This question is part of a homework assignment. Considering the hausdorff measure $\mathcal{H}_{2}$ on $\mathbb{R}^{3}$, I need to compute the measure of the unit cube:
$A = \{(x,y,z) \in \mathbb{R}^{3}: x\in [0,1], y \in[0,1], z\in[0,1]\}$
I am reasonably certain the answer is $\infty$, however in this case I would need to bound the approximating measure from below which I am unsure how to do.
Another idea I thought would be that $\cup_{k \in \mathbb{Q}\cap[0,1]} A_{k}$ is a subset of $A$ if for each $j \in \mathbb{Q}\cap[0,1]$ we define $A_{j} = \{(x,y,z):x\in[0,1], y\in[0,1], z = j\}$. Then if I could show each $A_{j}$ has positive measure this would prove it. However the next part of the question is to show that $A_{0}$ has positive measure so if I do it this way I would be skipping ahead a bit.
Suppose the $\mathcal H^2$ measure is finite: that is, there is $M<\infty$ such that for every $\delta>0$ the cube can be covered by family of sets $E_j$ such that $\operatorname{diam}E_j<\delta$ and $\sum_j (\operatorname{diam}E_j)^2\le M$.
The set $E_j$ is contained in a ball of radius $\operatorname{diam}E_j$. The three-dimensional Lebesgue measure of this ball is $$\frac{4\pi}{3}(\operatorname{diam}E_j)^3\le \frac{4\pi \delta}{3} (\operatorname{diam}E_j)^2$$ Summing over $j$ leads to a contradiction.