An undetermined measurable set and Borel Determinacy

Question

The existence of a measurable set which is not determined can be proved using the Axiom of Choice (or even without it). Also, we know that $\{ \text{Borel sets} \} \subset \{ \text{measurable sets} \}$. My questions are:

$\bullet$ Does this mean that we may find a Borel set which is not determined as a consequence of the existence of undetermined measurable sets?

$\bullet$ If so, does not this contradict the Borel Determinacy?

$\bullet$ If the answers of the above questions are no, do we only mean an undetermined measurable set $\in \{ \text{measurable sets} \} \setminus \{ \text{Borel sets} \}$?

$\bullet$ If there is a Borel set which is not determined for any other reason different from the one in the first question, can any one provide me with a reason, an example or a reference?

PS: Any reference which discusses the above questions and includes an example would be appreciated.

Answer 1

You've pretty much answered your own question, so let me confirm: All Borel sets are determined (the proof is due to Martin ~1975), a measurable, non-determined set hence is not Borel.

As far as references go, what precisely do you want a reference to? The proof that Borel sets are determined is covered in Kechris: Descriptive Set Theory. I don't know, off the top of my head, a reference for a non-determined, measurable set.

Answer 2

Stefan Mesken's answer addresses the bulk of your question; let me try to tie up the remaining loose end.

As noted in the comments, your parenthetical claim at the beginning of your question

the existence of a measurable set which is not determined can be proved using the Axiom of Choice (or even without it)

is false: it is consistent with ZF (under mild consistency assumptions) that every set is determined, measurable or non.

I think the issue here stems from a misreading of Hamkins' concluding comment. Hamkins states:

"The conclusion, therefore, which does not use the axiom of choice, is that if there is a non-determined set, then there is a non-determined set with measure $0$. In particular, [if there is a non-determined set, then] there is a non-determined set that is measurable."

(Emphasis/insertion mine.) The crucial bit here is the hypothesis, "if there is a non-determined set:" Hamkins is not saying that ZF proves "there is a measurable non-determined set," he's only saying that ZF proves $$(*)\quad\mbox{"If there is a non-determined set, then there is a measurable non-determined set."}$$

Your final claim

we can construct an undetermined set of zero measure without the AC, and so we have a measurable set which is not determined without the AC

is incorrect, since you're omitting the crucial hypothesis from $(*)$.