Let $f:[a,b]\rightarrow\mathbb{R}$ be a function. Suppose that there is a sequence of partitions $\{P_n\}_{n=1}^\infty$ with mesh tending to $0$, $P_n=\{a=t_0^n<t_1^n<\ldots<t_{r_n}^n=b\}$, such that, for any choice of interior points $s_i^n\in [t_{i-1}^n,t_i^n]$, we have that $\lim_{n\rightarrow\infty} \sum_{i=1}^{r_n} f(s_i^n)(t_i^n-t_{i-1}^n)$ exists.
Is it true that, in such a case, the limit must be unique? (In such a case, it would be $\int_a^b f(t)\,dt$).
Motivation: I have read the following definition for Riemann integrability: there is a number $I$ and a sequence of partitions $\{P_n\}_{n=1}^\infty$ with mesh tending to $0$, $P_n=\{a=t_0^n<t_1^n<\ldots<t_{r_n}^n=b\}$, such that, for any choice of interior points $s_i^n\in [t_{i-1}^n,t_i^n]$, we have $\lim_{n\rightarrow\infty} \sum_{i=1}^{r_n} f(s_i^n)(t_i^n-t_{i-1}^n)=I$. My question is whether we need to impose $\lim_{n\rightarrow\infty} \sum_{i=1}^{r_n} f(s_i^n)(t_i^n-t_{i-1}^n)$ to be always the same number $I$, or this fact is given for free.
Assume there are $s_i^n,q_i^n\in[t_{i-1}^n,t_i^n]$ such that $$\lim_{n\rightarrow\infty}\sum_{i=1}^{r_n}f(s_i^n)(t_i^n-t_{i-1}^n)=s\neq q=\lim_{n\rightarrow\infty}\sum_{i=1}^{r_n}f(q_i^n)(t_i^n-t_{i-1}^n).$$ Then consider $u_i^n\in[t_{i-1}^n,t_i^n]$ given by $$u_i^n=\begin{cases} s_i^n,&n\text{ even},\\ q_i^n,&n\text{ odd}. \end{cases}$$ By hypothesis, $\sum_{i=1}^{r_n}f(u_i^n)(t_i^n-t_{i-1}^n)$ converges, but it also has two convergent subsequences with distinct limits, namely $s$ and $q$. Since that can't be the case, the limits are independent of the choice of the $s_i^n$.