# generalized Riemann Integrability of $f\cdot g$

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

• What is the generalized Riemann integral? I do not recall encountering that concept, at least by that name. – Joonas Ilmavirta Jun 20 '18 at 20:24
• @JoonasIlmavirta I think it's mentioned in Bartle's texts. I don't think I've seen the terminology elsewhere. – Clarinetist Jun 20 '18 at 20:30
• @JoonasIlmavirta After some searching, I think it's more commonly known as the Henstock-Kurzweil integral. – Clarinetist Jun 20 '18 at 20:31
• Can you edit that into the question? It would be much easier to answer if there was a description of the less standard concept or a link to it. – Joonas Ilmavirta Jun 20 '18 at 20:34
• Sorry, I edited up. I'm talking about the Henztock Kurzweil Integral – Mateus Rocha Jun 20 '18 at 21:36

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.