A(n improperly) HK-integrable function $f$ is Lebesgue integrable on a (not necessarily bounded) measurable subset of $\mathbb{R}$ if and only if $|f|$ is also (improperly) HK-integrable on that subset.
Found the above claim at the following sources: (1)(2)(3)(4).

This means that if $f$ is a(n improperly) HK-integrable function whichi is not Lebesgue integrable, then $|f|$ is not (improperly) HK-integrable. Call such functions "conditionally" HK-integrable.

Question: Intuitively speaking$^*$, why are the integrals of conditionally HK-integrable functions well defined? Why don't we have to define something like the Cauchy principal value?

$^*$ I.e. feel free to skip over as many details as you want. I don't understand the HK integral well.

As you may have guessed from my decision to call such functions "conditionally HK-integrable", my intuition is that they should be analogous to conditionally convergent series. Therefore, intuitively I would expect a problem resulting from $\infty - \infty$ not being possible to define well, leading to an analog of the Riemann rearrangement theorem being applicable, and thus to some convention like the Cauchy principal value being necessary to ever assign any value to the integral at all.

If $|f| = f^+ + f^-$ is not HK-integrable, but $f = f^+ - f^-$ is HK-integrable (i.e. $f$ is "conditionally" HK-integrable), then this seems to require that both $HK\int f^+ = + \infty = HK \int f^-$. Not understanding the definition of HK integral well, however, I am not sure if this argument works.

Motivation: If the argument does work, then since the only HK-integrable functions which aren't Lebesgue integrable are conditionally HK-integrable, then we would have another answer to this popular question (see also this question) (and also this question).

The answer to the following question is related to this question: Can someone give an example of a function that is not Henstock-Kurzweil/gauge integrable?

Seemingly also related: Can conditionally convergent series be interpreted as a "generalized Henstock-Kurzweil integral"?

  • 1
    $\begingroup$ Did you try to see what happens in the gauge level when you try to prove, for example, that $\sum_n (-1)^n/n$ is convergent in the language of Henstock integral? $\endgroup$
    – Hugo
    Commented Jul 15, 2018 at 1:24
  • 1
    $\begingroup$ Reading T.Kunkle Rearrangements of Conditionally Integrable Functions 1994 can be useful. $\endgroup$ Commented Jul 19, 2018 at 16:27

1 Answer 1


The value of an infinite series is unique (when it exists), despite the Riemann Rearrangement Theorem. If you want to get a different sum, then you have to rearrange the terms, giving you a different sequence of terms (albeit consisting of the same values). Similarly, the value of a(n improper) Henstock–Kurzweil integral is unique result (when it exists), even if you can get a different integral by rearranging the values, giving you a different function.

For a specific example (following the comment that @Hugo made), let $ a _ n $ be $ ( - 1 ) ^ { n \! } / n $ (for a positive integer $ n $), and let $ f ( x ) $ be $ a _ { \lfloor x \rfloor } $ (where $ \lfloor x \rfloor $, also written $ [ x ] $, is the floor of $ x $: round $ x $ down to an integer) for real $ x \geq 1 $. Then $ \int _ { x = 1 } ^ \infty f ( x ) \, \mathrm d x = \sum _ { n = 1 } ^ \infty a _ n = - \! \ln 2 $. Now let the sequence $ b $ be a rearrangement of $ a $ such that $ \sum _ { n = 1 } ^ \infty b _ n = 3 $ (for example). We can similarly get a rearrangement $ g $ of $ f $ defined by $ g ( x ) = b _ { \lfloor x \rfloor } $, so that $ \int _ { x = 1 } ^ \infty g ( x ) \, \mathrm d x = 3 $. But $ b $ is not the same sequence as $ a $, and $ g $ is not the same function as $ f $.

The difference between $ g $ and $ f $ is that certain intervals are swapped around. From a measure-theoretic perspective, this should make no difference, since it's the same values defined on sets with the same measure, which is why the HK integral doesn't work as well as the Lebesgue integral in measure theory. But it doesn't stop the HK integral from being well defined.


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