I am looking for an example of a function that is not Henstock-Kurzweil integrable. Can anybody help me?

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    $\begingroup$ I do not know the answer, but I feel like a function that has something to do with a non Lebesgue measurable set would do. Since Lebesgue integral is "absolute convergent" and gauge integral is "conditional convergent"; the power of gauge integral is to integrate highly oscillating functions. $\endgroup$ – Xiao Jun 5 '14 at 9:02
  • $\begingroup$ How about $1/x$, say on $(-1,1)$ ? $\endgroup$ – user8268 Jun 5 '14 at 9:33
  • $\begingroup$ $1/x$ is Riemann integrable, so it is also Henstock-Kurzweil integrable, right? $\endgroup$ – Elke Jun 5 '14 at 11:12
  • $\begingroup$ No, a function must be bounded to be Riemann integrable, but I believe your second statement is true. $\endgroup$ – Ted Shifrin Jun 5 '14 at 11:38
  • $\begingroup$ Of course, it would be easy to come up with an unbounded functions that the integral sum is positive infinity. The goal would be finding a bounded function on $[a,b]$ that is not Henstock-Kurzweil integrable. $\endgroup$ – Xiao Jun 6 '14 at 11:21

Take any infinite series $a$, and let $f(x)$ be $a_{\lfloor{x}\rfloor}$ (where $\lfloor{x}\rfloor$ is the greatest integer that is not greater than $x$, that is $x$ rounded down to an integer). In general, $$ \int_i^j f(x) \,\mathrm{d}x = \sum_{n = i}^{j-1} a_n ,$$ whenever $i$ and $j$ are integers. As stated, this is a fine Riemann integral, but change $j$ to $\infty$ and we have:

  • $ \int_i^\infty f(x) \,\mathrm{d}x = \sum_{n = i}^\infty a_n $ as a Lebesgue integral iff $a$ is absolutely convergent;
  • $ \int_i^\infty f(x) \,\mathrm{d}x = \sum_{n = i}^\infty a_n $ as an improper Riemann integral iff $a$ is convergent (possibly conditionally);
  • $ \int_i^\infty f(x) \,\mathrm{d}x = \sum_{n = i}^\infty a_n $ also as a Henstock–Kurzweil integral iff $a$ is convergent (possibly conditionally).

So any divergent series gives a Henstock–Kurzweil nonintegrable function. The harmonic series, for example, gives $f(x) = 1/\lfloor{x}\rfloor$, which is not integrable.

OK, so just plain $1/x$ is an even simpler nonintegrable function, but I think that if you keep in mind the relationship between Lebesgue vs Henstock and absolute vs conditional convergence, then it will seem more obvious to you that, while Henstock–Kurzweil makes more functions integrable, it does not make all functions integrable.

  • $\begingroup$ For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question. $\endgroup$ – Toby Bartels May 2 '17 at 22:04

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