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Let $f, g: [a, \infty) \rightarrow \mathbb{R}$ be real functions s.t $f$ is continues at $[a, \infty)$ and $\lim_{x\to\infty}f(x) = 0$.
If $g$ is Riemann-integrable at $[a, N] \space\space, \forall N \in (a, \infty)$ and $G: [a, \infty) \rightarrow \mathbb{R}$ given by, $$ \begin{array}{c} G(x) = \int_{a}^{x} |g(\xi)| \space\space d{\xi} \end{array} $$ is bounded, then $$ \begin{array}{c} \int^{\infty}_{a} fg(\xi)\space\space d{\xi} \end{array} $$
converges absolutely.

Now if we let some $\alpha_1, \alpha_2 \in [a, \infty)$, then,

$\int_{\alpha_1}^{\alpha_2}|fg(x)|\space dx = \int_{\alpha_1}^{\alpha_2}|f(x)|\cdot|g(x)|\space dx$

I'm thinking that if I'd be able to equivalently express $\int_{\alpha_1}^{\alpha_2}|fg(x)|\space dx$ in terms of $G$ and $|f|$, it's done (by using the other assumptions and Cauchy's equivalence). The only idea I've came up with so far is using the MVT theorem for integrals, which I struggle to reason why I can do that.

I'd like to have some guidance. Thanks

X4J
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1 Answers1

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Hints:

(1) The function $f$ is bounded on $[a,\infty)$.

(2) The function $G$ is bounded by hypothesis and non-decreasing, so $\int_a^\infty |g(\xi)| \, d\xi = \lim_{x \to \infty}\int_a^x |g(\xi)| \, d\xi$ exists.

Now show that the Cauchy criterion for the improper integral of $fg$ is satisfied.

RRL
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  • So it is possible to generalize more by replacing continuity's assumption of $f$ with integrability at $[a, N]$ for each $N \in (a, \infty)$? – X4J Aug 26 '23 at 01:20
  • That is another question. But first we can relax the assumption that $f$ is everywhere continuous because $f(x) \to 0$. Given $\epsilon>0$, we have $|f(x)| < \sqrt{\epsilon}$ for all sufficiently large $x$ and we have $\int_{\alpha_1}^{\alpha_2}|g(x)|,dx < \sqrt{\epsilon}$ for all sufficiently large $\alpha_1,\alpha_2$ implying the CC is satisfied for the integral of $fg$. – RRL Aug 26 '23 at 01:28
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    Now are you asking if existence of $\int_a^\infty f(x) , dx$ and $\int_a^\infty |g(x)|, dx$ implies that $fg$ is integrable over $[a,\infty)$? Note that it is not necessary that $f(x) \to 0$ if the first integral exists. – RRL Aug 26 '23 at 01:31
  • Edit: sorry, I got confused with the counter-examples. I will try to think about it, thanks – X4J Aug 26 '23 at 01:37
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    For improper integrals on a finite interval, $f(x) = g(x) = 1/\sqrt{x}$ on $(0,1]$ is a counterexample for the existence of $\int_0^1 f(x)g(x)dx$. It is harder to produce a counterexample for an infinite interval. See this question with examples of a continuous function $f:[1,\infty)\to\Bbb R $ such that $f(x) >0 $, $\int_1^\infty f(x),dx $ converges and $\int_1^\infty f(x)^2,dx$ diverges. – RRL Aug 26 '23 at 02:29