Existence of non measurable set and ZF theory?

Question

Does statement: 'Existence of non measurable set' consistent with ZF theory. or if I throw Axiom of choice from ZFC theory. Can I prove or disprove existence of a countably additive measure function on all subsets of R which is translation invariant and in which measure of interval is equal to length of interval.

Answer 1

No. It most certainly does not.

This would mean that the fact there are non-measurable sets is equivalent to the axiom of choice. This cannot possibly be true. Measures live on $\Bbb R$. That's an extremely localized set. The axiom of choice affects the entire universe of sets.

Not to mention that the existence of non-measurable sets can be proved (independently) from $\sf DC+\aleph_1\leq2^{\aleph_0}$, $\sf BPI$, The axiom of choice for family of pairs, Banach-Tarski paradox, existence of free ultrafilters on $\omega$, Hahn-Banach theorem, and more and more. You can find some nice diagrams in Herrlich's book The Axiom of Choice.

Beyond that lies another point. While Solovay proved that it is consistent that every set is Lebesgue measurable, this proof requires the existence of an inaccessible cardinal, and as Shelah showed later on, this requirement is really needed (we can remove it, but then we have to give up sigma additivity).

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1. Robert M. Solovay, A model of set-theory in which every set of reals is Lebesgue measurable, Ann. of Math. (2) 92 (1970), 1--56.

2. Saharon Shelah, Can you take Solovay’s inaccessible away?, Israel J. Math. 48 (1984), no. 1, 1--47.

3. Horst Herrlich, The Axiom of Choice, Springer Lecture Notes in Mathematics 1876 (2006).