measure/Infinite

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 $$\mathcal{H^s}(F_i)<\infty$$

$$n=$$dim$$_H(\mathbb{R^n})=$$sup dim$$_H(F_i)\leq s
That looks fine, provided that you can prove $$\dim_H \left( \bigcup_i F_i \right) = \sup_i \dim_H(F_i).$$ It shouldn't be a difficult proof.