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While reading Analysis II- Terence Tao, he writes that almost all set we encounter are measurable (naturally), and if one wants to create non-measurable set, one need to consult the axiom of choice. Immediately, I think that can we create non-measurable set without the axiom of choice?

The question might seem stupid, I ask just out of curiosity.

Ross Millikan
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Micheal Brain Hurts
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    Curiosity isn't stupid. There are models of set theory without an axiom of choice, where all sets are measurable. They are... interesting, but since the proof of many theorems uses the axiom of choice, they may be... difficult. :D –  Jan 07 '18 at 16:52
  • https://mathoverflow.net/questions/73902/axiom-of-choice-and-non-measurable-set – Matthew Towers Jan 07 '18 at 16:52
  • Thanks a lot I'm sorry I just realized the duplication. – Micheal Brain Hurts Jan 07 '18 at 17:01

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