Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$

Fix $n\ge 1$ and let ${\lambda }_n^{\ast }$ be the Lebesgue outer set function on ${\mathbb{R}}^n$ (Lebesgue outer measure on ${\mathbb{R}}^n$). A set $E\subseteq {\mathbb{R}}^n$ is Lebesgue measurable when

${\lambda }_n^{\ast }(A)={\lambda }_n^{\ast }(A\cap E)+{\lambda }_n^{\ast }(A\setminus E)\text{for every }A\subseteq {\mathbb{R}}^n\mathrm{.}$

This formula makes sense before any outer-measure theorem is invoked. Under countable choice, Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume makes ${\lambda }_n^{\ast }$ an outer measure, and the displayed condition is then exactly Carathéodory measurability in the sense of Carathéodory measurable sets.

The family of Lebesgue measurable sets is written $\mathcal{L}({\mathbb{R}}^n)$, and Lebesgue measure is the restriction

${\lambda }_n:={\lambda }_n^{\ast }\text{\hspace{0.17em}⁣}{\upharpoonright }_{\mathcal{L}({\mathbb{R}}^n)}\mathrm{.}$

On the real line the subscript is dropped and $\lambda :={\lambda }_1$.

The quantifier over every test set $A$ is part of the condition, and no hypothesis on $E$ is imposed before it is tested. That $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra and that ${\lambda }_n$ is a complete measure on it are not part of this definition: they are proved, under the Axiom of Countable Choice, in Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume, which is recorded in this item's justified_by because it is a statement about the objects introduced here. Until that theorem the symbols $\mathcal{L}({\mathbb{R}}^n)$ and ${\lambda }_n$ name a family of sets and a restricted set function, nothing more.

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

Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume

Carathéodory measurable sets

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume ↗

• Why the Carathéodory criterion rather than the inner-outer criterion. Lebesgue's original definition compares the outer measure of $E$ with that of its complement inside a large box, and it is available only for bounded $E$; the criterion above is stated for every subset at once and is what makes the published Carathéodory machinery apply verbatim. The two agree, and the equivalence with the approximation criteria is Assuming countable choice, four equivalent descriptions of a Lebesgue measurable subset of ${\mathbb{R}}^n$.

• The definition is relative to ${\lambda }_n^{\ast }$ and to nothing else. Changing the outer measure changes the family; the family attached to a general outer measure is written ${\mathcal{M}}_{{\mu }^{\ast }}$ in Carathéodory measurable sets, and $\mathcal{L}({\mathbb{R}}^n)$ is the name reserved for the instance ${\mu }^{\ast }={\lambda }_n^{\ast }$.