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Currently the Lebesgue measure is defined by the outer measure $\lambda^*(A)$ by the criterion of Carathéodory: A set $A$ is Lebesgue measurable iff for every set $B$ we have $\lambda^*(B)=\lambda^*(B\cap A) + \lambda^*(B\cap A^C)$. However, before the criterion of Carathéodory the Lebesgue measure was defined by the outer and the inner measure $\lambda_*(A)$. For Lebesgue a set has a measure $\lambda(A)$ iff the inner and the outer measure is the same: $\lambda_*(A)=\lambda^*(A)$. Thus, we dropped the inner measure in the current definition of the Lebesgue measure.

Why is the inner measure currently not used in the definition of the measure? What are the problems which occur by using the inner measure?

My attempts so far: I read that the criterion of Carathéodory makes the extension theorem easier. Why is this the case. Is this the only reason?

Stephan Kulla
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2 Answers2

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Tao talks about this in his measure theory book, which last time I checked was still freely available.

He says roughly that there is an asymmetry here because when measuring finite unions of boxes, the measure is subadditive and not superadditive. This leads to the fact that the Jordan inner measure $$ m_{*, J}(U) := \sup \{m(A)\ :\ A \subset U\ \text{is a finite union of boxes}\} $$ doesn't care whether or not you put finite or countable in the definition (because you are taking a supremum and its subadditive). So a natural Lebesgue inner measure where you put countable here gives no increase in power/resolution of the measure

Thinking like this leads to the idea that Lebesgue inner measure should just be the outer measure of the complement.

SBK
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No comment on why C's version is "better", but it's not hard to show the two are equivalent - I never felt I understood the C definition until I realized this.

Hints/outline: First we wave our hands and claim that with both versions of the definition, $A$ is measurable if and only if $A\cap[n,n+1)$ is measurable for every integer $n$. So assume $A\subset[0,1]$.

It's easy to see from the definition of inner measure that $$\lambda_*(A)=1-\lambda^*([0,1]\setminus A).$$So $\lambda_*(A)=\lambda^*(A)$ if and only if $$\lambda^*([0,1])=\lambda^*(A)+\lambda^*([0,1]\setminus A) =\lambda^*([0,1]\cap A)+\lambda^*([0,1]\setminus A).$$

David C. Ullrich
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  • Thanks for your answer! I have a follow up question for your answer: https://math.stackexchange.com/questions/2806982/what-is-the-inutition-behind-the-caratheodory-criterion-for-measurable-sets – Stephan Kulla Jun 03 '18 at 20:07
  • Sir do you have time to answer me a question in the comments?..its about the exercise 1.5.6 in Donald Cohn measure theory.. My idea is that if we take as a counterexeample on the real line the sigma algebra that contains the sets that are countable or their complement is countable,with the counting measure and A=(0,1),...then the inner measure and outer measure (by definition in Cohn) are infinite,but the set does not belong to the completion of the sigma algebra (which is the sigma algebra itself,since the only null set is the empty set)...is my idea correct? – Marios Gretsas Feb 13 '20 at 15:55
  • @StephanKulla That "follow up" question was closed as a duplicate two years ago! So it's already been answered - asking for another answer without saying what your problem is with the existing answers is not a good idea. – David C. Ullrich Feb 13 '20 at 16:34
  • @MariosGretsas I might have time to try to answer your question, if you said what the question actually was. GIven that evidently you can't take the time to even state the question, no, sorry I don't have time to try to answer it. – David C. Ullrich Feb 13 '20 at 16:35
  • @DavidC.Ullrich it is the exercise 1.5.6 in Donald Cohn's Measure theory..we have as a propostion that for a set $A$ of finite measure in a measure space$ (X,S,\mu) $ the $A$ belongs to the completion $S_{\mu}$ if and only if the inner measure $\mu_$ and outer measure $\mu^$ of $A$ are equal..the exercise states that one direction of this proposition is not true if we assume that $\mu^*(A)=+\infty$..os i gave my counterexample in the previous comment..is it correct?...Thank you in advance :) – Marios Gretsas Feb 13 '20 at 22:21
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    @MariosGretsas Yes, that looks like a counterexample to me... – David C. Ullrich Feb 14 '20 at 12:41
  • Thank you for your answer..:) – Marios Gretsas Feb 14 '20 at 13:08
  • @DavidC.Ullrich: My solution for the question in the meanwhile is that the inner measure needs a union of disjoint boxes while the outer measure can have overlapping boxes in the definition. Thus, the inner measure is harder to determine. A definition which just uses the outer measure is therefore easier to use. What do you think about this intuitive explanation? – Stephan Kulla Feb 25 '20 at 17:54