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. $A \subset \R$ is measurable if and only if its complement $A^c = \R\setminus A$ is measurable. Indeed. the Caratheodory condition is the same for $A$ and $A^c$. For example, the set of irrational numbers is measurable because the set of rational numbers is by the previous example.
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.