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It is wellknown, that using the axiom of choice the Borel $\sigma$-algebra has cardinality $2^\mathbb{N}$, whereas the Lebesgue $\sigma$-algebra has cardinality $2^\mathbb{R}$. It immediately follows, that there are (many) Lebesgue-measurable sets, which are not Borel-measurable.

Now I know, that the existence of avmeasure on $\mathcal{P}(\mathbb{R})$ is consistent with ZF. Though this really doesn't say anything about the Borel algebra in the absence of choice it lead me to the question if $Bor_\mathbb{R}=Leb_\mathbb{R}$, where the first is the smallest $\sigma$-algebra containing all intervals and the second are the measurable sets of the (outer) Lebesgue measure.

Are there any sources on this?

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    $\begingroup$ Without the Axiom of Choice, you can prove that the Cantor set is an uncountable Borel subset of $ [0,1] $ with Borel measure $ 0 $. Any subset of the Cantor set is thus Lebesgue-measurable, and we know that there are $ 2^{\mathbb{R}} $-many of these. $\endgroup$ – Berrick Caleb Fillmore Oct 9 '17 at 22:06
  • $\begingroup$ Yes, but can you prove without Choice, that the Borel-algebra has cardinality $2^\mathbb{N}$? The proof I know uses transfinite induction on the Borel hierarchy and I think the latter one need choice. $\endgroup$ – Takirion Oct 9 '17 at 22:24
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    $\begingroup$ mathoverflow.net/questions/32720/… looks to be somewhat relevant here... $\endgroup$ – Steven Stadnicki Oct 9 '17 at 22:31
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    $\begingroup$ @Takirion: There is no such measure. Any measure in such a model would have to be the trivial measure as a countable subset of $ \mathbb{R} $ has null measure (as can be seen from its outer measure). $\endgroup$ – Berrick Caleb Fillmore Oct 10 '17 at 10:07
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    $\begingroup$ @Takirion: The moment you assume $ \mathsf{AC}_{\omega} $, it is no longer true that $ \mathbb{R} $ is a countable union of countable sets (because $ \mathsf{ZF} + \mathsf{AC}_{\omega} $ says that a countable union of countable sets is countable, while $ \mathsf{ZF} $ says that $ \mathbb{R} $ is uncountable), so measure theory becomes non-trivial. $\endgroup$ – Berrick Caleb Fillmore Oct 10 '17 at 10:12

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