Existence of non-Riemann-integrable bounded Baire class 1 function

Asked Jan 17 '25 at 22:04

Active Jan 18 '25 at 08:00

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Given a sequence $(f_n) \in C^0([a,b])^{\mathbb{N}}$ of continuous functions such that $f_n \to f$ (pointwise) for some bounded function $f$.

Can we conclude that $f$ a Riemann-integrable function?

I thought of this question a while ago, but could neither answer it myself nor find a solution online.

There are some results on the Baire class 1. So maybe this question is a weak version of a known classification.

I expect this question to be already known/answered in literature (given that it is so simple to state).

edited Jan 18 '25 at 08:00

asked Jan 17 '25 at 22:04

M.E.W.

1

In 1881 Volterra constructed a bounded derivative on $[0,1]$ which is not Riemann integrable, see Goffman, Casper: A bounded derivative which is not Riemann integrable. Amer. Math. Monthly 84 (1977), no. 3, 205--206. A derivative is a Baire class $1$ function; take $f_n(x)=n(f(x+1/n)-f(x))$ with $f$ suitable extended to $[0,1+\varepsilon]$ such $f(x+1/n)$ is always defined. Then $f_n \to f'$ pointwise on $[0,1]$. – Gerd Jan 18 '25 at 09:19
Thank you so much. – M.E.W. Jan 18 '25 at 11:26
In addition to the comment by @Gerd, see Maximum Baire class of a Riemann integrable function. – Dave L. Renfro Jan 18 '25 at 12:05

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