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I have several doubts about this exercise because one of the conditions a function must have to be Riemann integrable is to be continuous in that interval, condition $\lfloor x\rfloor$ does not meet. How is this exercise done? Or what does the approach have to be?

Prove that $f(x)=\lfloor x\rfloor$ is Riemann integrable on [0,5] and calculate $\int_0^5 \lfloor x\rfloor \,dx$, where $\lfloor x\rfloor = floor(x)$

Thanks in advance.

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    $\begingroup$ Where did you read that a Riemann integrable function needs to be continuous? $\endgroup$
    – Martin R
    Commented Jan 18, 2020 at 10:21
  • $\begingroup$ Yes, now I notice. If a function is continuous, then is Riemann integrable, but in this case, I don't know why is it integrable. $\endgroup$
    – user9867
    Commented Jan 18, 2020 at 10:24
  • $\begingroup$ Piece-wise continuity is sufficient for Riemann integrability. $\endgroup$
    – Doug M
    Commented Jan 18, 2020 at 10:37

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Hint: try to prove (or find a prove in your textbook) that a function is Riemann integrable if it has only finitely many discontinuities in the interval of integration.

More generally, a function is Riemann integrable if it has countable many discontinuities. You might also want to try to find a prove for this.

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  • $\begingroup$ More generally still, a function is Riemann integrable, if & only if the measure of its set of discontinuities is 0. The characteristic function of the Cantor set, for example, has uncountably many discontinuities - in particular, they lie at every point in the Cantor set - yet also has a well-defined Riemann integral of 0. $\endgroup$
    – indnwkybrd
    Commented Jan 18, 2020 at 11:44
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    $\begingroup$ I didn't say 'if and only if'. There is no point in wanting to explain lesbesgue null sets to someone who had difficulties with integrating floor function. $\endgroup$ Commented Jan 18, 2020 at 14:30
  • $\begingroup$ Oh yes, of course; I never thought you did ;) it was just a side comment. I've always been fascinated by the occasional tantalizing "sneak preview" of surprising results and strange pathologies that lay just a little bit ahead of my current understanding level. They tend to spur my interest & motivate me to learn. My thought was just to motivate measure theory, rather than explain it, by drawing attention to the curious notion that a function must have a set of discontinuities that is, in a loose sense, "even worse than uncountable", before it ceases to be Riemann integrable in general. $\endgroup$
    – indnwkybrd
    Commented Jan 21, 2020 at 13:13
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Monotone functions are integrable! To calculate the integral, integrate over [j,j+1] for j=0,1,2,3,4 and add.

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A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure).

https://en.wikipedia.org/wiki/Riemann_integral

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