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I was wondering about how Fourier series behaves in the setting of Henstock-Kurzweil integration. For example, the non-Lebesgue-integrable function $f(x) = \dfrac{1}{x}\mathrm{e}^{\mathrm{i}/x}$ can have a Fourier expansion since the Henstock-Kurzweil integrals $$(HK)\int^{\delta}_{0} \dfrac{1}{x}\mathrm{e}^{\mathrm{i}/x} {\rm d}x = (HK)\mathrm{i}\int^{\delta}_{0} ((x\mathrm{e}^{\mathrm{i}/x})' - \mathrm{e}^{\mathrm{i}/x}) {\rm d}x = \mathrm{i}\delta \mathrm{e}^{\mathrm{i}/\delta} - \mathrm{i}\int^{\delta}_{0} \mathrm{e}^{\mathrm{i}/x}{\rm d}x,$$ $$(HK) \int^{\delta}_{0} \dfrac{1}{x}\mathrm{e}^{\mathrm{i}/x} \mathrm{e}^{\mathrm{i}nx} {\rm d}x = (HK)\int^{\delta}_{0} \dfrac{1}{x}\mathrm{e}^{\mathrm{i}/x} {\rm d}x + \int^{\delta}_{0} \dfrac{\mathrm{e}^{\mathrm{i}nx}-1}{x}\mathrm{e}^{\mathrm{i}/x} {\rm d}x$$ are well defined.

So I'm interested in the question stated in the title. But in general, the product of two Henstock-Kurzweil integrable function need not be Henstock-Kurzweil integrable on $[a,b]$ (here is an example). Even $g(x)$ being smooth on $[a,b]$ would not guarantee that $f(x)g(x)$ is Henstock-Kurzweil integrable on $[a,b]$. So what can be said if $f$ is integrable and $g(x) = \mathrm{e}^{\mathrm{i}x}$?

Jianing Song
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First a short history lesson (the price one has to pay for an answer).

Some history. At the beginning of the 20th century Lebesgue defined an integral which quite revolutionized the theory. He had stated that his motivation was to produce an integral that included the Riemann integral and integrated all derivatives. Well, it did integrate some derivatives but not all derivatives.

A French mathematician Arnauld Denjoy (1884--1974) took on that program and gave a detailed analysis of how to integrate all derivatives, by extending Lebesgue's integral in a countable (but transfinite) sequence of steps. That was called, naturally, the Denjoy integral. Later a German named Perron gave a nonconstructive definition for the Denjoy integral. Denjoy severely (and dramatically) attacked his definition. In spite of that the world elected to call this now the Denjoy-Perron integral. Over the years there were many equivalent formulations but the name Denjoy-Perron persisted. A soviet mathematician, Tolstov, gave a very interesting characterization too in 1960. Nobody elected to change the name to the Denjoy-Perron-Tolstov integral.

Then, sometime in the 1970s, researchers became aware of another equivalent definition that had been presented in the 1950s by an Irish mathematician, Ralph Henstock and a Czech mathematician, Jaroslav Kurzweil. That version in time became known as the Henstock-Kurzweil integral and Denjoy and Perron somehow lost their patent rights to the name.

Many students now learn of this integral only by this new name and have no idea of the many contributions of mathematicians in the period 1910-1960. Many have never read Saks, Theory of the Integral, where that theory is very elegantly presented.

Basta. Thank you for your patience.

Your answer:

Sargent, W. L. C. On the integrability of a product. J. London Math. Soc. 23 (1948), 28–34.

From the Math Review by the Canadian mathematician Jeffery. "It is known that the product $f(x)k(x)$ is summable for every summable function $f(x)$ if and only if the function $k(x)$ is essentially bounded [H. Lebesgue, Ann. Fac. Sci. Univ. Toulouse (3) 1, 25–117 (1910), in particular, pp. 38–39]. It is also known that if $ f(x)$ is integrable in the Denjoy-Perron sense then it is sufficient for the integrability of $f(x)k(x)$ in this sense that $k(x)$ be equivalent to a function of bounded variation. This is a part of the general theory of Denjoy-Perron integrals. The author completes the latter result by proving the following theorem. If $k(x)$ is not equivalent to a function of bounded variation in $(a,b)$, there is a function $f(x)$ integrable in the Cauchy sense in $(a,b)$ and such that $f(x)k(x)$ is not integrable in the Cesàro-Perron sense of any order, nor in the general Denjoy sense in $(a,b)$. The Cesàro-Perron integral of order zero is the Denjoy-Perron integral."

Note: Winifred Lydia Caunden Sargent (8 May 1905 – October 1979) was an English mathematician. She studied at Newnham College, Cambridge and carried out research into Lebesgue integration, fractional integration and differentiation. For a full biography see: https://mathshistory.st-andrews.ac.uk/Biographies/Sargent/

POSTSCRIPT. For infomation on Fourier series using the Denjoy integral there is this Russian textbook from 1978 that has been ably translated by Peter Bullen who is himself an authority on this topic.

Čelidze, V. G.; Džvaršeĭšvili, A. G. The theory of the Denjoy integral and some applications. Translated from the Russian, with a preface and an appendix by P. S. Bullen. Series in Real Analysis, 3. World Scientific Publishing Co., Inc., Teaneck, NJ, 1989. xiv+322 pp. ISBN: 981-02-0021-8

B. S. Thomson
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  • Very detailed answer, thanks! What does it mean by "equivalent to a function of bounded variation"? – Jianing Song Apr 09 '22 at 01:11
  • @JianingSong "Equivalence" here is defined in terms of Lebesgue measure. A function $f$ is equivalent to a function of bounded variation $g$ if $f=g$ almost everywhere. The reason why this is needed is that integrability in the Lebesgue sense or the Denjoy-Perron sense ignores what is happening on a set of measure zero. – B. S. Thomson Apr 09 '22 at 02:35
  • I see. Thank you very much. – Jianing Song Apr 09 '22 at 03:55