I was wondering whether there is a heuristic simpler way to understand the saying that a compact $E\subseteq \mathbb{R}^d$ satisfies $\dim(E)=\alpha\in (0,d)$? Or maybe saying that $\mathcal{H}^{\alpha}(E)\in (0,\infty)$, where $\mathcal{H}^{\alpha}$ is the $\alpha$-Hausdorff outer measure?
For example, if $\lambda(E)=0$ where $\lambda$ is the Lebesgue measure, I would say that "$E$ has $0$ volume". This can also be done in other cases with somewhat complicated definitions. Is there a conceptually simple way to grasp what it means for $E$ to have Hausdorff dimension $\alpha$?