f is monotone and the integral is bounded. Prove that $\lim_{x→∞}xf(x)=0$ Question
$f : [0,\infty] \to \Bbb R $ is monotone and $\displaystyle∫^∞_0f(x)\,dx$ converges. 
Note: we also proved before $\lim_{x→∞}f(x)=0$
Show that even $\lim_{x→∞}xf(x)=0$
Thanks!
 A: Begin by noting that it's clear that $f$ is monotone decreasing.
Proceed by contrapositive: Suppose that there exists an $\epsilon > 0$ such that there are arbitrarily large $x$ for which
$$xf(x) > \epsilon$$
Without any loss of generality, $\epsilon = 2$. Choose a strictly increasing sequence $x_n$ converging to infinity such that the above inequality holds; in fact, we choose the sequence such that for every $n$,
$$x_n > 2x_{n - 1}$$
Now since $f$ is monotone decreasing, we have the estimate
$$\int_{x_{n - 1}}^{x_n}f \ge \int_{x_{n - 1}}^{x_n} f(x_n) = (x_n - x_{n - 1}) f(x_n)$$
But $x_n - x_{n - 1} > \frac{1}{2}x_n$ by construction, and
$$\int_{x_{n - 1}}^{x_n} \ge \frac{1}{2} x_n f(x_n) \ge 1$$
Now sum over $n$, and conclude that $f$ is not integrable.
A: As JLA began in this answer, there exists some $M >0$ such that for $x>M$ either $f(x)\geq 0$ or $f(x)\leq 0.$ WLOG assume the former. Then by the convergence of $\displaystyle\int_{0}^{\infty}f(x)dx,$ we must have that $f$ is monotone decreasing for sufficiently large $x.$ Then for large enough $x,$ $f(x)$ is non-negative and monotone decreasing and hence by the integral test, $\displaystyle \sum_{n=1}^{\infty}f(n)$ converges. Let $K >0$ be such that for $x>K$ we have $f(x)>0$ and $f$ is monotone decreasing.
Since $\displaystyle \sum_{n=1}^{\infty}f(n)$ converges and $0\leq f(n+1)\leq f(n)$ for $n > K$, by the Abel-Olivier-Pringsheim condition (that is, the theorem that Paramanand Singh notes in his first comment), we have $\displaystyle \lim_{n \to \infty} nf(n)=0$ 
Claim: $\displaystyle \lim_{x \to \infty} xf(x) =0$
It suffices to show that if $(a_n)$ is any sequence such that $a_n \to \infty$ then $a_nf(a_n)\to 0.$ Suppose not. Then there exists $\varepsilon_{0}>0$ such that for all $n >K$ we have $a_nf(a_n)\geq \varepsilon_{0}.$
For each $m \in \mathbb{N}$ set $n_m = \lfloor a_{m}\rfloor$ so that we have $n_m \leq a_{m}<n_m +1.$ For $m >K$ we have $f(a_m) \leq f(n_m)$ (as $f$ is monotone decreasing) and $\varepsilon_{0} \leq a_{m}f(a_m)$.
Since $nf(n) \to 0$ and $f(n) \to 0$ there exists $r \in \mathbb{N}$ such that for all $n \geq r$ we have $|nf(n)+ f(n)|< \dfrac{\varepsilon_{0}}{2}$
For $m \geq \max\{r, K\}$ consider:
$\dfrac{\varepsilon_{0}}{a_m}\leq f(a_m)\leq f(n_m)$
$\Rightarrow \varepsilon_{0} < a_mf(n_m) < (n_m +1)f(n_m)< \dfrac{\varepsilon_{0}}{2},$ 
a contradiction.
Therefore we must have $a_nf(a_n) \to 0$ and since $(a_n)$ is arbitrary it follows that $\displaystyle \lim_{x \to \infty}xf(x)=0$
