# Every monotonic function on $[a,b]$ is integrable?

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?

• 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
• 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
• 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
• @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