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I met these two statements in two different sources and a question came up which I cannot yet explain to myself.

Th1: If $f$ is defined and bounded in the interval $[a,b]$ then it is Riemann-integrable if and only if the set of points on which it is discontinuous has measure zero (i.e. is a null set).

Th2: If $f$ is defined and monotonic in the interval $[a,b]$ then it is Riemann-integrable.

So this made me thinking... If both statements are true, then in the setup of Th2, it must somehow follow that the set of points at which $f$ is discontinuous is a null set (has Lebesgue measure zero). But I can't explain to myself even intuitively why is that. I mean, what is preventing $f$ in the setup of Th2 of having discontinuities at a set of points $S \subset [a,b]$, which has measure greater than zero? This is my question #1.

Also, I need confirmation that both theorems are true, maybe I misread or misunderstood them. But Th2 I found in a book (so it must be reliable), and Th1 I found here (and the author seems to know very well what he's talking about; he has PhD in maths):

https://www.quora.com/What-are-necessary-conditions-for-the-Riemann-integral-to-exist

So I guess both theorems are true indeed, right? This is my question #2.

Finally, I would like to know if Th2 has some analogue in $\mathbb{R}^2$ or in $\mathbb{R}^3$ i.e. when we have a function $f$ of 2 or 3 variables. So this is my question #3 here. I do know that Th1 has analogues in $\mathbb{R}^2$ or in $\mathbb{R}^3$ but what about Th2?

Question 3 is kind of a side question. I am mostly interested in questions 1) and 2).

I've been thinking on all this in the last 2-3 days, and finally I decided to ask here since it seems I can't put together myself the pieces of the puzzle.

peter.petrov
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    Both are true. You can check that a monotonic function is regulated, hence it has at most a countable set of discontinuities (hence is a null set). Intuitively, if the set of discontinuities were larger, the function would diverge to $\infty$ (a sum of uncountably many positive numbers is $\infty$) – Jean-Claude Arbaut Dec 25 '21 at 10:11
  • See here – RRL Dec 25 '21 at 10:13
  • @Jean-ClaudeArbaut I am not sure I will be able to digest these regulated functions. Yeah, I was thinking that myself, that probably the function will have to go to $\infty$ in the setup of Th2 if it's discontinuous on a non-null set. But still... it's not intuitive at all to me why. Also, there's this example of https://en.wikipedia.org/wiki/Cantor_set which shows that things are far from intuitive there. So I cannot rely on my intuition much here. I guess there could be some patological case which I miss, I mean. – peter.petrov Dec 25 '21 at 10:15
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    @RRL Ah, that looks interesting and to the point. So in Th2, we can even claim the set of discontinuities is countable, yes? That's even stronger than having measure $0$, right? – peter.petrov Dec 25 '21 at 10:18
  • @peter.petrov: A monotonic function on $[a,b]$ can only have discontinuities of the first kind and therefore set of discontinuities of a monotonic function is countable. – Koro Dec 25 '21 at 10:23
  • @RRL This question you pointed me to seems to be exactly what I need. But I don't follow the proof which the OP provides that the set of discontinuities if countable. Could someone elaborate here? – peter.petrov Dec 25 '21 at 10:25
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    @peter.petrov: To prove the countability of the set of discontinuities of monotonic f defined on [a,b]: note that left hand and right hand limits exist at every interior point of the interval. So let $D$ be the set of discontinuities of f in (a,b). If $x\in D$ then LHL (at x) is not equal to RHL (at x). And hence between LHL and RHL, you can choose a rational number. Now using monotonicity of f, you can inject D into $\mathbb Q$, the set of rationals, which is another way of saying that $D$ is at most countable. – Koro Dec 25 '21 at 10:27
  • The key step is that a bounded monotone function must have finite right and left limits at every point. In that particular link it is stated without proof. That implies discontinuities are simple jumps which then implies countability. – RRL Dec 25 '21 at 10:30
  • @Koro Oh, I think I start to understand. Thank you and thanks everyone. – peter.petrov Dec 25 '21 at 10:39
  • Could anyone comment or provide some hints on question #3? – peter.petrov Dec 29 '21 at 17:45

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