1. If $f$ is Henstock-Kurzweil integrable $\Longrightarrow$ $f$ is measurable.
  2. $f$ is Lebesgue integrable $\Longleftrightarrow$ $|f|$ is Henstock-Kurzweil integrable.
  3. $|f|$ is Henstock-Kurzweil integrable $\Longrightarrow$ $f$ is Henstock-Kurzweil integrable.

Can we use the above three relations to show that $f$ is Lebesgue integrable $\Longrightarrow$ $f$ is Henstock-Kurzweil integrable, that is, can we drop the condition $f\ge 0$ as shown here ?

In a nutshell does there exits functions that are not Henstock-Kurzweil integrable but are Lebesgue integrable?

  • $\begingroup$ In your points 2 and 3, are you assuming that $f$ is measurable? They are false otherwise. $\endgroup$ – Arnaud D. Mar 1 '18 at 9:49
  • $\begingroup$ @ArnaudD. What do you mean? If $f$ is Lebesgue integrable then $f$ is obviously measurable and if $f$ is HK integrable, then it is measurable from 1. $\endgroup$ – yasir Mar 1 '18 at 9:54
  • $\begingroup$ But $|f|$ can be HK integrable without $f$ being so (if you accept the axiom of choice). $\endgroup$ – Arnaud D. Mar 1 '18 at 10:00
  • $\begingroup$ @ArnaudD.So, 3 does not hold? I know integrability of $f$ $\not \Rightarrow$ integrability of $|f|$, but the converse holds , right ? $\endgroup$ – yasir Mar 1 '18 at 10:04
  • 1
    $\begingroup$ If 3 was true, combining it with 1 would contradict my previous comment. For an explicit counterexample : consider a non-measurable subset $A$ of $[0,1]$ and the function that maps $x$ to $1$ if $x\in A$ and $-1$ otherwise. $\endgroup$ – Arnaud D. Mar 1 '18 at 10:34

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