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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.

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Nate Eldredge
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2 Answers2

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To complement your answer that there exist Riemann integrable functions in arbitrarily high Baire classes (incidentally, note that there are $2^c$ many Riemann integrable functions and only $c$ many Baire functions), there is a sense in which the set of Riemann integrable functions has a very, very tiny intersection with each of these Baire classes.

Among other results proved in the following paper, Marcus proves that for each countable ordinal $\alpha,$ the set of Riemann integrable functions in the normed space of bounded Baire-$\alpha$ functions from $[a,b]$ to $\mathbb R$ with the sup norm is a nowhere dense set in this space.

Solomon Marcus, Remarques sur les fonctions intégrables au sens de Riemann [Remarks on functions integrable in the sense of Riemann], Bulletin Mathématique de la Société des Sciences Mathématiques et Physiques de la République Populaire Roumaine (N.S.) 2(50) #4 (1958), 433-439. MR 22 #12180; Zbl 93.05901

I think there are several stronger and/or more general versions of this result, but I'm not in a position to look into it now.

Dave L. Renfro
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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.

Nate Eldredge
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