# Why is the inner measure problematic?

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?

• @Watson ??? What's "awkward" about the inner measure of the irrationals? Empty interior, so what? The inner measure is defined in terms of compact subsets, making the inner measure of the irrationals $\infty$. – David C. Ullrich May 24 '18 at 13:45
• You stated the Cartheodory condition incorrectly. I'm going to go ahead and change it - feel free to rollback to the wrong version if you prefer... – David C. Ullrich May 24 '18 at 14:02
• @DavidC.Ullrich Thanks a lot :-) – Stephan Kulla May 25 '18 at 19:12
• How is inner measure defined on the real line? – zhw. Jun 10 '18 at 18:32

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 $$\lambda^*([0,1])=\lambda^*(A)+\lambda^*([0,1]\setminus A) =\lambda^*([0,1]\cap A)+\lambda^*([0,1]\setminus A).$$
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