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Working in $\mathsf{ZF}+\mathsf{DC}$ (that is, we are allowed to use Dependent Choice but not full choice), suppose that there exists a non-measurable subset of the unit interval $[0,1]$ (just non-measurable, so we're not allowed to assume that this is a Berstein set, or a Vitali set, or any other "concrete" kind of non-measurable set). How can one prove that there must exist a non-measurable subset $X\subseteq[0,1]$ such that $\mu_*(X)=0$ and $\mu^*(X)=1$?

For context: I am attempting to understand the proof that the Axiom of Determinacy implies every subset of $\mathbb R$ is Lebesgue-measurable. The strategy (as it appears in various sources) seems to be, given an $X\subseteq[0,1]$, to define a game such that a winning strategy for one player implies $\mu_*(X)>0$ while a winning strategy for the other player implies $\mu^*(X)<1$. So a proof like this necessarily uses the fact that the existence of some non-measurable set implies that there exists another with inner measure zero and full outer measure; all the proofs I've read simply mention this as if it was obvious, but I am finding it nontrivial to come up with a proof. Also, I am guessing that assuming Dependent Choice should not be so central to the argument, rather it's there mostly just to ensure that one can utilize the usual definitions of Lebesgue measure, measurable, etc. without much trouble.

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  • $\begingroup$ This may be even trickier than you thought, because the definition itself of Lebesgue measure needs to be thought out carefully in the absence of the axiom of choice. We discussed this in our article Kanovei, V.; Katz, M. "A positive function with vanishing Lebesgue integral in Zermelo-Fraenkel set theory." Real Analysis Exchange 42(2), 2017, 385-390. See arxiv.org/abs/1705.00493 and doi.org/10.14321/realanalexch.42.2.0385 and mathscinet.ams.org/mathscinet-getitem?mr=3721807 $\endgroup$
    – Mikhail Katz
    Commented Feb 29 at 12:08
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    $\begingroup$ Some of the arguments in this great answer by Elliot Glazer may or may not be relevant to this question. $\endgroup$
    – Emil Jeřábek
    Commented Feb 29 at 12:39
  • $\begingroup$ @MikhailKatz you're right, the question as I stated it is quite tricky. Given the context of why I'm asking it, though, one should really be allowed to assume at least Dependent Choice. I will edit the question accordingly... $\endgroup$
    – David Fernandez-Breton
    Commented Feb 29 at 13:30

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I'll start as in Monroe Eskew's answer: Assume $X$ isn't measurable and get Borel sets $A,B$, with $A\subseteq X\subseteq B$, such that the measure of $A$ (resp. $B$) is the inner (resp. outer) measure of $X$. So $B-A$ is a Borel set of positive measure. Consider $B-A$ as a measure space with Lebesgue measure, scaled so that its total measure is $1$. Then this $B-A$ has a measure-preserving Borel isomorphism to the standard space $[0,1]$ (with Lebesgue measure). The image of $X-A$ under this isomorphism has inner measure $0$ (because $A$ has been subtracted) and outer measure $1$ (because of the scaling of the measure on $B-A$).

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    $\begingroup$ Perhaps I should have asked explicitly whether there's a more "elementary" proof of this fact. Using the fact that every standard Borel space has a measure-preserving Borel isomorphism to $[0,1]$ seems akin to "using a bazooka to kill an ant". I had suspected that, this being in the one-dimensional space $[0,1]$ where the definition of Lebesgue measure has a clear intuitive underpinning in geometry, there should be some clever but elementary argument somewhere. Anyway, if after some time it seems no such argument is available, I'll accept this answer as the best one... $\endgroup$
    – David Fernandez-Breton
    Commented Feb 29 at 17:06
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    $\begingroup$ @DavidFernandez-Breton I agree that the isomorphism to thhe standard $[0,1]$ is like a bazooka and should rather be used to destroy a whole ant hill. The particular ant here should allow a simpler attack. I'd suggest defining a function $f:[0,1]\to[0,1]$ by setting $f(x)=\mu([0,x]\cap B-A)$, where $\mu$ is the scaled mesure as in my answer. The idea is that the restriction of $f$ to a slight modification of $B-A$ should give the desired isomorphism. The modification referfs to worries about the map being literally a bijection; $f$ itself should suffice for a.e. bisection. $\endgroup$
    – Andreas Blass
    Commented Feb 29 at 17:19
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    $\begingroup$ I hope that the modification in my previous comment isn't another bazooka. It might become simpler if one first observes that $A$ (resp. $B$) can be taken to be $F_\sigma$ (resp. $G_\delta$), so $B-A$ is $G_\delta$. Also, it might help to observe that the measure on $B-A$ (extended by $0$ to the rest of $[0,1]$) is absolutely continuous wwith respect to Lebesgue measure. $\endgroup$
    – Andreas Blass
    Commented Feb 29 at 17:25

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