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Can you think of a function that is neither improper Riemann nor Lebesgue integrable, but is Henstock-Kurzweil integrable?

I'd like to put a bounty on this question, but my reputation is not nearly enough yet. Translated to math, find $f$ such that

$$ f \notin \mathscr{L,R^*} $$ but $$ f\in \mathscr{HK} $$ where $\mathscr{HK}$ denotes the set of Henstock-Kurzweil integrable functions.

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I could be mistaken, but doesn't the first paragraph of the wiki article on the gauge integral give your desired function? – JSchlather Oct 24 '11 at 7:06
Good pickup! Is that integral improper Riemann integrable? I have modified the question. – Samuel Tan Oct 24 '11 at 7:17
'Translated to math' as opposed to what? – Alexei Averchenko Oct 25 '11 at 6:33
@AlexeiAverchenko opposed to fluffy English. I'm just having too much fun typesetting math in LaTeX, being new to this forum. I'll get over it in a few weeks. – Samuel Tan Oct 25 '11 at 6:42
up vote 7 down vote accepted

This is a blatant cheat, but anyway, here goes:

Take $f(x) = \frac{\sin{x}}{x}$, which is well-known to be improperly Riemann integrable, but not Lebesgue integrable.

Take the characteristic function $g$ of $[0,1] \cap \mathbb{Q}$ which is Lebesgue integrable but not improperly Riemann integrable.

The KH-integral integrates both, hence it integrates $h(x) = f(x) + g(x)$.

Clearly, $h(x)$ cannot be either, improperly Riemann integrable or Lebesgue integrable, because this would force $g$ or $f$ to have a property it doesn't have.

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I'm not sure what $\chi_{[0,1]\cap \mathbb{Q}}$ is. Some kind of characteristic function? But yes, that is an answer. – Samuel Tan Oct 25 '11 at 5:11
@Samuel: Yes, that's the intention. – t.b. Oct 25 '11 at 5:14
In the spirit of t.b.'s answer, here is another example: take $D$ the Dirichlet function (not improper Riemann integrable) and add it to the Dirichlet integral, $$\int_0^{\infty} D(x) + \frac{sin(x)}{x} dx$$ This belongs to $\mathscr{HK}$ though – Samuel Tan Oct 25 '11 at 5:16

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