I would like to prove the following:

Let g be a monotone increasing function on [0,1]. Then the set of points where g is not continuous is at most countable.

My attempt:Let g(x^-)~,g(x^+) denote the left and right hand limits of g respectively. Let A be the set of points where g is not continuous. Then for any x\in A, there is a rational, say, f(x) such that g(x^-)\lt f(x)\lt g(x^+). For x_1\lt x_2, we have that g(x_1^+)\leq g(x_2^-). Thus f(x_1)\neq f(x_2) if x_1\neq x_2. This shows an injection between A and a subset of the rationals. Since the rationals are countable, A is countable, being a subset of a countable set.

Is my work okay? Are there better/cleaner ways of approaching it?

**Answer**

This looks beautiful to me: or, more truthfully, it looks like exactly what I would write.

If anything else can be asked of this argument, maybe it is a justification that monotone functions have discontinuities as you have described. I happen to have recently written this up in lecture notes for a “Spivak calculus” course: see \S 3 here. Although the fact is quite well known, many texts do not treat it explicitly. I think this may be a mistake: in the the same section of my notes, I explain how this can be used to give a quick proof of the Continuous Inverse Function Theorem.

**Attribution***Source : Link , Question Author : AKM , Answer Author : AKP2002*