Prove or disprove:
a) $f$ is Reimann integrable in the interval $[1,b]$ $\forall$ $b>1$, $f(x) > 0$ $\forall x$, and $\int_1^\infty f(x)dx $ is convergent. Then, $$\lim_{x\rightarrow\infty}f(x)=0$$ b) Does your conclusion remain the same if $f(x)$ can take any value?
I am struggling with this question. Here is what I know but I can't bring the pieces together. Need a rigorous proof.
For part a, since $\int_1^\infty f(x)dx $ is convergent, $\int_1^\infty f(x)dx = \lim_{x \to \infty}\int_1^x f(u)du $ exists.
Intuitively, I think I should show that since the function is positive and the integral is convergent, it must be that there exist a large enough $N$ such that $\forall n >N$, $f(x_n)-f(x_{n-1})<\epsilon$ for any arbitrary $\epsilon>0$ which should imply that the limit of the function is zero. But I am not sure about this.
For part b, my guess is that the conclusion will not hold as we need the function to be positive to prove the statement in part a. But I don't know how can I show it.
Thanks for your help.
If $f$ is positive (and even continuous) , then we can have $\int_1^\infty f(x) \, dx$ converge and $f(x) \not\to 0$ as $x \to \infty$.
Take $f(x) = x^{-2} + \sum_{k=2}^\infty \phi_k(x)$ where
$$\phi_k(x) = \begin{cases}k^2(x - k + k^{-2}) & k- k^{-2} \leqslant x \leqslant k \\ k^2(k + k^{-2}-x) & k < x < k + k^{-2}\\ 0& \text{otherwise} \end{cases}$$
In this case $\limsup f(x) = 1$ and $\liminf f(x) = 0$ so the limit does not exist, but $$\int_1^\infty f(x) \, dx = 1 + \sum_{k=2}^\infty \frac{1}{k^2} < \infty$$