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Define

$f(x)= \begin{cases} 3 & 0 \le x \le 2 \\ 2 & 2<x<3 \\ 4 & 3 \le x \le 6 \end{cases} $

over $[0,6]. $ Is this function Riemann integrable over $[0,6]$?

I believe that it is, but I'm not sure if my solution is correct.

Basically, I considered the partition of the interval: $\{0, 2 - \delta, 2 +\delta, 3 -\delta, 3 + \delta, 6 \}$ and showed that over this partition, the lower sum is $20 - 3 \delta$ and the upper sum is $20 + 3 \delta$.

Since the difference is only $6 \delta$, we can make delta arbitrarily small, meaning the function is integrable.

Is this correct?

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  • $\begingroup$ Your method reminds of the proof of a more general result. You can have the same conclusion for any function that is bounded on $[0,6]$ having only a finite number of discontinuities within $(0,6)$. $\endgroup$
    – 311411
    Dec 6 '21 at 2:36
  • $\begingroup$ another route: it is monotonic on $[0,3]$, it is monotonic on $[3,6]$ $\endgroup$
    – 311411
    Dec 6 '21 at 2:45
  • $\begingroup$ Riemann integrable if a function is piecewise continuous. $\endgroup$
    – stephenkk
    Dec 6 '21 at 2:45
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Yes, your answer is correct by the Cauchy criterion of Riemann Integrability. Your function $f$ is bounded over $[0,6]$, and you can take $\delta = \dfrac{\epsilon}{6}> 0$. The partition you used works !

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