# Why do non-Decreasing Functions have countable discontinuities [duplicate]

I was reading some notes and one of the results in it implicitly used a result which fell along the lines of "non-decreasing functions have countable discontinuities". I don't completely understand why.

The notes were essentially describing the properties of a CDF of a "weird" random variable. They made the above statement and then concluded (after some more proofs) that the CDF is "cadlag".

## marked as duplicate by Rahul, mrf, Did, Stefan Hansen, Old JohnFeb 8 '13 at 8:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Short answer: Each discontinuity jumps across at least one rational, and rationals are countable. Long answer: Surely this is a duplicate of a previous question; that question will have a long answer. – Rahul Feb 8 '13 at 6:55
• @ℝⁿ Yes, I agree this is duplicate. It didn't turn up somehow. Thanks! – Inquest Feb 8 '13 at 7:13

## 2 Answers

A non-decreasing function can only have jump discontinuities; that is, at a discontinuity the left and right limits still must exist (since bounded, increasing sequences of numbers have limits), but are different. Each jump means there is an interval missing from the range of the function, and these intervals are all disjoint (since the function is non-decreasing, the skipped intervals march up the y-axis). There can be at most countably many discontinuity points since there are at most countably disjoint intervals (to see this last bit, just pick a rational number in each each interval).

• Why must the limits be different for a nondecreasing function. It's clear if the function is strictly increasing, but a nondecreasing function with a discontinuity could be a horizontal line with a bunch of holes poked in it. Are you implicitly assuming that the function is total? – Jonathan Gallagher Oct 26 '15 at 1:22
• @JonathanGallagher We're assuming the function is defined in an interval (which could be whole $\mathbb{R}$). Hence, you cannot have a horizontal line with holes poked in it. – EA304GT Oct 4 '16 at 23:00

If you look at how many jumps are larger than $\frac{1}{k}$ in the interval $[n,n+1)$ for any integer $n$ and positive integer $k$, this must be finite as otherwise the function becomes infinite in that interval.

So the number of jumps in the interval $[n,n+1)$ must be finite or countable as the limit of a finite sequence. So the total number of jumps on the real line must be finite or countable as a countable sum.