Lebesgue inner measure on the real line

For a set $E\subseteq \mathbb{R}$, the Lebesgue inner measure of $E$ is

${\lambda }_{\ast }(E):=\sup \{{\lambda }^{\ast }(K):K\subseteq E\text{ and }K\text{ is compact}\},$

the supremum being taken in $[0,+\mathrm{\infty }]$.

This is well defined without a measurability assumption on $E$ or on its compact subsets: Lebesgue outer measure is defined on every subset of $\mathbb{R}$, the family contains $\mathrm{\varnothing }$, and ${\lambda }^{\ast }(\mathrm{\varnothing })=0$. Assuming the Axiom of Countable Choice, compact subsets of $\mathbb{R}$ are Borel and hence Lebesgue measurable, so ${\lambda }^{\ast }(K)=\lambda (K)$ and the displayed definition agrees with the usual compact-inner-approximation formula.

• The definition uses compact subsets, not merely closed ones, because compact subsets of $\mathbb{R}$ always have finite Lebesgue measure.

• Assuming the Axiom of Countable Choice, for bounded sets For bounded subsets of $\mathbb{R}$, Lebesgue measurability is equivalent to equality of inner and outer measure shows that equality ${\lambda }_{\ast }(E)={\lambda }^{\ast }(E)$ is exactly Lebesgue measurability.