# Showing that monotone functions have at most countable discontinuities.

I want to show that a map $$F: \mathbf{R} \to \mathbf{R}$$ has at most countable discountinuities, if $$F(x) \leq F(y)$$ whenever $$x \leq y$$.

Here's the idea. Let's use standard notation $$F(x^+), F(x^-)$$ for upper and lower limits of $$F$$ around $$x$$. And let $$D(F) = \{x : F(x^+) \neq F(x^-)\}$$, which is the set of discountinuities for a monotone function. Now suppose that $$D(F)$$ is uncountable. Suppose $$x, y \in D(F)$$ are distinct. Then $$(F(x^-), F(x^+)) \cap (F(y^-), F(y^+)) = \emptyset$$. This is because with $$r = d(x, y)$$, and assuming $$x < y$$, if $$F(x^+) > F(y^-)$$, this means that $$\inf_{x \leq s < x + r/2} F(x) > \sup_{y - r/2 < t \leq y} F(t)$$, which happens if and only if for some particular $$x \leq s < x + r/2$$ and $$y - r/2 < t \leq y$$, we have $$F(s)> F(t)$$. But this is a contradiction since as stated, $$s \leq t$$. Hence we have $$F(x^+) \leq F(y^-)$$, which means the intervals are disjoint as required.

We have shown that $$D(F)$$ uncountable implies the existence of uncountably many disjoint intervals $$(F(x^-), F(x^+))$$, where $$x$$ ranges over $$D(F)$$. But this can't happen because each open interval contains a distinct rational, which is a countable set.

• Your proof looks good. :) – Tki Deneb Dec 8 '18 at 20:59
• Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right. – user25959 Dec 8 '18 at 20:59

To be clearer, you should probably point out that $$x + r/2 = y - r/2$$, which is how you get $$s < x + r/2 = y - r/2 < t$$ and therefore $$s \le t$$ (you say this is true "as stated", but I don't see you stating it).
You might also want to say more about why a discontinuity in a monotone function must have $$F(x^+) \ne F(x^-)$$.