$ \lim_{t\to\infty}\int_{0}^{t}|g(s)|ds=0?$, when $\lim_{t\to\infty} |g(t)|=0$

Question

I was proving something with about differential equations and I stumbled on a problem which was cited in my book as trivial, but I don´t see it clearly:

So we have some function $g$ with: $$\lim_{t\to\infty} |g(t)|=0$$

How to show that:

$$ \lim_{t\to\infty}\int_{0}^{t}|g(s)|ds=0?$$

* EDIT: * The question is related to the answer to the following post: Diff equations-Duhamels formula for $t\to\infty$ What was meant here by "Then it's easy to show that the r.h.s. converges to zero."

Answer 1

If $g$ (measurable) satisfies $\lim_{t\rightarrow\infty}\int^t_0|g(s)|\,ds=0$ then $\int^A_0|g(s)|\,ds\leq\lim_{t\rightarrow\infty}\int^t_0|g(s)|\,ds=0$ which means that $g=0$ a.s on any interval $[0,A]$ and so over on the whole real line. Bottom line, the only solution to your problem is $g\equiv0$ (a.s).

Answer 2

Having $ g\left(t\right)\underset{t\to +\infty}{\longrightarrow}0 $, it is possible for $ \int_{0}^{+\infty}{\left|g\left(t\right)\right|\mathrm{d}t} $ to converge to some value other than $ 0 $, which is the case for $ g:t\mapsto\mathrm{e}^{-t^{2}} $. It is also possible for $ \int_{0}^{+\infty}{\left|g\left(t\right)\right|\mathrm{d}t} $ to diverge, taking, for example, $ g:t\mapsto\frac{\sin{t}}{t} $.