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!
 A: 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).
