The product of a content-zero set and a compact interval has content zero

If $A=\mathrm{\varnothing }$, the empty family covers $A\times [c,d]$. If $d=c$, use [F1] with tolerance $\min \{\epsilon ,1\mathrm{/}2\}$, obtaining a finite cube cover $Q_i$ with side lengths ${\mathrm{\ell }}_i\le 1$ and total base volume at most $\epsilon$; then the cubes $Q_i\times [c,c+{\mathrm{\ell }}_i]$ cover $A\times \{c\}$ and have total $(m+1)$-volume at most the base total.

Suppose $A\mathrm{\ne }\mathrm{\varnothing }$ and $L:=d-c>0$. Use [F1] with tolerance $\delta :=\min \{1\mathrm{/}2,\epsilon \mathrm{/}(2(L+1))\}$; enlarge any zero-side cubes slightly, using the unused half of this tolerance, so that the resulting finite cover has $0<{\mathrm{\ell }}_i\le 1$ and total base volume below $2\delta \le \epsilon \mathrm{/}(L+1)$. For each $i$, let $N_i:=1+\lfloor L\mathrm{/}{\mathrm{\ell }}_i\rfloor$. Fact [F2] gives $N_i{\mathrm{\ell }}_i>L$ and $N_i{\mathrm{\ell }}_i\le L+{\mathrm{\ell }}_i$, and the $N_i$ consecutive $(m+1)$-cubes of side ${\mathrm{\ell }}_i$ above $Q_i$ cover $Q_i\times [c,d]$.

The total volume of the cubes in step 1.2 is ${\sum }_i{N}_i{\mathrm{\ell }}_i^{m+1}\le {\sum }_i(L+{\mathrm{\ell }}_i){\mathrm{\ell }}_i^m\le (L+1){\sum }_i{\mathrm{\ell }}_i^m<\epsilon$, the first two inequalities being the bounds $N_i{\mathrm{\ell }}_i\le L+{\mathrm{\ell }}_i$ and ${\mathrm{\ell }}_i\le 1$ of step 1.2 and the last the strict bound on the base total. Together with step 1.1 this supplies an arbitrarily small finite cube cover in every case, so $A\times [c,d]$ has content zero.