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.