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?
Best Answer
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.