Showing the Cumulative distribution only has countable plateaus

Question

Let $F$ be a cumulative distribution function. Show that $F$ only has countably many plateaus.

My idea: Define $A_{n}:=\{[a,b]\subseteq \mathbb R: [a,b]$ is a plateau of length $\geq \frac{1}{n}\}$

I want to prove $|A_{n}|<\infty$, so I assume $|A_{n}|=\infty$

This means there exists $([a_{i},b_{i}])_{i\in \mathbb N}$ so that $\lambda([a_{1},b_{1}])\leq...\leq\lambda([a_{k},b_{k}])$

But I do not know whether I am on the right track here, and how to continue

The Poisson distribution CDF has $\infty$ plateaus of length $1$, so you won’t be proving $|A_n| < \infty$. Perhaps you should begin by reviewing the definition of countable? — Anders Kaseorg, Feb 10 '19 at 20:18
https://math.stackexchange.com/questions/1413874/can-we-add-an-uncountable-number-of-positive-elements-and-can-this-sum-be-finit — BGM, Feb 10 '19 at 20:26

score 1 · Accepted Answer · 2019-02-10T22:42:06.103

1

Let $G(q):=\inf\{x\in \mathbb{R}:F(x)\ge q\}$, $0<q<1$ be the corresponding quantile function. A flat region of $F$ corresponds to a jump of $G$. Since $G$ is non-decreasing on $(0,1)$, it can have at most countable number of jumps.

edited Feb 10 '19 at 22:42

answered Feb 10 '19 at 22:18

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That last result is because whenever $L_+=\lim_{x \to x_0^+} G(x)$ is different from $L_-=\lim_{x \to x_0^-} G(x)$, there is a rational number in the interval $(L_-,L_+)$ which isn't in the range of $G$. And the monotonicity means this rational number can't be in any of the other "jump intervals" either. – Ian Feb 10 '19 at 22:46

score 1 · Answer 2 · answered Feb 10 '19 at 22:50

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To each plateau you can associate a rational number (namely, some rational number in the interval where your function takes a constant value). These rational numbers are different for each plateau. Therefore, there is an injective mapping from the set of plateaus into the rational numbers. As a consequence, the number of plateaus is at most countable.

Actually, this property does not have anything to do with the non-decreasing nature of the cdf.

answered Feb 10 '19 at 22:50

GReyes

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Yes it does, because you used the monotonicity to ensure that the rational numbers are different for all the plateaus. It takes a different argument to show that $\mathbb{R}$ cannot be decomposed into an uncountable union of disjoint intervals. – Ian Feb 11 '19 at 00:35
The rational numbers I am choosing are not the values of $F$, but some rational number on the intervals where $f$ takes a constant value. It is obviously true that you cannot represent $\mathbb{R}$ as an uncountable union of intervals because all intervals contain a rational number so you cannot have more intervals than rational numbers in $\mathbb{R}$. – GReyes Feb 11 '19 at 03:05
Sorry, I was stuck in the idea for the case of countably many jumps, where the monotonicity is needed in order to ensure that the relevant rational numbers are distinct. – Ian Feb 11 '19 at 03:36

Showing the Cumulative distribution only has countable plateaus

2 Answers2