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!
This doesn't really help in the understanding but I love finding explicit examples. The function
$$f(x) = x(\cos^2 x)^{x^5}$$
is Lebesgue integrable on $[0,\infty)$ and $f(2n\pi) =2n\pi, n = 1,2,\dots $