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The definition of Riemann integral isn't this simple:

$$\int_a^bf(x)dx = \lim_{n\to\infty}\sum_{i=0}^nf(x_i)\Delta x \tag 1$$

Where $\Delta x = \frac{b-a}n$ and $x_i = a + i\Delta x$.

Rigorously, you have to define a partition of the domain of integration, define the upper and lower Riemann sums and check if they are equal. If they are, the function is said to be Riemann integrable and their common value is the value of the integral.

But this simpler definition is often used. My question is: being established that $f(x)$ is Riemann integrable, does $(1)$ always hold?

Thanks in advance.

Elvis
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  • I think this is the left sum, In which we take the left most point of the interval. – Sam May 12 '24 at 22:00
  • yes, if $f$ is Riemann integrable, then (1) holds. The question of how much justification is required depends on your precise definition (Riemann vs Darboux). See my remarks here. If you actually mean Riemann’s definition, then it’s a trivial consequence, as I showed in the link. If you mean Darboux’s definition, then it could be a little longer, though for this uniform partition, you probably don’t even need to go through the full effort of proving the equivalence of the Darboux and Riemann integrals. – peek-a-boo May 12 '24 at 22:00
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    @hamam_Abdallah I don't think that function is Riemann integrable. – Elvis May 12 '24 at 22:12

1 Answers1

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Yes. In fact, an even stronger statement is true. Recall that if $t_0,\dots,t_n$ is a partition of $[a,b]$ (i.e. $t_0,\dots,t_n$ is a finite sequence of real numbers satisfying $a=t_0<\dots<t_n=b$), then the mesh length of the partition is the maximum value of $t_i-t_{i-1}$ for $i\in\{1,\dots,n\}$. We have the following theorem:

Suppose $f$ is Riemann integrable on $[a,b]$. Then, for every $\varepsilon>0$ there is a $\delta>0$ such that, if $t_0,\dots,t_n$ is a partition of $[a,b]$ with mesh length $<\delta$, and $x_i\in[t_{i-1},t_i]$ for each $i$, then $$ \left|\sum_{i=1}^{n}f(x_i)(t_i-t_{i-1})-\int_{a}^{b}f(x) \, dx\right|<\varepsilon \, . $$

A proof can be found in Michael Spivak's Calculus, fourth edition, chapter 13, p. 282. It's an interesting exercise to derive your statement from the above theorem.

Joe
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