Let $E\subset \mathbb R^d$ a borelien s.t. $m_\alpha (E)<\infty $ where $m_\alpha $ is that $\alpha -$Hausdorff measure. Show that for all $\varepsilon>0$ there is a compact $K\subset E$ s.t. $m_\alpha (E\backslash K)<\varepsilon$.
In fact, I don't know how to prove the claim. I tried to prove it as we prove this result with the Lebesgue measure, but it's not conclusif. There is probably a subtlety I don't see.