Question: let $a\in \mathbb R$ and $f: [a,\infty)\to \mathbb R$ in every interval $[a,b]$ with $a<b$ Riemann-Integrable and let $\lim_{x\to\infty} f(x) = 0$. Show that $\int_a^\infty f(x)dx$ exists only if $\lim_{n\to\infty} \int_a^nf(x)dx$ exists with $n \in N$.
I have a problem to show that from $\lim_{x\to\infty} f(x) = 0$ follows
$$\tag{1}\lim_{n\to\infty} \int_a^nf(x)dx \le M \text{ for some }M > 0.$$
Is it the right way to proof the Problem? If yes how can I show (1)? If no, how can I prove it?
hint
Assuming that $\lim_{x\to+\infty}f(x)=\color{red}{0}$,
You need to prove the implication
$$\lim_{n\to+\infty}\int_a^nf \in \Bbb R (=L)\implies$$ $$ \lim_{X\to+\infty}\int_a^Xf \in \Bbb R (=L).$$ Given $\epsilon>0$.
For $ N $ large enough, you have $$|\int_a^Nf -L|<\frac{\epsilon}{2}.$$ For $ X $ great enough, $N=\lfloor X\rfloor \text { is also large}$ and $$\Bigl |\int_a^Xf-L\Bigr |\le$$ $$|\int_a^{\lfloor X \rfloor}f-L|+|\int_{\lfloor X \rfloor}^X(f-\color{red}{0})|$$ $$< \frac{\epsilon}{2}+(X-\lfloor X\rfloor)\frac{\epsilon}{2}<\epsilon)$$ because $$X-\lfloor X \rfloor <1$$