Measure Theory "Incompleteness"

Question

So, i have been studying some measure theory , and from what i was introduced we can have sets that are jordan-mensurable and lebesgue-mensurable. In both of these cases we can find sets that we cant classify, we cant give a measure to, if we use the axiom of choice in the case of lebesgue. My question is, is it possible to create a theory that can classify in terms of measure to every set? That can complete lebesgue-measure? Or is it just impossible?(can we prove that is impossible?). Keep in mind im thinking of a theory that can work with the axiom of choice.

Answer 1

Not if you want all the "reasonable" properties of a measure to hold. You should look up the Vitali set.

Answer 2

It depends on what properties you want your measure to have. You can define "outer" measure for every set, and it agrees with Lebesgue measure for measureable sets, but it's not additive for all sets. In fact you can pretty much define Lebesgue measurable sets as the ones for which outer measure behaves additively.

You can also define inner measure, and a bounded set is Lebesgue measurable if and only if its outer and inner measures are equal.

If you want a true measure (countably additive) that assigns a finite measure to every bounded set and a non-zero measure to every open interval, it's certainly possible, but it won't be translation invariant. For example, define a measure that assigns measure $\frac{1}{q^3}$ to each rational with denominator $q$ and to any set the sum of the measures of the rationals in that set.

As for why the axiom of choice guarantees that any translation-invariant countably-additive measure must make some sets non-measurable, look up "Vitali set" in Wikipedia.

It may also be worth noting that even without the axiom of choice, the contiuum hypothesis implies the existence of non-Lebesgue-measurable sets, as see https://mathoverflow.net/questions/191262/sierpinskis-construction-of-a-non-measurable-set , so to get a set theory with all sets Lebesgue measurable requires that both the axiom of choice and the continuum hypothesis fail.