If monotone decreasing and $\int_0^\infty f(x)dx <\infty$ then $\lim\limits_{x\to\infty} xf(x)=0.$ 
Let $f:\mathbb{R}_+ \to \mathbb{R}_+$ be a monotone decreasing function defined on the positive real numbers with $$\int_0^\infty f(x)dx <\infty.$$ Show that $$\lim_{x\to\infty} xf(x)=0.$$

This is my proof: Suppose not. Then there is $\varepsilon$  such that for any $M>0$ there exists $x\geq M$ such that $xf(x)\geq \varepsilon$. So we can construct a sequence $(x_n)$ such that $x_n \to \infty $ and $x_n f(x_n ) \geq \varepsilon$. So $$\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq  \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.$$ So we get a contradiction.  I feal like I have the correct idea but some details are wrong.  Any help would be appreciated.
 A: For every $c>0$, there exists $R$ such that for $x>0, |\int_R^xf(x)dx|<c/2$ and there exists $R'$ such that $x>R'$ implies that $f(x)<{c\over {2R}}$. Let $x>\sup(R,R')$, we have $c/2\geq \int_R^xf(x)dx\geq (x-R)f(x)$  since $f$ decreases and is positive. We deduce that $xf(x)\leq c/2+Rf(x)\leq c$.
A: Notice that, since $f$ is monotone decreasing, you have for each $x$,
$$0\leq f(x) (x - \frac{x}{2}) \leq \int_{\frac{x}{2}}^{x} f(t) \, dt$$
Therefore,
$$0\leq xf(x) \leq 2\int_{\frac{x}{2}}^{x} f(t) \, dt$$
The right hand side goes to zero since the integral converges.
Added: You should convince yourself that the last sentence is true.  You could do this by writing the integral as a sum of terms of the form $\int_{x_i/2}^{x_i} f(t) \,dt$, for an appropriate sequence $\{x_i\}$.
A: For all $x>0$, in the interval $[\frac{1}{2}x ; x]$ the minimum of $f$ is $f(x)$, since $f$ is decreasing. Thus,
$$\frac{1}{2} x f(x) = \int_{\frac{1}{2}x}^{x} f(x) \mathrm dt \le \int_{\frac{1}{2}x}^{x} f(t) \mathrm dt \le \int_{\frac{1}{2}x}^{+ \infty} f(t) \mathrm dt \to 0$$
as $x \to + \infty$. Thus, by comparison,
$$\lim_{x \to + \infty} \frac{1}{2}x f(x) = 0$$ which is equivalent to 
$$\lim_{x \to + \infty} x f(x) = 0$$
A: You can solve this using BCT. First, note that for the integral to exist, we must have
$$\lim_{x \rightarrow \infty} f(x) = 0$$
Now we know that the original limit is of indeterminate form. Now from here, since the improper integral converges, we know for sure that $f(x) < \frac{1}{x^a}$ where $a > 1$. This is by BCT since the improper integral of $\frac{1}{x}$ doesn't converge. Thus we know that $xf(x) < x(\frac{1}{x^a})$
