I know there exists a non-measurable subset in [0,1). Call it P
I was thinking in the intersection maybe? I don't know how to proceed.
As far as I know, this is quite a tricky result. There is a proof of it in Halmos's Measure Theory (Theorem E, p.70).
Given a set M of positive measure, it's a very natural idea to think that would provide an answer to the problem, but unfortunately that won't work. The difficulty is that a general non-measurable set P might be concentrated in one half of the interval, and M in the other half. The standard construction of a non-measurable set P does appear to give a set that is somehow spread right through the interval, but Halmos's proof requires a strengthening of that construction.
Actually, what Halmos proves is that if is a "finite, sigma additive, translation invariant, non-trivial" measure, like Lebesque measure, then there exist a non-measurable subset of A. In fact, he exhibits that set- which can be taken to be the same for all such measures.
The proof in Halmos involves saying that two number in [0,1] are "equivalent" if and only if x- y is rational, forming the "equivalence classes" and forming the "non-measurable set", A, by taking one member of each equivalence class to be in A.
You then define "translation by p modulo 1" as "x+ p if x+p< 1, x+p-1 if . Then define "translation of A by p modulo 1" as
If I remember correctly, Halmos shows that, for any "finite, sigma additive, translation invariant measure" (such as Lebesque measure), the measure of "translation of A by p modulo 1" is equal to the "translation of A by q modulo 1" where p and q are any two rational numbers.
He also shows that those two sets are disjoint, for any two rational numbers, and that every number in [0, 1] is in one of them. That is, [0, 1] is the countable union of such sets where the union is over all rational numbers.
Since the measure is sigma additive, the measure of countable sum of same number. That will be finite only if the common measure of all those sets is 0 in which case [0, 1) (and every subset of [0, 1)) has measure 0. And that is only true for the trivial measure for every measureable A.
That proof is a little to abstract for me and uses a concept (inner measure) that my prof explicitly rejected.
However, I believe I have a possible solution, but first I need to show that a measurable subset of P must have measure 0. I have posted this in a new topic.