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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.

Makoto Kato
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  • Caratheodory is pretty intuitive. Basically, all the condition says is that if you measure the volume of a set from the outside and if you measure it from the inside, you get the same volume. See this MathOverflow link: http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability – Batman Mar 30 '15 at 00:51
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    "I don't like this. I think it's unintutive. It seems to come out of nowhere." - I'm afraid with that attitude you won't get very far in analysis... – Math1000 Mar 30 '15 at 01:23
  • @Math1000 What's wrong with looking for a different proof other than a traditional one? – Makoto Kato Mar 30 '15 at 22:47
  • Nothing wrong with proving something in more than one way. Just that a lot of analysis isn't exactly "intuitive." Some texts are better than others about this...I'll agree that e.g. Rudin has a lot of proofs that "seem to come out of nowhere." – Math1000 Mar 31 '15 at 01:19
  • @Math1000 I believe from my experiences that most theorems have natural proofs. – Makoto Kato Mar 31 '15 at 04:54

1 Answers1

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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

For every bounded open interval $A$ of $\mathbb R^n$, $\mu^*(A) = \mu^*(A\cap M) + \mu^*(A - M)$

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.

GEdgar
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