Inner measure

Remarks

The outer measure induced by $\mu$ is $\mu^*(A) = \inf \set{ \mu(B) : B \in \Sigma, \ A \subseteq B }$, and $\mu_*(A) \le \mu^*(A)$ for all $A$. If $A \in \Sigma$, then $\mu_*(A) = \mu(A) = \mu^*(A)$. For Lebesgue measure $\lambda$ on $\R^n$, the supremum may be taken over compact subsets of $A$. For Lebesgue measure and a set $A$ with $\lambda^*(A) < \infty$, the set $A$ is Lebesgue measurable if and only if $\lambda_*(A) = \lambda^*(A)$. Example: for $A = \Q \cap [0, 1]$ and Lebesgue measure, $A$ is measurable with $\lambda_*(A) = \lambda^*(A) = 0$. Non-example of measurability: a Vitali set $V \subseteq [0, 1]$ has $\lambda_*(V) = 0 < \lambda^*(V)$. Unlike an outer measure, an inner measure is superadditive: $\mu_*(A \cup B) \ge \mu_*(A) + \mu_*(B)$ for disjoint $A$ and $B$. Some authors define inner measures abstractly by such axioms, without a given measure.