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I'm trying to obtain an example of a Jordan measurable set in $\Bbb R$ that is not a Borel set. I already know the construction of a Lebesgue measurable set that is not Borel, but I can't see (despite all my efforts) how to use this to obtain the desired example (I can't imagine other way of constructing the example without using this set).

Previous search in MSE: In this question, the construction of such an example is provided. However, I'm not able to understand what the guy who answered means with sterograpic representation, among other things.

Any tip or hint would be extremely appreciated, I've been thinking on this for hours.

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Daniel
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  • I think he meant Stereographic projection (https://en.wikipedia.org/wiki/Stereographic_projection) – Renart Nov 03 '15 at 21:22
  • Hint: Every subset of the Cantor middle thirds set is Jordan measurable (why?), and there are ___ many subsets of the Cantor middle thirds set but only ___ many Borel sets. – Dave L. Renfro Nov 03 '15 at 21:25
  • @DaveL.Renfro Thanks for the hint Dave, I already know that proof. I forgot to add that I'm looking for a construction of such a set, if possible. Is it so difficult to construct? – Daniel Nov 04 '15 at 03:00
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    I don't know enough about the foundations of logic, set theory, recursion theory, etc. to give much of an answer about how constructive such an example could be, but I suspect that any "construction" would not be very explicit. I'll ask someone who knows a lot more about this aspect than I do. – Dave L. Renfro Nov 04 '15 at 15:05
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    Any "explicit" construction actually requires the use of at least a modicum of the axiom of choice. The silly reason is that it is consistent with set theory without choice that every set of reals is a countable union of countable sets, so every set of reals is Borel (but, of course, not every countable union of countable sets is countable). Once we assume the usual amount of choice to carry out classical analysis, these silly examples disappear. We can then produce explicit examples (in an admittedly also silly way): As pointed out by Dave, any subset of the Cantor set is Jordan measurable, – Andrés E. Caicedo Nov 04 '15 at 15:16
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    so we just need to "exhibit" a non-Borel subset of the Cantor set. The standard way of doing this is by describing a complete analytic or co-analytic set. Analytic sets are the projections of Borel sets, and I suspect that this may be what is behind the "stereographic" suggestion in the question linked to above (I haven't looked). A good reference with examples of these sets, and aimed at an audience of analysts (as opposed to logicians) is MR0655599 (83j:26001). van Rooij and Schikhof. A second course on real functions. CUP, 1982. – Andrés E. Caicedo Nov 04 '15 at 15:21
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    Another excellent reference is MR1321597 (96e:03057). Kechris, Alexander S. Classical descriptive set theory. Graduate Texts in Mathematics, 156. Springer-Verlag, New York, 1995. Neither of these two suggestions requires a background in set theory or logic. – Andrés E. Caicedo Nov 04 '15 at 15:23

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