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Is it possible to prove the non-completeness of the Borel-Lebesgue measure on $\mathbb{R}$ (restricted to the Borel $\sigma$-algebra) without the full axiom of choice, but still with Countable Choice ? It seems to be the case when I read Non-Borel sets without axiom of choiceNon-Borel sets without axiom of choice, but I was unable to really prove it.

Is it possible to prove the non-completeness of the Borel-Lebesgue measure on $\mathbb{R}$ (restricted to the Borel $\sigma$-algebra) without the full axiom of choice, but still with Countable Choice ? It seems to be the case when I read Non-Borel sets without axiom of choice, but I was unable to really prove it.

Is it possible to prove the non-completeness of the Borel-Lebesgue measure on $\mathbb{R}$ (restricted to the Borel $\sigma$-algebra) without the full axiom of choice, but still with Countable Choice ? It seems to be the case when I read Non-Borel sets without axiom of choice, but I was unable to really prove it.

just adding [lo.logic] tag
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edit approved Mar 25, 2016 at 5:59
Ed Dean
edit approved Mar 25, 2016 at 5:59
Ed Dean
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Michael
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Non-completeness of the Borel-Lebesgue measure and countable choice

Is it possible to prove the non-completeness of the Borel-Lebesgue measure on $\mathbb{R}$ (restricted to the Borel $\sigma$-algebra) without the full axiom of choice, but still with Countable Choice ? It seems to be the case when I read Non-Borel sets without axiom of choice, but I was unable to really prove it.