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

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

score 4 · Accepted Answer · answered Jun 21 '18 at 01:22

4

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.

answered Jun 21 '18 at 01:22

Eric Wofsey

342,377

Old answer, but I will comment anyway since I added an answer. In fact the function $f(x)=x^{-1}\sin x^{-1}$ has an improper Riemann integral on $[0,1]$. The Henstock-Kurzweil integral includes all improper Riemann integrals. So the example shows that the product of two functions, one improperly Riemann integrable and one Riemann integrable, need not be improperly Riemann integrable. In particular it is an easy example for this question about a much larger integral. – B. S. Thomson Apr 07 '22 at 17:31

score 0 · Answer 2 · answered Apr 07 '22 at 17:18

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.

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].

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.

generalized Riemann Integrability of $f\cdot g$

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