Why is the inner measure problematic?

Solution 1:

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.

Solution 2:

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