For a nonzero real $c$, dilation by $c$ multiplies Lebesgue outer measure by $\mathrm{\mid }c{\mathrm{\mid }}^n$, and reflection in the origin preserves it

For reals $u_i\le v_i$ one has $c[u,v]=[cu,cv]$ when $c>0$ and $c[u,v]=[cv,cu]$ when $c<0$, in both cases a closed rectangle whose $i$-th side length is $\mathrm{\mid }c\mathrm{\mid }(v_i-u_i)$; its volume is therefore ${\prod }_{i<n}(\mathrm{\mid }c\mathrm{\mid }(v_i-u_i))=\left({\prod }_{i<n}\mathrm{\mid }c\mathrm{\mid }\right){\prod }_{i<n}(v_i-u_i)=\mathrm{\mid }c{\mathrm{\mid }}^n\mathrm{vol}[u,v]$.

Put $q:=\mathrm{\mid }c{\mathrm{\mid }}^n>0$. If $s_0=\inf S$ for a nonempty $S\subseteq [0,+\mathrm{\infty }]$, then $q{s}_0$ is a lower bound of $qS:=\{qs:s\in S\}$: for every real $s\in S$ the inequality $s_0\le s$ gives $q{s}_0\le qs$ by [F6], while the claim is automatic when $s=+\mathrm{\infty }$. Conversely, let $t$ be a lower bound of $qS$. If $t=+\mathrm{\infty }$, then every element of $qS$ is $+\mathrm{\infty }$, hence every element of $S$ is $+\mathrm{\infty }$ and therefore $s_0=+\mathrm{\infty }$. If $t$ is real, then $q^{-1}>0$ by [F6], so $t\le qs$ implies $q^{-1}t\le s$ for every real $s\in S$, and again the claim is automatic when $s=+\mathrm{\infty }$; thus $q^{-1}t$ is a lower bound of $S$, so $q^{-1}t\le s_0$ and therefore $t\le q{s}_0$. Hence $\inf (qS)=q\inf S$.

The assignment $[u,v]\mapsto c[u,v]$ is a bijection from the countable closed-rectangle covers of $E$ onto those of $cE$, with inverse given by multiplication by $c^{-1}$. For one such cover, let $(a_k)$ be its sequence of rectangle volumes and $(s_n)$ the partial sums of ${\sum }_{k=0}^{\mathrm{\infty }}{a}_k$ in the sense of Series in the nonnegative extended real line; let $(t_n)$ be the partial sums of the transformed cover cost. By step 1.1 each transformed term is $q{a}_k$, and the shared recursion of nonnegative extended series gives $t_n=q{s}_n$ for every $n$. Therefore the transformed cover cost is $q{\sum }_{k=0}^{\mathrm{\infty }}{a}_k$ by step 1.2. So step 1.2 turns the infimum of all transformed cover costs into ${\lambda }^{\mathrm{c}\mathrm{l}}(cE)=\mathrm{\mid }c{\mathrm{\mid }}^n{\lambda }^{\mathrm{c}\mathrm{l}}(E)$, and [L1] then gives the same identity for ${\lambda }_n^{\ast }$.

For a test set $A$ one has $A\cap cE=c((c^{-1}A)\cap E)$ and $A\setminus cE=c((c^{-1}A)\setminus E)$, so step 2.1 turns the Carathéodory identity for $cE$ tested against $A$ into $\mathrm{\mid }c{\mathrm{\mid }}^n$ times the identity for $E$ tested against $c^{-1}A$; multiplication by the positive real $\mathrm{\mid }c{\mathrm{\mid }}^n$ is injective on $[0,+\mathrm{\infty }]$, and $A\mapsto c^{-1}A$ is a bijection of the power set, so $cE$ is Lebesgue measurable exactly when $E$ is, and then ${\lambda }_n(cE)={\lambda }_n^{\ast }(cE)=\mathrm{\mid }c{\mathrm{\mid }}^n{\lambda }_n^{\ast }(E)=\mathrm{\mid }c{\mathrm{\mid }}^n{\lambda }_n(E)$.