Countable Union of Compact Null Sets Is Null

Let $K\subseteq {\mathbb{R}}^m$ be compact and null, and let $\eta >0$. If $K=\mathrm{\varnothing }$, the empty family covers it. Otherwise [F1] gives a countable closed-cube cover $(C_j)$ with total volume at most $\eta \mathrm{/}8$. Write $C_j={\prod }_{r<m}[a_{jr},a_{jr}+{\mathrm{\ell }}_j]$ and let $x_{jr}=a_{jr}+{\mathrm{\ell }}_j\mathrm{/}2$ be its center coordinates. Set ${\delta }_j=\min \{1,\eta 2^{-j-5}\mathrm{/}(3^m(1+{\mathrm{\ell }}_j{)}^{m-1})\}$ and let $C_j^{\mathrm{\prime }}={\prod }_{r<m}(x_{jr}-({\mathrm{\ell }}_j+2{\delta }_j)\mathrm{/}2,x_{jr}+({\mathrm{\ell }}_j+2{\delta }_j)\mathrm{/}2)$, an open concentric enlargement. By [F11], $({\mathrm{\ell }}_j+2{\delta }_j{)}^m-{\mathrm{\ell }}_j^m={\sum }_{r=1}^m\iota \left(\genfrac{}{}{0px}{}{m}{r}\right){\mathrm{\ell }}_j^{m-r}(2{\delta }_j{)}^r\le {\delta }_j(1+{\mathrm{\ell }}_j{)}^{m-1}{\sum }_{r=1}^m\iota \left(\genfrac{}{}{0px}{}{m}{r}\right)2^r<3^m{\delta }_j(1+{\mathrm{\ell }}_j{)}^{m-1}\le \eta 2^{-j-5}$, because ${\delta }_j^r\le {\delta }_j$, ${\mathrm{\ell }}_j^{m-r}\le (1+{\mathrm{\ell }}_j{)}^{m-1}$ and ${\sum }_{r=1}^m\iota \left(\genfrac{}{}{0px}{}{m}{r}\right)2^r=3^m-1$. Define the covering cost of an open cube of side $s$ to be $s^m$; this agrees with [F5] for its closed counterpart. Put $d_j=\eta 2^{-j-5}$ and $b_j=({\mathrm{\ell }}_j+2{\delta }_j{)}^m={\mathrm{\ell }}_j^m+{\mathrm{\Delta }}_j$, where $0\le {\mathrm{\Delta }}_j\le d_j$. For every $N\ge 0$, induction using [F9] gives ${\sum }_{j<N}{d}_j=\eta (1-2^{-N})\mathrm{/}16\le \eta \mathrm{/}16$: the formula is $0$ at $N=0$, and the successor step adds $\eta 2^{-N-5}$. By [F7], the original nonnegative volume series has every partial sum at most $\eta \mathrm{/}8$. Finite-sum additivity and monotonicity [F10] therefore give ${\sum }_{j<N}{b}_j\le \eta \mathrm{/}8+{\sum }_{j<N}{d}_j\le 3\eta \mathrm{/}16$ for every $N$. Since $b_j\ge 0$, [F7] gives a convergent total covering-cost series with sum at most $3\eta \mathrm{/}16<\eta \mathrm{/}4$. The open cubes cover $K$, and compactness [F2] gives a finite subcover.

Order the finite subcover by its original indices and write $L_r$ and $x_r$ for the side and center of its $r$th open cube. Its finite total cost is at most the partial sum of $(b_j)$ through its largest original index, hence at most $3\eta \mathrm{/}16<\eta \mathrm{/}4$ by [F7, F10, step 1.1]. The finite polynomial sum ${\sum }_r(L_r+\zeta {)}^m$ is continuous at $\zeta =0$, so choose $\zeta >0$ for which it is less than ${\sum }_r{L}_r^m+\eta \mathrm{/}8<3\eta \mathrm{/}8$. By [F4], choose rational $q_r$ with $L_r<q_r<L_r+\zeta$ and a rational center $y_r\in {\mathbb{Q}}^m$ whose coordinates differ from $x_r$ by less than $(q_r-L_r)\mathrm{/}4$. The closed rational cube of side $q_r$ centered at $y_r$ contains the open cube of side $L_r$ centered at $x_r$, since each coordinate displacement is less than $L_r\mathrm{/}2+(q_r-L_r)\mathrm{/}4<q_r\mathrm{/}2$. These finitely many rational cubes cover $K$, and their total volume is less than $3\eta \mathrm{/}8<\eta \mathrm{/}2$ by [F5].

Fix a bijection $\nu :\mathbb{N}\to \mathbb{Q}$ and a pairing bijection $\pi :{\mathbb{N}}^2\to \mathbb{N}$ from [F3]. A list of $t$ rational cubes is determined by $(m+1)t$ rationals: its side and its $m$ center coordinates for each cube. Encode a tuple $(a_0,\dots ,a_{r-1})\in {\mathbb{N}}^r$ recursively by $c_0=0$ and $c_{r+1}(a_0,\dots ,a_r)=\pi (a_0,c_r(a_1,\dots ,a_r))$. Since $\pi$ and $\nu$ are injective, pairing the list length $t$ with the code of its $(m+1)t$ rational coordinates gives an injection from all finite rational cube lists into $\mathbb{N}$. For each natural code, use its unique list preimage if it has one, and otherwise use the empty list; this defines an enumeration $({\mathcal{C}}_s{)}_{s\ge 0}$ that includes every list. By step 2.1, every compact null $K$ has at least one enumerated list covering it with total volume less than $\eta \mathrm{/}2$. The set of qualifying codes is nonempty, so it has a least member by [F6]. The least-code rule assigns a unique cover to each pair $(K,\eta )$ without choosing from an arbitrary family.

Fix $\epsilon >0$. For each $i\ge 0$, apply the unique rule of step 3.1 to $(K_i,{\eta }_i)$ with ${\eta }_i=\epsilon 2^{-i-1}$. Let $t_i$ be the finite length of its selected list, whose total volume is less than $\epsilon 2^{-i-2}$. Define $N_0=0$ and $N_{i+1}=N_i+t_i$ recursively as in [F8]. For each $n\in \mathbb{N}$, if $N_i\le n<N_{i+1}$ for some $i$, take the cube at position $n-N_i$ in the $i$th list; otherwise take the degenerate zero-side cube at the origin. The half-open index intervals are disjoint: if $i<j$, then $N_{i+1}\le N_j$, so no $n$ lies in both. Each cube of the $i$th list, at position $j<t_i$, occurs at index $N_i+j$; thus every selected list occurs in the sequence.

Any finite initial segment meets only finitely many blocks. If it meets none, its total volume is zero. Otherwise, its finite set of block indices has a largest member $r$: scan the finite list and retain the larger at each comparison using the linear order in [F8]. Each partial block costs at most its full-list bound from step 4.1. By [F9], the budget sum $B_r={\sum }_{i=0}^r\epsilon 2^{-i-2}$ satisfies $B_0=\epsilon \mathrm{/}4=\frac{\epsilon }{2}(1-2^{-1})$ and $B_{r+1}=B_r+\epsilon 2^{-r-3}=\frac{\epsilon }{2}(1-2^{-r-1})+\epsilon 2^{-r-3}=\frac{\epsilon }{2}(1-2^{-r-2})$. So induction gives $B_r=\frac{\epsilon }{2}(1-2^{-r-1})<\epsilon \mathrm{/}2$. Thus every partial sum of the nonnegative volume series is at most $\epsilon \mathrm{/}2$, and [F7] gives convergence with total at most $\epsilon \mathrm{/}2<\epsilon$. The sequence covers ${\bigcup }_i{K}_i$, so [F1] shows that this union is null. Since every selected list is assigned by the unique least-code rule of step 3.1, the proof uses no choice principle.