Lebesgue outer measure on ${\mathbb{R}}^n$

Fix $n\ge 1$. Lebesgue outer measure ${\lambda }_n^{\ast }$ on ${\mathbb{R}}^n$ is the outer set function induced by the premeasure ${\mu }_0$ of Elementary volume is a sigma-finite premeasure on the algebra of elementary sets on the algebra ${\mathcal{E}}_n$ of elementary sets (Elementary sets: the finite unions of half-open boxes in ${\mathbb{R}}^n$), in the sense of The outer set function induced by a premeasure:

${\lambda }_n^{\ast }(E):=\inf \{{\sum }_{k=0}^{\mathrm{\infty }}{\mu }_0(A_k):A_k\in {\mathcal{E}}_n\text{ for every }k\in \mathbb{N}\text{ and }E\subseteq {\bigcup }_{k\in \mathbb{N}}{A}_k\}$

for $E\subseteq {\mathbb{R}}^n$, the series being the nonnegative extended sum of Series in the nonnegative extended real line. The family of covering costs is nonempty, because ${\mathbb{R}}^n\in {\mathcal{E}}_n$ and the sequence $({\mathbb{R}}^n,\mathrm{\varnothing },\mathrm{\varnothing },\dots )$ covers every $E$, so the infimum is a well-determined element of $[0,+\mathrm{\infty }]$. On the real line the subscript is dropped and ${\lambda }^{\ast }:={\lambda }_1^{\ast }$.

The values ${\lambda }_n^{\ast }(E)$ are defined for every subset of ${\mathbb{R}}^n$, with no measurability hypothesis. That the resulting set function is an outer measure, and that it agrees with ${\mu }_0$ on ${\mathcal{E}}_n$, are proved in Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume; until then the name outer measure is not claimed, exactly as The outer set function induced by a premeasure stipulates.

• Why the covers are by elementary sets and not by boxes. Both give the same value, since an elementary set is a finite union of boxes and a countable family of finite lists reindexes to a countable family of boxes; taking elementary sets is what makes the definition an instance of the published construction, so that the Carathéodory theory applies with nothing reproved. The comparison with covers by closed, open and cubic boxes is Countable covers by closed boxes, by open boxes and by closed cubes all compute Lebesgue outer measure.

• The definition itself spends no choice principle; it is an infimum of a nonempty subset of $[0,+\mathrm{\infty }]$. Countable choice enters only when the infimum is shown to be countably subadditive, and that is recorded where it happens.