Elementary lower and upper bounds on a unit cube

If a nonempty bounded $U$ has diameter $r$, its $i$th coordinate ranges between an infimum $a_i$ and supremum $b_i$ with $b_i-a_i\le r$. Hence $U\subseteq {\prod }_i[a_i,b_i]$ and ${\lambda }_n^{\ast }(U)\le {\prod }_i(b_i-a_i)\le r^n$. If $r=0$ this is a zero-volume singleton box.

For every finite-scale cover of $A$, countable subadditivity gives ${\lambda }_n^{\ast }(A)\le {\sum }_j(\mathrm{diam}{U}_j{)}^n$. Infimising and then taking the small-scale supremum gives the lower bound; absent covers give the same inequality with infinity on the right. The empty set has both values zero.

Partition $(0,1{]}^n$ into $m^n$ half-open cubes of side $1\mathrm{/}m$. Each has diameter $\sqrt{n}\mathrm{/}m$ (the supremum of corner distances), so the total cost is $m^n(\sqrt{n}\mathrm{/}m{)}^n=n^{n\mathrm{/}2}$. Choose $m$ large for any prescribed positive scale. Thus the upper bound holds, while the lower bound is the unit box volume one. For $n=1$ both bounds equal one.