A Nonempty Euclidean Open Set Is Not Null

Choose $x\in O$ and $r>0$ with the Euclidean ball $B(x,r)\subseteq O$. Put $s=r\mathrm{/}(2\sqrt{m})>0$ and let $Q={\prod }_{j=1}^m[x_j-s\mathrm{/}2,x_j+s\mathrm{/}2]$. For every $y\in Q$, $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2\le \sqrt{m}s\mathrm{/}2=r\mathrm{/}4<r$, so $Q\subseteq O$ and $\mathrm{vol}(Q)=s^m>0$.

Suppose $O$ were null. By [F1], with $\epsilon =s^m\mathrm{/}4$ there is a sequence of closed cubes $C_i$ covering $Q$ whose side lengths ${\mathrm{\ell }}_i\ge 0$ have ${\sum }_i{\mathrm{\ell }}_i^m\le s^m\mathrm{/}4$. For each $i\ge 0$, set ${\delta }_i=\min \{1,s^m{2}^{-i-5}\mathrm{/}(3^m(1+{\mathrm{\ell }}_i{)}^{m-1})\}$ and enlarge $C_i$ concentrically to an open cube $C_i^{\mathrm{\prime }}$ of side ${\mathrm{\ell }}_i+2{\delta }_i$. By [F6] and ${\delta }_i\le 1$, $0\le {\mathrm{\Delta }}_i:=({\mathrm{\ell }}_i+2{\delta }_i{)}^m-{\mathrm{\ell }}_i^m\le d_i:=s^m{2}^{-i-5}$. For every $N\ge 0$, induction using the finite-sum recursion gives ${\sum }_{i<N}{d}_i=s^m(1-2^{-N})\mathrm{/}16\le s^m\mathrm{/}16$. Thus every partial sum of the nonnegative enlarged-cost sequence $b_i=({\mathrm{\ell }}_i+2{\delta }_i{)}^m={\mathrm{\ell }}_i^m+{\mathrm{\Delta }}_i$ satisfies ${\sum }_{i<N}{b}_i\le s^m\mathrm{/}4+s^m\mathrm{/}16=5s^m\mathrm{/}16$, by [F4] and [F5]. A second application of [F5] gives a convergent total enlarged-cost series with sum at most $5s^m\mathrm{/}16<s^m\mathrm{/}2$.

The open cubes $C_i^{\mathrm{\prime }}$ cover the closed box $Q$, so [F2] gives a finite subcover $C_{i_1}^{\mathrm{\prime }},\dots ,C_{i_t}^{\mathrm{\prime }}$. Order its indices increasingly and take $N$ larger than the largest one. By [F4], the subcover's total cost is at most ${\sum }_{i<N}{b}_i$, which is at most $5s^m\mathrm{/}16<s^m\mathrm{/}2$ by step 2.1.

Write $\overline{C_i^{\mathrm{\prime }}}$ for the closed cube with the same center and side as $C_i^{\mathrm{\prime }}$, so $\mathrm{vol}(\overline{C_i^{\mathrm{\prime }}})=b_i$. Cut $Q$ and these finitely many closed covering cubes at all their coordinate endpoints. The positive-side grid cells inside $Q$ partition it; the interior of each cell lies in one covering cube because no covering-cube boundary crosses the cell. Assign each cell to the least index of a covering cube containing its interior. By finite cut-additivity from [F3], the assigned cells for each cube have total volume at most that closed cube's volume, while all cells sum to $\mathrm{vol}(Q)=s^m$. Thus $s^m\le {\sum }_{j=1}^t\mathrm{vol}(\overline{C_{i_j}^{\mathrm{\prime }}})<s^m\mathrm{/}2$, a contradiction. Therefore $O$ is not null.