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Let $f:[0,1]\to\mathbb R$ be an arbitrary function. It is well known that if $f$ is of Baire Class 1, then the points at which $f$ is continuous form a dense subset of $[0,1]$. My question now is, if the converse is true:

Let $f$ be continuous at each point of a dense subset of $[0,1]$. Is then $f$ necessarily of Baire Class 1?

I guess not, and I would like to see some references showing a counter example or, even better, some book/paper answering this and the following question:

If the answer to the above question is negative, what additional property on $f$ is needed to guarantee that $f$ is of Baire Class 1?

I suppose that the set of continuity in a certain sense has to be bigger than just dense.

Any help is highly appreciated. Thank you in advance!

sranthrop
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1 Answers1

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See the first example at Examples of Baire class 2 functions; you put a continuous "sawtooth" of height 1 in each of the intervals in the complement of the Cantor set $C$. Then it is continuous at every point of $C^c$ and discontinuous at every point of $C$. However, a Baire class 1 function restricted to any perfect set has a comeager continuity set, so this function is not Baire class 1. (It is Baire class 2.)

Also see Maximum Baire class of a Riemann integrable function for a function which is continuous at each point of a dense set, but is not even Borel: just consider $1_B$ where $B$ is any non-Borel subset of the Cantor set. (Thanks to MathematicsStudent1122 for suggesting this in a comment.)

For a result in the other direction, it seems to be true that if the continuity set is co-countable (i.e. the discontinuity set is countable) then the function is Baire class 1. This appears as Exercise 24.17 in Kechris, Classical Descriptive Set Theory, but I can't solve it off the top of my head. None of the other notions of "big set" seem to help; the examples above show that it is not sufficient for the continuity set to be comeager or full measure (or both).

Nate Eldredge
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