Show that if \(f:\mathbb R\to\mathbb R\) is monotone increasing, then \(f\) has at most a countable set of jump discontinuities.
Solution
1. Every discontinuity of a monotone increasing \(f\) is a jump.Proof: at each \(x\), the one-sided limits \(f(x-) = \sup_{y<x}f(y)\) and \(f(x+) = \inf_{y>x}f(y)\) exist (boundedness on compact intervals follows from monotonicity; for the extended real line they exist in \([-\infty,\infty]\)). Continuity at \(x\) holds iff \(f(x-) = f(x) = f(x+)\), so a discontinuity is exactly a point where \(f(x-) < f(x+)\), a jump of positive size.
2. Associate to each jump point a distinct rational.Proof: for a jump at \(x\), pick a rational \(q(x) \in (f(x-), f(x+))\) (possible since the interval is nonempty and \(\mathbb{Q}\) is dense).
3. Distinct jump points get distinct rationals.Proof: if \(x_1 < x_2\) are jump points, then \(f(x_1+) \le f(x_2-)\) (as \(f\) is increasing), so the intervals \((f(x_1-), f(x_1+))\) and \((f(x_2-), f(x_2+))\) are disjoint; hence \(q(x_1) \ne q(x_2)\).Thus \(x \mapsto q(x)\) is injective from the set of jump discontinuities into \(\mathbb{Q}\).
4. Q.E.D. Proof: the jump set injects into the countable set \(\mathbb{Q}\), so it is at most countable.