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.

  • $\begingroup$ The definition is usually that there exists some $S$ such that the various limits converge to $S$. $\endgroup$ – copper.hat Apr 10 at 18:12
  • $\begingroup$ @copper.hat Yes, that is what I mention in the motivation with $I$ being your $S$. But I do not know if this $S$ is given by simplicity, or it is actually necessary... $\endgroup$ – jxm Apr 10 at 18:14
  • $\begingroup$ I would suspect that it is true, possibly along the following vague lines: If the limits are not the same then the Darboux limits would be different, hence not Riemann integrable. If the limits are all the same then so are the Darboux limits. $\endgroup$ – copper.hat Apr 10 at 20:03

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 $s_i^n$.

  • $\begingroup$ Good answer (+1). Not only must all Riemann sums converge to the same limit for a specific sequence of partitions $(P_n)$ with $\|P_n\| \to 0$, they must converge to the same limit for any other sequence of partitions. This is not easy to prove. $\endgroup$ – RRL Apr 11 at 6:45
  • $\begingroup$ @RRL: one can use tags which lead to approximation of upper Darboux sum and another set of tags for lower Darboux sum. This makes the Darboux integrals equal and the function Riemann integrable and then any Riemann sum converges to the Riemann integral $\endgroup$ – Paramanand Singh Apr 11 at 14:04

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