This is a part of my proof:
$$\begin{align} \left| \sum^{n}_{i=1} V(r(\tau_i)) \cdot \int_{t_{i-1}}^{t_i} r'(t) dt - \int_{a}^{b} V(r(t)) \cdot r'(t) dt \right| &\leq \sum^{n}_{i=1} \int_{t_{i-1}}^{t_i} \| V(r(\tau_i))-V(r(t)) \| \; \| r'(t) \| dt \\ &\le \frac{\varepsilon}{l} \displaystyle \sum^{n}_{i=1} \int_{t_{i-1}}^{t_i} \| r'(t) \| dt \\ &\le \frac{\varepsilon}{l} \int_{a}^{b} \| r'(t) \| dt \\ &= \varepsilon \end{align}$$
My only question is how $ \displaystyle \sum^{n}_{i=1} \displaystyle \int_{t_{i-1}}^{t_i} $ $ = $ $ \displaystyle \int_{a}^{b} $ ?
Assuming $a = t_0 \le t_1 \le t_2 \ldots \le t_n = b$ then by induction we can show $$ \sum_{i=1}^n \int_{t_i-1}^{t_i} f(x) \, \mathrm{d} x = \int_{t_0}^{t_1} f(x) \, \mathrm{d} x + \int_{t_1}^{t_2} f(x) \, \mathrm{d} x + \cdots + \int_{t_{n-1}}^{t_n} f(x) \, \mathrm{d} x = \int_{t_0}^{t_n} f(x) \, \mathrm{d} x = \int_a^b f(x) \, \mathrm{d} x $$ at least for the Riemann integral.