I am working with some "measure" (I think it is not a measure since it is not subadditive) $$m(E)=\inf \left\{ \sum\limits_{k=1}^\infty \operatorname{diam} F_k:E\subset\bigcup\limits_{k=1}^\infty F_k \right\}.$$ Now I want to prove it is not subadditive. So I consider two disjoint segments $I =\{x\in \mathbb{R}^d : x_i=0\text{ if }i\neq 1\text{ and }0\le x_1 \le 1\}$ and $J=I+h$ where $h\in\mathbb{R}^d$ and $|h|<\varepsilon$. Now I have to show that $m(I)=m(J)=1$ and $m(I\cup J)\le 1+\varepsilon$. First of all, I can show $m(I)\le 1$ because I can just recover it by I which has diameter $1$. But since we don't have subadditivity, how to show $m(I)\ge 1$?
To show $m(I\cup J)\le 1+\varepsilon$ I am not sure how to proceed. The way I tryed it, is to cover the set $I\cup J$ by the parallelogram generated by $e_1=(1,0,0,\ldots,0)$ and $h$ and it's greater diagonal gives the diameter of this set but I think this is not the way to go.
$m(\cdot)$ is subadditive- You are just checking that it is not additive on disjoint closed sets. This is easiest if you take $h= \epsilon e_2=(0,\epsilon,0,...,0).$