generalized Riemann Integrability of $f\cdot g$

Question

Let $\mathcal{R}([a.b])$ the set of all Riemann-integrable functions in $[a,b]$. Let $\mathcal{R}^{*}([a,b])$ the set of all Generalized Riemann-Integrable functions in $[a,b]$ (I'm talking about the Henstock-Kurzweil Integral).

We know that, if $f,g\in\mathcal{R}([a,b])$, so $f\cdot g\in\mathcal{R}^{*}([a,b])$. Can we say the same for Generalized Riemann- Integral?

Saying different, $f,g\in\mathcal{R}^{*}([a,b])\Rightarrow f\cdot g\in\mathcal{R}^{*}([a,b])$?

Answer 1

No, this is not true. For instance, let $f(x)=\frac{\sin (1/x)}{x}$ and let $g(x)=\operatorname{sgn}(f(x))$. Then $f$ and $g$ are both Henstock-Kurzweil integrable on $[0,1]$, but $f(x)g(x)=|f(x)|$ is not. The point is that the Henstock-Kurzweil integral allows for a kind of "conditional" (rather than absolute) convergence, which can turn into divergence when you multiply by a function that makes the product always have the same sign.

Answer 2

Another (recent, i.e., 2022) question pointed to this one. So even at this late date I should point out these facts in case someone else is directed here and wants the definitive answer, not just an isolated counterexample to a naive question.

1. The product $f(x)k(x)$ is Lebesgue integrable on an interval $[a,b]$ for every integrable function $f(x)$ if and only if the function $k(x)$ is essentially bounded

[First proved in H. Lebesgue, Ann. Fac. Sci. Univ. Toulouse (3) 1, 25–117 (1910), in particular, pp. 38–39].

2. The product $f(x)k(x)$ is Denjoy-Perron integrable on an interval $[a,b]$ for every Denjoy-Perron integrable function $f(x)$ if and only if the function $k(x$) is equivalent to a function of bounded variation.

[The "if" part is ancient history (I don't know the source). The "only if" is in Sargent, W. L. C. On the integrability of a product. J. London Math. Soc. 23 (1948), 28–34.]

The question is stated for "the generalized Riemann integral" or "the Henstock-Kurzweil integral." The Denjoy-Perron integral is equivalent to these and was first by about half a century. Let's not forget that.