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Monotone functions are continuous except countably many points. If function is Riemann integrable it has only a finite number of discontinuity points. So how monotone functions are Riemann integrable on closed interval always?

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    $\begingroup$ "If function is Riemann integrable it has only a finite number of discontinuity points." That's not true; consider the function $f(x)=\begin{cases}1/2&\text{if $1/2<x\le1$,}\\1/4&\text{if $1/4<x\le1/2$,}\\1/8&\text{if $1/8<x\le1/4$,}\\&\vdots\\0&\text{if $0=x$}\end{cases}$ on the interval $[0,1]$. $\endgroup$ – Rahul Sep 5 '14 at 20:31
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    $\begingroup$ Using the measure of the set of discontinuities is fairly high tech. One can prove that monotone functions are Riemann integrable using the definition. $\endgroup$ – Ayman Hourieh Sep 5 '14 at 20:35
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Suppose $f$ is nondecreasing. For any partition $a = x_0 < x_1 \ldots < x_n = b$ of your interval $[a,b]$, any Riemann sum is between the left Riemann sum $L = \sum_{j=1}^n f(x_{j-1})(x_j - x_{j-1})$ and the right Riemann sum $R = \sum_{j=1}^n f(x_{j})(x_j - x_{j-1})$. The difference between them is at most $(f(b) - f(a)) \delta$ where $\delta = \max_j (x_j - x_{j-1})$.

Proof without words:

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  • $\begingroup$ Seems like one should do the proof without words using a discontinuous function, otherwise the most general phenomena is not observed. $\endgroup$ – RghtHndSd Sep 5 '14 at 21:09
  • $\begingroup$ Like it better now? $\endgroup$ – Robert Israel Sep 5 '14 at 21:18
  • $\begingroup$ Would upvote again if I could. $\endgroup$ – RghtHndSd Sep 5 '14 at 21:24
  • $\begingroup$ By the way, the argument (and the picture) go back to Newton who, of course, did not note that this was really the theorem he was establishing. $\endgroup$ – Andrés E. Caicedo Sep 5 '14 at 21:36
  • $\begingroup$ (See here for references.) $\endgroup$ – Andrés E. Caicedo Sep 5 '14 at 23:54

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