1

I have several questions here.
First, does the fact that $f$ is integral means the integral is finite? I was wondering if I can prove that $L(f,P)$ goes to infinity for some partition $P$ if $\lim\limits_{x\to\infty}f(x)\neq0$.
Secondly, I think I might need to use the definition of improper integrals here, but I'm not sure how the definition helps.

WinnieXi
  • 91

1 Answers1

1

Hint

First: yes, saying that $f$ is integrable means that its improper Riemann integral over $[0, \infty)$ is finite.

Then, use a proof by contradiction and Cauchy criterion for improper integral convergence.

If $f$ doesn’t converge to zero, it exists $a\gt 0$ and a sequence $\{x_n\}$ converging to $\infty$ such that $f(x_n) \ge a$ and $x_{n+1}-x_n \ge 2$ for all $n \in \mathbb N$.

If $\int_0^\infty f(x) \ dx$ exists, then (using Cauchy criterion for improper integral convergence) $$\lim\limits_{n \to \infty} \int_{x_n-1}^{x_n+1} f(x) \ dx =0.$$

Derive a contradiction with the fact that $f$ is supposed to be uniformly continuous.

Angelo
  • 12,328
mathcounterexamples.net
  • 70,018
  • Is the hypothesis of nonnegativeness necessary ? Is it possible to prove the theorem without that hypothesis ? – Angelo Jun 06 '21 at 11:22
  • @Angelo You're right. The nonnegativeness hypothesis can be dropped. – mathcounterexamples.net Jun 06 '21 at 12:09
  • Could you modify your proof in order that it works even without the nonnegativeness hypothesis ? – Angelo Jun 06 '21 at 13:54
  • @Angelo It is immediate. As if $f$ doesn't converge to zero, either $f$ or $-f$ satisfies all the steps of the proof. – mathcounterexamples.net Jun 06 '21 at 16:06
  • I'm not sure how the why we need $x_{n+1}-x_n \geq 2$ here. My takeaway from your hint is that if $\lim\limits_{x\to\infty}f(x)\neq 0$, then there exists some $x_n\to \infty$ with $x_n\geq a$. Since f is uniformly continuous, for $a/2$, there is some $\delta$ such that $f(x)\geq a/2$ for all $x$ with $x-x_n < \delta$. Thus we have $\int_{x_n+\delta}^{x_n-\delta} f(x) \ dx \geq a/\delta$, which is a contradiction of the Cauchy criterion. But I'm not sure if my understanding is right. – WinnieXi Jun 07 '21 at 12:58