Lebesgue measure, Borel sets and Axiom of choice I cannot proceed my study on measure theory since it seems my measure theory is really unstable. I desperately need someone to briefly answer below 3 questions...
**For convenience, i will write Lebesgue Measurable set to mean the usual Lebesgue Measurable set, and write Codable-Lebesgue Measurable set to mean the Lebesgue Measurable set defined by Codable-Borel sets.
$1$. It is well-known that "Existence of Non-Lebesgue measurable set is unprovable in ZF". WHAT Lebesgue measurable set? 
My definition for Lebesgue measurable set is defined by using Riesz Representation Theorem (Whose existence is guranteed by Axiom of Countable Choice, hence it is undefinable without choice). However, I heard that the usual construction of Lebesgue measure is by using Caratheodory's existence theorem. In that way, can Lebesgue measurable set be definable without Axiom of Choice? That is, it really doesn't make sense to me, 'existence of non-Lebesgue Measurable set is unprovable in ZF' since Lebesgue measure cannot be defined without choice (in my way of construction of Lebesgue measure).
$2$. Can "Existence of a set that is Lebesgue measurable but non-Borel" be proved without choice? I saw the standard example (Luzin's example using continued fraction), but i'm not sure what Lebesgue measurable set stated there. Moreover, if it is not true in ZF, what about "Existence of a set that is Codable-Lebesgue Measurable but non-Borel"?
$3$. (This is closely related to 1) Why Lebesgue measure is unique? Assuming Axiom of Choice, it is a theorem, that "If $\mu_1$ and $\mu_2$ are translation-invariant measures on sigma algebras $\mathfrak{M}_1$ and $\mathfrak{M}_2$ repectively, containing all Borel sets, and $\mu_1(K), \mu_2(K) <\infty$ for every compact set $K$, then there exists a constant $c$ such that $\mu_1(E)=c\mu_2(E)$ for all the borel sets $E$". You can see that this theorem does not imply that $\mathfrak{M}_1=\mathfrak{M}_2$ in the hypothesis.
Thank you in advance.
 A: For the first question, note that if we assume Dependent Choice then all $\sigma$-additivity arguments can be carried out perfectly. In such setting using the Caratheodory theorem makes perfect sense, and indeed the Lebesgue measure is the completion of the Borel measure.
It was proved by Solovay that $\mathsf{ZF+DC}$ is consistent with "Every set is Lebesgue measurable", relative to an inaccessible cardinal. This shows that assuming large cardinals are not inconsistent, we cannot prove the existence of a non-measurable set (Vitali sets, Bernstein sets, ultrafilters on $\omega$, etc.) without appealing to more than $\mathsf{ZF+DC}$.
Under the assumption of $\mathsf{DC}$ it is almost immediate that there are only continuum Borel sets, but if we agree to remove this assumption then it is consistent that every set is Borel, where the Borel sets are the sets in the $\sigma$-algebra generated by open intervals with rational endpoints. For example in models where the real numbers are a countable union of countable sets; but not only in such models. Do note that in such bizarre models the Borel measure is no longer $\sigma$-additive.
For sets with Borel codes the proof follows as in the usual proof in $\mathsf{ZFC}$. There are only $2^{\aleph_0}$ possible codes, but there are $2^{2^{\aleph_0}}$ subsets of the Cantor set, all of which are codible-Lebesgue measurable.
Lastly, we have a very good definition for the Lebesgue measure, it is simply the completion of the Borel measure. The Borel measure itself is unique, it is the Haar measure of the additive group of the real numbers. The Lebesgue measure is simply the completion of the Borel measure which is also unique by definition. This is similar to the case where the rational numbers could have two non-isomorphic algebraic closures, but there is always a canonical closure. Even if you can find two ways to complete a measure, "the completion" would usually refer to the definable one (adding all subsets of null sets).
