# Equivalence of Axiom of choice and all subsets of $\mathbb{R}$ are Lebesgue Measurable?

In my Measure theory course, we proved Vitali's Theorem which stated that there exist a subset of $\mathbb{R}$ which is not Lebesgue Measurable. We assumed axiom of choice to show that there exist a set which will not be lebesgue measurable by contradiction.

then my professor made the statement that Axiom of choice is equivalent to saying that all subsets of $\mathbb{R}$ are Lebesgue Measurable.

My confusion is Axiom of choice helped us to find a non-measurable set therefore how it is equivalent to saying all subsets are measurable.

My thinking of equivalence here is if we assume one we should be able to prove the other.

EDIT:

Okay so I discussed this with my professor again based on the answers. He said that By equivalent he did not mean Mathematical Equivalence but equivalent in a sense of axioms

More precisely, If we look at the contrapositive of the above theorem it says that, All subsets of $\mathbb{R}$ are lesbegue measurable then axiom of choice is not true. He meant that One can take "all subsets of $\mathbb{R}$ are Lebesgue Measurable" as an Axiom and it will be independent of Axiom of Choice and the existing ZF.

This statement is very wrong (regardless to whether you misunderstood your professor, or if your professor misspoke, or if he or she actually believe this claim). Not only the axiom of choice implies there are non-measurable sets, the axiom of choice does not follow from the statement "there is a non-measurable set".

The axiom of choice can fail in many ways, while the real numbers can still be well-orderable, which will imply that there are Vitali sets and other non-measurable sets.

The axiom of choice is a global statement about the entire universe of sets. The Lebesgue measurable sets live somewhere at the bottom of this universe, where there are only real numbers, sets of real numbers and sets of sets of real numbers. But the grim reality is, those are very very small sets.

But what is true is that it is consistent that the axiom of choice fails, and all the sets are Lebesgue measurable. So the most you can say is that indeed without assuming some fragment of the axiom of choice, we cannot prove that there are non-measurable sets.

• "Your professor is very wrong", or maybe, just maybe, the student is very mistaken as to what the professor said. Oct 17, 2016 at 8:52
• Your comment is very correct. I should be more careful. Oct 17, 2016 at 8:54
• I have edited the question based on responses. your views will be valuable if you could add something. Oct 20, 2016 at 7:34
• @Abhinav: Admittedly, I'm not sure what your teacher had meant. Yes, as I wrote, the axiom of choice implies the existence of non-measurable sets, and therefore the axiom "all sets are Lebesgue measurable" implies the negation of the axiom of choice. But they are not even close to being equivalent in any meaningful sense. Not philosophically, and certainly not mathematically. Also, it should be an equivalence over ZF, rather than ZFC, since ZFC includes the axiom of choice, and therefore proves the existence of non-measurable sets. Oct 20, 2016 at 7:38
• No, ZFC is exactly ZF+AC. Oct 20, 2016 at 7:43

The statement is, indeed, false. In ZFC -- the most common set theory in which most mathematicians work* -- which in particular includes the axiom of choice, you can prove that there exists a non-Lebesgue measurable set.

Perhaps your professor misspoke (or you misheard), and they intended to say the opposite: the existence of a non-measurable set is equivalent to the axiom of choice. However, that is also very much false; the existence of a non-measurable set is much weaker than the full axiom of choice.

* In so far as most mathematicians can be said to "work in" a foundational theory.

I can't write this as a comment above because I'm new. Like the others said, the Axiom of Choice (AC) does not implies all sub sets of reals are Lebesgue measurable. There is a way to do that though.

Using the Axiom of Determinacy (AD), one can proof with infinite (topological) games that every subset of real numbers is Lebesgue measurable in particular Vitali set and Bernstein set (rare sets) are measurable. The thing is AD implies a weaker form of AC that is the Axiom of Countable Choice. But many people prefer AC because AD is sort of too good to be true.

If anyone wants to read more about infinite games you should see Thomas Jech.

• "The thing is AD implies a weaker form of AC that is the Axiom of Countable Choice. Therefore, AD is stronger that AC." Just because AD implies a fragment of choice (not even all of countable choice, incidentally!) does not mean that AD is stronger than AC. AC and AD are incomparable: each implies the negation of the other. May 23, 2022 at 17:23
• @CesarIvanGuillenAlmanza: Maybe you should edit your answer and remove that sentence. May 23, 2022 at 17:56
• Done. Thanks for the comments guys. May 23, 2022 at 18:23