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  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?

Martin Sleziak
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miyagi_do
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  • In your points 2 and 3, are you assuming that $f$ is measurable? They are false otherwise. – Arnaud D. Mar 01 '18 at 09:49
  • @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. – miyagi_do Mar 01 '18 at 09:54
  • But $|f|$ can be HK integrable without $f$ being so (if you accept the axiom of choice). – Arnaud D. Mar 01 '18 at 10:00
  • @ArnaudD.So, 3 does not hold? I know integrability of $f$ $\not \Rightarrow$ integrability of $|f|$, but the converse holds , right ? – miyagi_do Mar 01 '18 at 10:04
  • No, it doesn't. $|f|$ can be integrable even without $f$ being measurable. – Arnaud D. Mar 01 '18 at 10:16
  • @ArnaudD.No I am not asking about that. I am saying 3 holds without any mention of $f$ being measurable or not. If $|f|$ is HK integrable then $f$ is HK integrable ?. – miyagi_do Mar 01 '18 at 10:25
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    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. – Arnaud D. Mar 01 '18 at 10:34

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