It can be shown that axioms of choice allows us to construct non-measurable sets but can I say the converse i.e. $ZF + non \; measure \; sets \; exist$ contains $ZFC$.
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1I don’t think so, although someone more qualified can answer. This is because I know you don’t quite need the full axiom of choice to get non-measurable sets, there are some slightly weaker conditions (almost choice) that give you non measurable sets. – C Bagshaw Feb 11 '22 at 11:04
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@Bagggggggs That's right. If you know it's consistent that ZF + not AC + there are non-measurable sets, then you know that ZF + there are non-measurable sets does not imply AC. – Zhen Lin Feb 11 '22 at 11:29