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.

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.
