Compact null sections imply a compact set is null

If $a=b$, compactness bounds $K$ inside $\{a\}\times [-R,R{]}^n$ for some $R>0$. Divide the $n$-dimensional base into a finite grid of cubes of side at most $\delta$ and cover each corresponding point of $K$ by an $(n+1)$-cube of side $\delta$ centred in the first coordinate at $a$. The number of grid cubes is at most $(2R\mathrm{/}\delta +2{)}^n$, so their total $(n+1)$-volume is at most $(2R+2\delta {)}^n\delta \to 0$. By [F1], $K$ is null. Assume henceforth that $a<b$.

Fix $\epsilon >0$, put $L:=b-a+1$, and set $B:=\epsilon \mathrm{/}(2^{n+2}L)$. For a fixed $t$, nullity [F1] gives a sequence of closed cubes covering $K_t$, with sides ${\mathrm{\ell }}_i$ and total volume below $B\mathrm{/}4$. Enlarge cube $i$ concentrically to an open cube of side ${\mathrm{\ell }}_i+2h_i$, where $h_i=\min \{1,B{2}^{-i-4}\mathrm{/}(3^n(1+{\mathrm{\ell }}_i{)}^{n-1})\}$ for $i\ge 0$. The binomial formula gives $({\mathrm{\ell }}_i+2h_i{)}^n-{\mathrm{\ell }}_i^n\le 3^n{h}_i(1+{\mathrm{\ell }}_i{)}^{n-1}\le B{2}^{-i-4}$. Thus the enlarged open cover has total covering cost below $B\mathrm{/}4+B\mathrm{/}8<B$, where an open cube of side $s$ has covering cost $s^n$. Compactness of $K_t$ supplies a finite subcover, still of cost below $B$. This proves existence separately for each $t$; no family of chosen covers indexed by $t$ is formed.

Let $\mathcal{P}$ be the set of all pairs $(I,\mathcal{Q})$ such that $I$ is an open interval, $\mathcal{Q}$ is a finite list of open $n$-cubes of total covering cost below $B$, and $K\cap (I\times {\mathbb{R}}^n)\subseteq I\times O_{\mathcal{Q}}$, where $O_{\mathcal{Q}}$ is their union. Every $t\in [a,b]$ lies in the interval of at least one such pair. Indeed, take just one finite cover of $K_t$ supplied by step 1.2 and write $O$ for its union. The open sets $\mathbb{R}\times O$ and $\{(x,y):\mathrm{\mid }x-t\mathrm{\mid }>1\mathrm{/}r\}$ for positive integers $r$ cover $K$: points over $t$ lie in the first, and every other point lies in one of the latter. A finite subcover exists. Taking $r$ at least as large as its finitely many indices shows that $K\cap ((t-1\mathrm{/}r,t+1\mathrm{/}r)\times {\mathbb{R}}^n)$ lies in $\mathbb{R}\times O$. Thus this interval and the finite cube list belong to $\mathcal{P}$. This construction proves coverage by the set of all witnessed intervals without selecting one pair for every $t$.

By [L1], the open cover $\{I:(I,\mathcal{Q})\in \mathcal{P}\}$ of $[a,b]$ has a Lebesgue number. Take a finite partition into closed intervals $J_1,\dots ,J_r$ of positive length smaller than that number. For each of these finitely many intervals take a witnessing pair $(I_j,{\mathcal{Q}}_j)\in \mathcal{P}$ with $J_j\subseteq I_j$. Write ${\tilde{Q}}_{j,1},\dots ,{\tilde{Q}}_{j,m_j}$ for its list. Only finitely many witnesses are used, which needs no choice axiom. Then ${\sum }_j\mathrm{\mid }{J}_j\mathrm{\mid }=b-a<L$. For each prism $J_j\times {\tilde{Q}}_{j,\mathrm{\ell }}$, use the following subdivision. [L1, step 2.1, choose] Let ${\lambda }_j:=\mathrm{\mid }{J}_j\mathrm{\mid }$ and let $s_{j,\mathrm{\ell }}$ be the side length of ${\tilde{Q}}_{j,\mathrm{\ell }}$. If ${\lambda }_j\ge s_{j,\mathrm{\ell }}$, partition the interval direction into at most $\lceil {\lambda }_j\mathrm{/}{s}_{j,\mathrm{\ell }}\rceil$ pieces of length at most $s_{j,\mathrm{\ell }}$; if ${\lambda }_j\le s_{j,\mathrm{\ell }}$, partition each of the $n$ base directions into at most $\lceil s_{j,\mathrm{\ell }}\mathrm{/}{\lambda }_j\rceil$ pieces of length at most ${\lambda }_j$. In either case the prism is covered by finitely many closed $(n+1)$-cubes of total volume at most $2^n{\lambda }_j{s}_{j,\mathrm{\ell }}^n=2^n\mathrm{\mid }{J}_j\mathrm{\mid }{\mathrm{vol}}_n({\tilde{Q}}_{j,\mathrm{\ell }})\mathrm{.}$ These cubes cover $K$, and their total volume is at most $2^n{\sum }_{j=1}^r\mathrm{\mid }{J}_j\mathrm{\mid }{\sum }_{\mathrm{\ell }=1}^{m_j}{\mathrm{vol}}_n({\tilde{Q}}_{j,\mathrm{\ell }})<2^{n}L\cdot \frac{\epsilon }{2^{n+2}L}<\epsilon \mathrm{.}$ By [F1], $K$ is null.

Therefore compact null sections imply the whole compact set is null.