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My book says that the statement in the title is true. What is so special about a function being monotonic? When I think of a non (Riemann) integrable function, I think of $f$, where $f$ is $1$ on rationals and $0$ on irrationals. If we were to translate each point upwards to make this function monotonic (but not continous), why would it be integrable now?

Ovi
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  • A key part of that result is that the function is monotonic and bounded (usually at that point in the development of theory we work under the umbrella assumption that all considered functions are bounded). How would you move each point upwards to make that function monotonic and bounded? – Jyrki Lahtonen Dec 12 '16 at 22:35
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    For what it's worth, in my first-year analysis course, Theorem 53 was "Let $f: [a, b] \to \mathbb{R}$ be increasing. Then $f$ is Riemann integrable." No boundedness imposed. – Patrick Stevens Dec 12 '16 at 22:41
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    A monotonic function has only jump discontinuities, and only countably many of them. You can't translate the points upwards making it monotonic without also making it much more continuous. – Daniel Fischer Dec 12 '16 at 22:41
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    @PatrickStevens The boundedness follows from the closedness of the domain, $\min { f(a), f(b)} \leqslant f(x) \leqslant \max { f(a), f(b)}$ for all $x \in [a,b]$. – Daniel Fischer Dec 12 '16 at 22:42
  • Anyway a monotonic bounded function is Riemann integrable because we can make the upper and lower sums as close to each other as we wish simply by using a division to equal length but short enough subintervals. This is because the difference between the upper and lower sums is then the total variation times the common length of the subintervals. – Jyrki Lahtonen Dec 12 '16 at 22:43
  • @DanielFischer Apologies, you're right. – Patrick Stevens Dec 12 '16 at 22:43
  • Yeah, sorry about bringing up boundedness. That was a bit misleading here (even though boundedness is tacitly assumed because otherwise we cannot find suprema/infima on all subintervals). – Jyrki Lahtonen Dec 12 '16 at 22:44
  • Anyway, see here for a discussion of why monotonic functions are Riemann integrable. What was the question again? – Jyrki Lahtonen Dec 12 '16 at 22:52
  • In some sense it is that back-and-forth wiggling of that example function that makes it non integrable. Some other related results/observations: 1) a function of bounded variation is integrable ('cause it is the difference of two monotonic functions), 2) a function not of bounded variation can be integrable, if the wiggle is limited to e.g. a neighborhood of a single point (think: topologist's sine curve), 3) a somewhat similar function; $f(x)=0$ if $x$ is irrational, $f(x)=1/n$ if $x=m/n, \gcd(m,n)=1$, again is integrable over $[0,1]$. Not a full picture, but may shed some light? – Jyrki Lahtonen Dec 12 '16 at 23:03
  • And we have the result that a function is Riemann integrable if the points of incontinuity form a Lebesgue null set. You probably haven't studied measure theory yet, so that may be all Greek to you. – Jyrki Lahtonen Dec 12 '16 at 23:07
  • @JyrkiLahtonen and others thank you for your illuminating comments! – Ovi Dec 13 '16 at 00:35

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