# Question about Riemann integrability: do we need to specify that all Riemann sums converge to the same number in the definition?

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 $$0$$, $$P_n=\{a=t_0^n, such that, for any choice of interior points $$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 $$0$$, $$P_n=\{a=t_0^n, such that, for any choice of interior points $$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.

• The definition is usually that there exists some $S$ such that the various limits converge to $S$. – copper.hat Apr 10 at 18:12
• @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... – jxm Apr 10 at 18:14
• 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. – 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 $$s_i^n$$.
• 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. – RRL Apr 11 at 6:45