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Is every Baire one function continuous a.e.? I guess it has a positive answer. Because

  1. $f:\mathbb{R} \to \mathbb{R}$ is a Baire one function iff $f$ continuous everywhere except for meagre set.

  2. A subset $A$ of $X$ is meagre that it is negligible. So, we can assume that $A$ has a measure zero (I am not sure in this part).

Therefore, $f:\mathbb{R} \to \mathbb{R}$ is a Baire one function iff the discontinuity set of $f$ has measure zero.

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    $\begingroup$ Not every meagre set has measure zero, and you can't just "assume" it is. $\endgroup$ – Wojowu Mar 30 '17 at 7:20
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    $\begingroup$ math.stackexchange.com/questions/102482/… $\endgroup$ – Jonas Meyer Mar 30 '17 at 7:27
  • $\begingroup$ @JonasMeyer. Thanks for the link and the answer $\endgroup$ – flourence Mar 30 '17 at 7:46
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    $\begingroup$ Note that @Jonas Meyer shows that a Baire one function on $\mathbb R$ can be discontinuous a.e.! $\endgroup$ – Dave L. Renfro Mar 30 '17 at 14:13
  • $\begingroup$ Some types of meager closed sets of positive measure are called "fat Cantor sets". $\endgroup$ – DanielWainfleet Mar 30 '17 at 21:48

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