$f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$ Trying to solve

$f(x)$ is uniformly continuous in the range of $[0, +\infty)$ and $\int_a^\infty f(x)dx  $ converges.

I need to prove that:
$$\lim \limits_{x \to \infty} f(x) = 0$$
Would appreciate your help!

For all $\delta>0$, the fact that $\int_0^\infty f(x)dx=\sum\limits_{k=0}^\infty\int\limits_{k\delta}^{(k+1)\delta}f(x)dx$ converges implies that $\lim\limits_{k\to\infty}\int\limits_{k\delta}^{(k+1)\delta}f(x)dx=0$.  Let $\varepsilon>0$ be given.  Let $\delta>0$ be such that $|x-y|<2\delta$ implies $|f(x)-f(y)|<\varepsilon/2$.  Let $N>0$ be such that $k\geq N$ implies $\left|\int\limits_{k\delta}^{(k+1)\delta}f(x)dx\right|<\frac{\varepsilon\delta}{2}$.  Let $x_0\geq (N+1)\delta$.  Then there exists $k\geq N$ such that $x_0$ is in $[k\delta,(k+1)\delta]$.  It follows that $|f(x)-f(x_0)|<\varepsilon/2$ for all $x\in[k\delta,(k+1)\delta]$, and this implies that $\left|\int\limits_{k\delta}^{(k+1)\delta}f(x)dx\right|\geq \delta(|f(x_0)|-\varepsilon/2)$.  Since $\left|\int\limits_{k\delta}^{(k+1)\delta}f(x)dx\right|<\frac{\varepsilon\delta}{2}$, this implies that $|f(x_0)|<\varepsilon$.   Therefore $\lim\limits_{x\to\infty}f(x)=0$.
