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I know that it's established that non-measurable sets cannot be constructed without the axiom of choice but I was wondering if countable choice was enough? Since the ones that I have seen use uncountable choice.

My intuition tells me no. Since I tend to think of measurable sets as being closed under "countable operations". Although if we're being precise the operations in question are unions, intersections and complements; so i was wondering if we can use countable choice in some way to construct a non-measurable set.

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Countable choice is not enough. See Solovay example here https://en.m.wikipedia.org/wiki/Non-measurable_set#:~:text=In%20mathematics%2C%20a%20non-measurable,assigned%20a%20meaningful%20"volume".&text=Solovay%20constructed%20Solovay's%20model%2C%20which,of%20the%20reals%20are%20meas

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The axiom of countable choice is strictly weaker than the axiom of dependent choice, which is not strong enough to prove the existence of non-measurable sets. However, with dependent choice and $\aleph_1\le 2^{\aleph_0}$ you do get a non-measurable set. The axiom of choice for pairs is also sufficient.

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    $\begingroup$ Hi professor, to solve this my question I tried to use the finite induction but unfortunately I suspect either that it is not possible to apply it or that I did it wrongly so could I ask for your assistance please? $\endgroup$ Commented Aug 14, 2020 at 18:36
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    $\begingroup$ @AntonioMariaDiMauro: I’m sorry, but I’m afraid that I’m the wrong person to ask about diffeomorphisms: that’s an area that I’ve never studied at all. $\endgroup$ Commented Aug 14, 2020 at 18:52
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    $\begingroup$ Okay, don't worry. I'm sorry for the trouble. $\endgroup$ Commented Aug 14, 2020 at 18:55
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    $\begingroup$ @AntonioMariaDiMauro: No trouble! $\endgroup$ Commented Aug 14, 2020 at 18:59
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It is not possible to construct a non-measurable set using only countable choice. Solovay showed that (assuming the existence of an inaccessible cardinal) it is consistent to have $ZF+DC+$every set of reals is Lebesgue measurable. Here $DC$ is dependent choice, which is strictly stronger than countable choice.

It's worth mentioning that Shelah has shown that the inaccessible cardinal is essential for getting Solovay's result.

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