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Suppose that $f(x) \geq 0$ for all $x\in \mathbb{R}$. The improper Riemann integral $\int_{-\infty}^{\infty}f(x)dx = C_1$, where $C_1$ is a finite real number. $|g(x)| < C_2$ for all $x\in\mathbb{R}$ where $C_2$ is a finite real number.

Under these conditions, is it true that $f(x)g(x)$ is improper Riemann integrable? That is, is $\int_{-\infty}^{\infty}f(x)g(x) dx$ finite?

I have seen this question, which deals with the product of improper Lebesgue integrable functions - specifically that this product is not necessarily Lebesgue integrable. This question is a case of the other question when one function is both Lebesgue integrable and bounded, which is not what I want. I have that $g(x)$ is not necessarily absolutely integrable and want to know whether the product $f(x)g(x)$ is integrable.

sharkbait
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1 Answers1

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It is true, assuming $fg$ is Riemann integrable on finite intervals.

WLOG consider integrals over $[0,\infty)$.

Since $f$ is improperly integrable and nonnegative, given any $\epsilon > 0$, we have for $b > a$ and $a$ sufficiently large,

$$\int_{a}^{b} f(x) \, dx = \left|\int_{a}^{b} f(x) \, dx\right| < \epsilon/C_2$$

Thus,

$$\left|\int_{a}^{b} f(x) g(x) \, dx\right| \leqslant \int_a^b f(x) |g(x)| \, dx \leqslant C_2 \int_{a}^{b} f(x) \, dx < \epsilon$$

We then have convergence of $\int fg$ by the Cauchy criterion.

It is not true if $f$ changes sign infinitely often. The counterexample here is $f(x) = \frac{\sin x}{x}$ and $g(x) = \sin x$.

RRL
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