# Is a function with a dense set of points of continuity of Baire Class 1?

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!

• It's automatically somewhat "bigger"; the set of continuity is always $G_\delta$, so if it is dense then it is comeager and in particular (by Baire category) not meager. – Nate Eldredge Sep 8 '17 at 19:39
• See this. It's just a stronger result. – MathematicsStudent1122 Sep 8 '17 at 20:19

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.)