In this answer (see example 1), Andrés Caicedo gives an example of a function $f : [0,1] \to \mathbb{R}$ which is Riemann integrable and is (strictly) of second Baire class.
Are there Riemann integrable functions of higher Baire class? Arbitrarily high? Are there Riemann integrable functions which are not in any Baire class?
To be precise, recall that for an ordinal $\alpha$, the Baire class $B_{\alpha}$ is defined inductively as follows. $B_0$ is the set of all continuous functions $f : [0,1] \to \mathbb{R}$. Then, once $B_\beta$ are constructed for all $\beta < \alpha$, we say $f \in B_\alpha$ if there exists a sequence $f_n \in \bigcup_{\beta < \alpha} B_\beta$ with $f_n \to f$ pointwise. Note that this stabilizes at $\alpha = \omega_1$.
If $\mathcal{R}$ is the set of all Riemann integrable functions, my question is: do we have $\mathcal{R} \subset B_\alpha$ for some $\alpha$? If so, what is the least such $\alpha$? If not, what is the least $\alpha$ such that $\mathcal{R} \cap B_{\omega_1} \subset B_\alpha$?
Of course, it may be helpful to recall that $f$ is Riemann integrable iff the set of discontinuities of $f$ has Lebesgue measure zero.
Actually I realized this is rather trivial. Let $C$ be the Cantor set, and let $A \subset C$ be any set which is not Borel. Then $f = 1_A$ is not Borel, hence not in any Baire class. But $f=0$ on the open set $C^c$, so $f$ is continuous at every point of $C^c$, which has full measure. Hence $f$ is Riemann integrable.
Since $C$ is an uncountable Polish space, we should also be able to do the same with sets $A$ that are of sufficiently high complexity that $1_A$ is in an arbitrarily high Baire class.