Let $f:[a,b] \longrightarrow \mathbb{R}$ and $\displaystyle\sum(f; P^{\ast})$ a Riemann sum. Prove that, if $\displaystyle \lim_{|P|\to 0}\sum(f; P^{\ast}) = L$, then $f$ is a limited function
$\textbf{My idea:}$ Suppose that $f$ is unlimited, so $f$ is unlimited in a partition interval $[t_{s-1}, t_{s}]$ of $P = \lbrace t_{0}, ..., t_{n} \rbrace$. Thus, for all $A > 0$ theres exist a $c_{s} \in [t_{s-1}, t_{s}]$ such that $$|f(c_{s})(t_{s} - t_{s-1})| > \Bigg|\sum\limits_{k=1, k\neq s}^{n}f(c_{k})(t_{k} - t_{k-1})\Bigg| + A.$$ Therefore $$\Bigg|\sum(f;P^{\ast})\Bigg| = \Bigg|f(c_{s})(t_{s} - t_{s-1}) + \sum\limits_{k=1, k\neq s}^{n}f(c_{k})(t_{k} - t_{k-1})\Bigg| > A.$$ So, $\displaystyle \lim_{|P|\to 0}\sum(f; P^{\ast})$ doesn't exist.
Is this a correct idea?
Note: This answer corresponds to the original version of the question.
The second last inequality should be $$|f(c_{s})(t_{s} - t_{s-1})| > \Bigg|\sum\limits_{k=1, k\neq s}^{n}f(c_{k})(t_{k} - t_{k-1})\Bigg| \color{red}{+A},$$ so that \begin{align} \left|\sum(f;P^{\ast})\right| &= \Bigg|f(c_{s})(t_{s} - t_{s-1}) + \sum\limits_{k=1, k\neq s}^{n}f(c_{k})(t_{k} - t_{k-1})\Bigg| \\ &\ge \left||f(c_{s})(t_{s} - t_{s-1})| - \Bigg|\sum\limits_{k=1, k\neq s}^{n}f(c_{k})(t_{k} - t_{k-1})\Bigg|\right| \\ &= |f(c_{s})(t_{s} - t_{s-1})| - \left|\sum\limits_{k=1, k\neq s}^{n}f(c_{k})(t_{k} - t_{k-1})\right| > A \end{align} for an arbitrarily large $A>0$.