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This question is based on this post, where in the comments, Toby Bartels conjectures that every Henstock-Kurzweil (gauge) integrable function $f\in\mathcal{HK}$ can be expressed as $f= g + h$ for a Lebesgue integrable function $g\in\mathcal{L}$ and improper Riemann integrable function $h\in\mathcal{R}^*$

Is this true? Intuitively to me this seems obviously false, since $\mathcal{HK}$ is strictly larger than $\mathcal{L}$ and $\mathcal{R}^*$, but I can’t think of any counterexamples or a way to prove a counterexample exists. If a counterexample exists, I know from properties of the HK integral that it must not have compact support, $|f|$ must not be HK integrable, and $f$ is not non-negative, as any of these conditions would imply that it would be Lebesgue integrable.

If this isn’t true, as I suspect, can we prove a more general statement, let’s say for $f=g\circ h$. How would one go about proving this true or false?

  • On what are you integrating? How do you define improper Riemann integral? – Jakobian Nov 12 '23 at 14:45
  • @Jakobian Sorry if this was not clear from my post. The HK integral and the Riemann integral can be generalized to any n-dimensional interval, so we work in $\mathbb{R}^n$. The use of "improper Riemann integral" is standard as a generalization of the regular Riemann integral (which supersedes the class of Lebesgue integrable functions). And then of course $\mathcal{L}$ refers to $L^1([a,b],\lambda)$. I'd recommend reading this article and the linked post for more information. –  Nov 12 '23 at 17:32
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    I'm asking about improper Riemann integrals only. Please give a proper definition, say on $[a, b]$ with $a < b$. – Jakobian Nov 12 '23 at 17:39
  • Are you unfamiliar with the improper Riemann integral? Read the Wikipedia article. With the Riemann integral, we cannot integrate over an unbounded interval or a function that has a vertical asymptote. Thus we define the improper Riemann integral as the corresponding limit of the bounds. For example, $\int_0^1 \frac{1}{\sqrt{x}} ,dx$ is not Riemann integrable, but it is improper Riemann integrable, and we define it $\int_0^1 \frac{1}{\sqrt{x}} ,dx = \lim_{a\to 0^+}\int_a^1 \frac{1}{\sqrt{x}} ,dx = \lim_{a\to 0^+}(2 - 2\sqrt{a}) = 2$ –  Nov 12 '23 at 17:53
  • Of course I'm familiar with the improper Riemann integral. But the definition I know is informal, as a limit at the singularity. I want you to give a proper formal definition so that we know what precisely do you mean. – Jakobian Nov 12 '23 at 17:57
  • The definition seems informal because the difference in definition between the regular and improper Riemann integrals is that in the later we give a precise value to the integral of a function that would be undefined. The article I linked gives a nice formal definition for the Riemann integral (as there are many equivalent definitions), and the improper Riemann integral is achieved by defining it in the "improper" cases (outlined in the Wikipedia article, which I again suggest you read). There isn't enough room in this comment to me to give each definition, but it is indeed well defined. –  Nov 12 '23 at 18:07
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    If I'm going to answer your question I need to know what you count as a counter-example. So please explain in the answer what would you consider as an improper Riemann integral on say, $[0, 1]$. – Jakobian Nov 12 '23 at 18:19
  • Here is an equivalent definition from this post. Let $f : (a,b] \rightarrow \mathbb{R}$ be a function such that $f|{[\tilde a,b]}$ is Riemann integrable on every closed interval $[\tilde a, b]$ with $a < \tilde a < b.$ $f$ is improperly Riemann integrable if, for any sequence $(a_n)_n \subseteq (a,b]$ that converges to $a$, the sequence $\Big( \int{a_n}^b f(x), \mathrm{d}x \Big)_{n \in \mathbb{N}}$ converges. In that case, the limit is written as just $\int_a^b f(x) , \mathrm{d}x$. –  Nov 12 '23 at 18:26
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    In the future refrain from asking multiple questions in one post, if you have to ask them, do it in separate answer. I've answered your first question but I won't be answering the second one - if you're curious about it, you have to ask another question. – Jakobian Nov 18 '23 at 03:21

1 Answers1

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Not every gauge integrable function is a sum of Lebesgue integrable and improper Riemann integrable function (for definition of improper Riemann integrable function on $[a, b]$, see below).

Comparison theorem. If $f, g$ are gauge integrable on $[a, b]$ and $|f|\leq g$, then $f$ is Lebesgue integrable.

Hake's theorem. If $f$ is gauge integrable on $[c, b]$ for all $c\in (a, b)$ and $\lim_{c\to a^+} \int_c^b f$ exists, then $f$ is gauge integrable on $[a, b]$ with $\int_a^b f = \lim_{c\to a^+} \int_c^b f$.

Let $f$ be defined as $f(2^{-n}(1+x)) = (-1)^{n+1}2^n/n$ for $x\in (0, 1]$ and $n = 1, 2, ...$ and $f(0) = 0$.

Define $g(2^{-n}(1+x)) = f(x)$ for $x\in (0, 1]$, $n = 1, 2, ...$ and $g(0) = 0$.

From Hake's theorem it can be seen that $f$ is gauge integrable and $\int_0^1 f = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \ln(2)$.

Similarly, Hake's theorem implies that $g$ is gauge integrable and $\int_0^1 g = \sum_{n=1}^\infty 2^{-n}\ln(2) = \ln(2)$.

Suppose that we could write $g = h+k$ where $h$ is Lebesgue integrable and $k$ is improper Riemann integrable. Here I call a function $k:[0, 1]\to\mathbb{R}$ improper Riemann integrable if there exists a partition $x_0 < ... < x_n$ of $[0, 1]$ such that $k$ is Riemann integrable on $[x, y]$ for $x_i < x < y < x_{i+1}$ and the limit $$\lim_{x\to x_i^+, y\to x_{i+1}^-} \int_x^y k$$ exists for all $i = 0, ..., n-1$. Now fix such partition, and let $x = 2^{-n}, y = 2^{-n+1}$ where $n$ is large enough so that $[x, y]\subseteq (x_0, x_1)$. Then $k$ is Riemann integrable on $[x, y]$. Since $|g|\leq |k|+|h|$ and $|k|+|h|$ is gauge integrable on $[x, y]$, it follows from comparison theorem that $g$ is Lebesgue integrable on $[x, y]$. This implies that $f$ is Lebesgue integrable. However, $\int_0^1 |f| \geq \sum_{n=1}^N \frac{1}{n}\to \infty$, so $f$ can't be Lebesgue integrable. The obtained contradiction shows that $g$ is not a sum of Lebesgue integrable and improper Riemann integrable function.

Jakobian
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  • Thank you! I had tried applying Hake’s theorem but didn’t really get anywhere, so I appreciate your answer. I’m confident the second question I asked is also false, but I also suspect it would be significantly harder to prove. –  Nov 18 '23 at 05:36