Non-Lebesgue Measurable Sets: Understanding Measurement

[dimitri151]

In analysis we were shown the existence of non-Lebesgue measurable sets (eg a choice function over the rational equivalence partition of an interval). From the proof it seems that this means you can't assign number to the Lebesgue measure of this set i.e. if you say its measure is zero it's not enough and if you say it's some finite number then it is too much. However the way we learned Lebesgue measure was that Lebesgue measure was Lebesgue outer measure restricted to a certain family if sets. But Lebesgue outer measure is defined for all sets in P(R). So my question is what is the Lebesgue outer measure (what number) of a non-Lebesgue measurable set like the one above?

 

[micromass]

See http://math.stackexchange.com/questions/14591/vitali-type-set-with-given-outer-measure

If all you know is that you take one number $x\in [0,1]$ for each equivalence class, then the outer measure is not uniquely determined. You will know that the outer measure is an element of ]0,1], but that's all you'll know. The outer measure could be each element in ]0,1], depending on how you made the choice...

 

[dimitri151]

I see. But if you specify unambiguously how the elements $x\in[0,1]$ are chosen then the set E (the non-measurable set) will have a specific outer measure. But then there will be some set A not disjoint from E such that the sum of the outer measure of the part of A in E and the outer measure of the part of A not in E will be greater than the outer measure of A. It's hard to imagine how this could be.