Carathéodory’s lemma

In measure theory, Carathéodory’s lemma is used for constructing measures and, for example, can be applied to the construction of the Lebesgue measure and is used in the proof of Carathéodory’s extension theorem.The idea is that to define a measure on a measurable space $(X,𝒜)$ we would first construct an outer measure (http://planetmath.org/OuterMeasure2), which is a set function defined on the power set of $X$. Then, this outer measure is restricted to $𝒜$ and Carathéodory’s lemma is applied to show that this restriction does in fact result in a measure. For an example of this procedure, see the proof of Carathéodory’s extension theorem.

Given an outer measure $\mu$ on a set $X$, the result first defines a collection of subsets of $X$ — the $\mu$-measurable sets.A subset $S\subseteq X$ is called $\mu$-measurable (or Carathéodory measurable with respect to $\mu$) if the equality

$$\mu (E)=\mu (E\cap S)+\mu (E\cap S^c)$$

holds for every $E\subseteq X$. Then, Caratheodory’s lemma says that a measure is obtained by restricting $\mu$ to the $\mu$-measurable sets.

Lemma (Carathéodory).

Let $\mu$ be an outer measure on a set $X$, and $\mathrm{A}$ be the class of $\mu$-measurable sets. Then $\mathrm{A}$ is a $\sigma$-algebra (http://planetmath.org/SigmaAlgebra) and the restriction of $\mu$ to $\mathrm{A}$ is a measure.

It should be noted that for any outer measure $\mu$ and sets $S,E\subseteq X$, subadditivity of $\mu$ implies that the inequality $\mu (E)\le \mu (E\cap S)+\mu (E\cap S^c)$ is always satisfied. So, only the reverse inequality is required and consequently $S$ is $\mu$-measurable if and only if

$$\mu (E)\ge \mu (E\cap S)+\mu (E\cap S^c)$$

for every $E\subseteq X$.