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.

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 $\{x_i\}$.

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 $xf(x)\leq c/2+Rf(x)\leq c$.