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Given function $f$ that is continuous and defined on the closed, finite inverval $[a,b]$

Also given any two partitions $P_1$ and $P_2$ and their common refinement $P$ which consists of all of the points of $P_1$ and $P_2$

The Lower Riemann Sum is given by $L$ and the Upper Riemann Sum is given by $U$

Thus:

$$L(f,P_1)\le L(f,P)\le U(f,P)\le U(f,P_2)$$

Now my calculus-book states that:

Hence, every lower sum is less than or equal to every upper sum. Since the real numbers are complete, there must exist at least one real number $I$ such that $L(f,P)\le I\le U(f,P)$ for every partition $P$. If there is only one such number, we will call it the definite integral of $f$ on $[a,b]$

Question: What are scenario's where there are indeed more than one numbers $I$ that satisfy these conditions?

I don't understand why there would be more than one number, since it is explicitly stated that $L(f,P)\le I\le U(f,P)$ for every partition $P$

Thanks!

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    $\begingroup$ If $f$ is continuous, then $f$ is Riemann integrable, and there's only one such number. For discontinuous $f$, there can be more. Consider the Dirichlet function $$f(x) = \begin{cases} 1 &, x \in \mathbb{Q} \\ 0 &, x \notin \mathbb{Q}\end{cases}.$$ Then for $[a,b] = [0,1]$, for every partition $P$ we have $L(f,P) = 0$ and $U(f,P) = 1$. $\endgroup$ – Daniel Fischer Jun 23 '17 at 11:29
  • $\begingroup$ Then what is the "at least one" statement from my calculus-book about? $\endgroup$ – GambitSquared Jun 23 '17 at 11:40
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    $\begingroup$ It's about discontinuous functions. Discontinuous functions can be Riemann integrable too. $\endgroup$ – Daniel Fischer Jun 23 '17 at 11:46

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