07/10/24

Reference: Lecture notes section 2.2

Since we cannot define $\mu$ satisfying the three properties above for all sets, we can try to define in for a smaller class of subsets. In fact, the Lebesgue outer measure does it for us. Remember that the failure of $\mu^*$ to satisfy our properties was that we did not have additivity. So the solution is to consider only those subsets for which such additivity holds.

Definition. A subset $A \subset \mathbb{R}$ is Lebesgue measurable if for every subset $E \subset \R$ we have $\mu^(E) = \mu^(E\cap A) + \mu^*(E \setminus A)$.

This property of $A$ is called the Caratheodory condition.

Remark. Since $E\cap A$ and $E\setminus A$ are disjoint, by subbaditivity property we always have

$\mu^(E) \leq \mu^(E\cap A) + \mu^*(E \setminus A)$

so the definition of measurability is equivalent to the reverse inequality.

This definition remedies the problem with disjoint unions.

Lemma. If $A$ is measurable and $B$ is any set disjoint from $A$ then

$\mu^(A\cup B)=\mu^(A)+\mu^*(B)$

Proof. Apply the definition of measurability to $E=A\cup B$.∎

Example. $\emptyset$ and $\R$ are measurable.

Example. If $\mu^*(A)=0$ then $A$ is measurable. In particular, countable subsets and the ternary Cantor set are measurable.

Proof. Since $E\cap A$ is a subset of $A$ and $E\setminus A$ is a subset of $E$, by monotonicity

$\mu^*(E\cap A)=0$ and

$\mu^(E\setminus A) + \mu^(E\cap A) = \mu^(E\setminus A) \leq \mu^(E)$

which proves the reverse inequality. ∎

Example. The proof discussed earlier shows that the set constructed using the equivalence relation $x\sim y$ if $x-y\in\mathbb{Q}$ and the axiom of choice is not measurable.