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The standard proof considers the equivalence relation $x\sim y\iff x-y\in\Bbb Q$, and from the quotient set $[0,1]/\sim$ we select one representative from each equivalence class, and place it in a set $A$, and then proceed. However, the selection process invokes the axiom of choice.

My resource for measure theory says that it has been proven that the axiom of choice is necessary to construct non-Lebesgue measurable sets; does this mean that if one works in $ZF$, without $C$, every set is comfortably Lebesgue-measurable? I am really asking if there is a proof that no geometric measure exists on all of $\mathcal{P}(\Bbb R^n)$, "geometric" meaning shifting and rotating sets doesn't affect their measure, without invoking the $AC$; if no such proof exists, doesn't this pose problems...?

I personally believe in the axiom of choice as a sensible axiom to take in the process of mathematics - but I know that it is a touch sensitive, and many authors try to exclude it from proofs - so in my learning of measure theory should I assume that all resources and theorems are talking in the context of $ZFC$?

FShrike
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    It means what it says... that there is no proof that a nonmeasurable set exists without using the axiom of choice. In other words, it is consistent, in the absence of the axiom of choice, that every set is measurable. Not sure what you mean by "comfortably measurable" but the result means assuming this won't generate any contradictions. (Side note: Really we can't do away with choice entirely since something like countable choice is needed for Lebesgue measurability to make sense at all. The real result is that the existence of nonmeasurable sets is not provable in ZF + dependent choice.) – spaceisdarkgreen Nov 07 '21 at 18:49
  • Why would the nonexistence of such a proof pose problems? – spaceisdarkgreen Nov 07 '21 at 18:50
  • @spaceisdarkgreen I’m no set theorist it just seems interesting and a bit weird. Nice to know that set theory assumes ZFC though – FShrike Nov 07 '21 at 18:51
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    It doesn't really make sense IMO to be concerned (on philosophical grounds) that a reference might not be assuming axiom of choice, since any proof in ZF is a proof in ZFC. But as a practical matter though this hypothetical reference might be harder to use since even a result that doesn't require choice might have a simpler proof with choice, or possibly there might be notable results that require choice that it omits. – spaceisdarkgreen Nov 07 '21 at 19:18
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    But yes, it's probably safe to assume the background theory is ZFC, though in basic measure theory of the reals one really only needs countable (or maybe dependent) choice, except for the result that nonmeasurable sets exist, which is really only a curiosity in this context since everything else is generally just results about the measurable sets, and it's immaterial whether those consist of all sets of reals or not. – spaceisdarkgreen Nov 07 '21 at 19:20

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