How to show that a set of discontinuous points of an increasing function is at most countable

Solution 1:

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.

Solution 2:

Just for slight variation, another proof.

Assume $f:[a,b]\to \mathbb R$ is monotonically increasing and let $D$ be its set of discontinuities. For every $x\in D$ let $c_x=\lim_{t\to x+}f(t)-\lim_{t\to x-}f(t)$ (since $f$ is monotone the one-sided limits exist (and are finite)). As $x$ was a point of discontinuity it follows that $c_x>0$. Now, let $S$ be the sum of all $c_x$. More formally, consider the set $T$ of all finite sums of elements of the form $c_x$, and let $S$ be its supremum.

Now, for every finite sum $s=c_{x_1}+ \cdots +c_{x_n}$ (we may assume the points $x_i$ appear in their natural order in $[a,b]$) using monotonicity of $f$ it follows that the total variation of $f$, that is $f(b)-f(a)$, is not less than the sum $s$ (intuitively, because that sum is the sum of total variations at points along the way from $a$ to $b$). In symbols: $s\le f(b)-f(a)$. It follows that the supremum also satisfies $S\le f(b)-f(a)$. But a sum of infinitely many positive elements can be bounded only if there are countably many elements. Thus, $D$ is countable.