Suppose $f$ is a monotonically increasing function. $f$ can be $1-1$, in which case $f$ is strictly increasing. $f$ can be stair-step, in which case the jumps are at most countable. What are some other cases?
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The jumps form an at most countable set for any montonic function. – Kavi Rama Murthy Aug 19 '21 at 09:44
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@Kavi Rama Murthy: The following is rather advanced (which is why I pinged K. R. Murthy, in case he's interested), but of possible interest is my answer to Points of differentiability of $f(x) = \sum\limits_{n : q_n < x} c_n$ (which I might one day greatly expand, because at the time I had recently moved and the material I have on this topic was still scattered among different boxes) and my 4 November 2000 sci.math post on the non-differentiability behavior of monotone functions. – Dave L. Renfro Aug 19 '21 at 13:23
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Conversely to Kavi Rama Murthy's comment, every countable set is the set of discontinuities of a monotonic function. – TheSilverDoe Aug 19 '21 at 13:53
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$f$ could be a combination of continuous increase and jump increase. And the continuous increase might be like $x\mapsto \int_b^x g(t) dt $ (for a non-negative integrable $g$) or it might be singular, like the Cantor function ("devil's staircase"). In general a monotone increasing function $f$ can be uniquely (modulo an additive constant) decomposed as the sum of three terms of the type just described. – John Dawkins Aug 19 '21 at 18:29
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1Regarding the comment by @John Dawkins see Two questions on Lebesgue Decomposition of an increasing function?, and for the (more advanced) analogous version for measures see Lebesgue decomposition $\nu =\nu_{cont}+\require{cancel} \cancel{\nu {sing}}+\nu{pp}$. – Dave L. Renfro Aug 20 '21 at 06:55