# Help with a proof that $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = 0$

The question is

Let f be a continuous Lebesgue integrable function on $[0,+∞)$, show $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = 0$.

My attempt:

Suppose $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) \neq 0$, then $\exists \epsilon>0 \forall x_0\in[1,+\infty)\exists x>x_0$ st. $|f(x)|>\epsilon$. Thus we can construct a sequence $x_0 < x_1 < x_2 < ...$ st. $f(x_k)>\epsilon$ for all $k=1,2,3...$.

Also by continuity, $\exists \delta_1,\delta_2,\delta_3,...$ st. that $|f(x)|>\frac{\epsilon}{2}$ for each $(x_k-\delta_k,x_k+\delta_k)$.

Thus $\int_1^{ + \infty } f > \sum\limits_k {\varepsilon {\delta _k}} = \varepsilon \sum\limits_k {{\delta _k}}$. However, it is possible for $\sum\limits_k {{\delta _k}}$ to converge, so I am not able to conclude $\int_1^{ + \infty } f = + \infty$.

Thank you!

• There are continuous positive integrable functions such that $\limsup\limits_{x\to\infty} f(x) = +\infty$. You need further assumptions to deduce $\lim\limits_{x\to\infty} f(x) = 0$. – Daniel Fischer Aug 14 '15 at 21:05
• Do you mean that $f$ is Lebesgue integrable? (Not Lebesgue measurable) – Michael Burr Aug 14 '15 at 21:08
• Yes. I am sorry. That's my mistake. Thank you! – Tony Aug 14 '15 at 21:09
• Something like uniformly continuous or bounded variation would be helpful in this problem, otherwise, you can use your construction to create a counterexample. – Michael Burr Aug 14 '15 at 21:16
• Being integrable does not help much – A.Γ. Aug 14 '15 at 21:20

$$f(x) = x(\cos^2 x)^{x^5}$$
is Lebesgue integrable on $$[0,\infty)$$ and $$f(2n\pi) =2n\pi, n = 1,2,\dots$$