We suppose that $y(t) \not\rightarrow 0$ as $t \rightarrow \infty$ There exists a sequence $(t_{n})$ in $\mathbb{R}_{+}$ such that $t_{n} \rightarrow \infty$ as $n \rightarrow \infty$ and $|y(t_{n})| \geq \varepsilon$ for all $n \in \mathbb{N}$ By the uniform continuity of $y$ there exists a $\delta > 0$ such that, for all $n \in \mathbb{N}$ and all $t \in \mathbb{R}$ $$ |t_{n} - t| \leq \delta \; \Rightarrow \; |y(t_{n}) - y(t)| \leq \frac{\varepsilon}{2}. $$ So, for all $t \in [t_{n}, t_{n}+\delta]$ and for all $n \in \mathbb{N}$ we have \begin{eqnarray*} |y(t)| & = & |y(t_{n}) - (y(t_{n})- y(t))| \geq |y(t_{n})| - |y(t_{n})- y(t)| \geq \\ & \geq & \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}. \end{eqnarray*}Therefore, $$ \left| \int_{0}^{t_{n}+\delta} y(t) dt - \int_{0}^{t_{n}} y(t) dt \right| = \left| \int_{t_{n}}^{t_{n}+\delta} y(t) dt \right| = \int_{t_{n}}^{t_{n}+\delta} |y(t)| dt \geq \frac{\varepsilon \delta}{2} > 0 $$ for each $n \in \mathbb{N}$ By the hypothesis, the improprer Riemann integral $\int_{0}^{\infty} y(t) dt$ exists, and thus the left hand side of the inequality converges to 0 as $n \rightarrow \infty$ yielding a contradiction.