Why non-measurable sets exist?

Question

What Is the reason behind the fact that $\Bbb R$ is not a complete measure space? Is it only due to the cardinality of $\Bbb R$? Or other structures on $\Bbb R$ also play a role? Can one make every subset of $\Bbb R$ trivially measurable by assigning them to a number of a complete ordered field of the cardinality $\aleph_2$ ?

Answer 1

The reason would depend on the particular argument you give but it all boils down to the fact that (with the axiom of choice) one can construct very bizarre sets of real numbers.

You also have to be more precise. There are measures on the set of all subsets of $\mathbb R$, e.g., counting measure. What does not exist is a translation invariant extension of Lebesgue measure to all subsets of $\mathbb R$. For this certainly the additive structure of $\mathbb R$ is important.

For an interesting discussion of this matter together with two proofs of the existence of non-measurable sets as well as what happens in other dimensions see "The joys of Haar measure".

Answer 2

There is a lot of inaccuracy in what you are saying. As Ittay notes, you are talking about the Lebesgue measure only, otherwise, for example, pick any singleton (say $\{0\}$) and say that $A\subseteq\Bbb R$ has measure $1$ if and only if $0\in A$. You can check that this is a measure and that it is $\sigma$-additive, and that every set is measurable.

In fact, under additional set theoretic hypotheses you can show it might be possible that there is a measure extending the Lebesgue measure and measures all the subsets of $\Bbb R$. But it is important to note that this measure will not be translation invariant in general.

So why are there non-Lebesgue measurable sets? This is because there are a lot of sets of real numbers. And every Lebesgue measurable set is "essentially Borel" (in the sense that it is equal to a Borel set modulo a null set), but there are very few Borel sets.

Moreover, the axiom of choice allows us to carry all sort of crazy constructions. It will allow us to choose representatives from $\Bbb{R/Q}$ to construct a Vitali set, and it will allow us to create ultrafilters on $\Bbb N$ which can be translated (in a canonical way) to subsets of $\Bbb R$ which are not measurable in general; and it will allow us to do much much more.

Yes, we like to think about sets of real numbers in terms of open intervals, and their intersections, unions and complements. But that much won't get you very far in understanding all the subsets of $\Bbb R$. Even less so when assuming the axiom of choice.

So what happens without the axiom of choice? It can be the same situation if a "sufficient" fragment of the axiom of choice is assumed to hold (for example assume that $\Bbb R$ can be well-ordered, or that free ultrafilters on $\Bbb N$ exist, or so on). Or it can be that all those things fail, and in fact all sets are measurable.

But now the term "measurable" needs to be handled with care. It might be that there are no non-atomic $\sigma$-additive measures, or it might be that there is an extension of the Lebesgue measurable which is not invariant under translations, or it might be that really the Lebesgue measure is $\sigma$-additive and measures every set (which would require additional set theoretic assumptions to achieve). All those things are much more complicated to handle and understand than just understanding where non-measurable sets come from.

And the axiom of choice is far too useful for mathematicians to just drop it.