$f$ 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!

If the integral converges than you have: $\lim_{M\to\infty} \int_a^M f(x) dx = c\ $ for some $c$. Assuming that f does not converge to 0 we have:
$\forall M>0\  \exists p>M: |f(p)|>\varepsilon\ \ $ for some $\varepsilon$.
Without loss of generality we have $\forall M>0\  \exists p>M: f(p)>\varepsilon\ \ $. So there exists $(p_n)_{n=1}^\infty$ s.t. $p_n \to \infty\ $.
Using uniform continuity we have that for $q: |p-q|<\delta\ \ $ the inequality holds: $f(q)>\varepsilon/2\ $. So $\int_a^{p+\delta} f(x) dx - \int_a^{p-\delta} f(x) dx > \delta \cdot \epsilon \ \ $, which means that the sequence $(\int_a^{q_n}f(x)dx)_{n=1}^\infty$ (where $q_0=p_0-\delta,\ q_1=p_0+\delta,\ q_2=p_1-\delta \ldots)\ \ $ is not a Cauchy sequence so it can not converge.
