Let $U$ be an open subset of $\mathbb{R}$. $U$ is a countable disjoint union of open intervals $I_n$. We denote $m(U) = \sum l(I_n)$, where $l(I_n)$ is the length of $I_n$. Let $A$ be a subset of $\mathbb{R}$. We denote by $\mu^*(A)$ the infimum of $m(U)$ where $U$ is an open subset such that $A \subset U$. We say a subset $M$ of $\mathbb{R}$ is measurable if it satisfies the following condition.
For every subset $A$ of $\mathbb{R}$, $\mu^*(A) = \mu^*(A\cap M) + \mu^*(A - M)$.
This is Caratheodory's definition of measurability. I don't like this. I think it's unintutive. It seems to come out of nowhere. Now I'll state my question. Let $\mathcal{B}$ be the $\sigma$-algebra generated by the set of open subsets of $\mathbb{R}$. Can we prove the following assertion without using Caratheodory's method?
There exists a unique measure $\mu$ on $\mathcal {B}$ with the following properties.
1) $\mu([0, 1]) = 1$
2) For $a \in \mathbb{R}$ and $M \in \mathcal{B}$, $a + M \in \mathcal{B}$(we need to prove this) and $\mu(a + M) = \mu(M)$.
I'm almost sure the answer is affirmative, because it's very unlikely that Lebesgue's original paper used Caratheodory's method.
It is true that Caratheodory's condition is not needed for Lebesgue measure. In fact, Caratheodory writes that his condition was based on the previously known condition: $M \subseteq \mathbb R^n$ is Lebesgue measurable if and only if
Caratheodory wanted a condition that could be applied more generally. For example, for "arc length" meausure in $\mathbb R^n, n \ge 2$, where every open interval has infinite measure.