Existence of a weird cumulative function

Question

Does it exist a non discrete random variable with CDF $F$ with the following hypothesis : $F$ is constant in a neighborhood of each point of continuity

I had the idea of Cantor function but it does not work in my case...

Answer 1

A valid answer can depend on your definition of "discrete random variable." Here are two non-equivalent definitions:

Def 1 (image based): A random variable $X:\Omega\rightarrow\mathbb{R}$ is "discrete" if its image set $X(\Omega)$ is a finite or countably infinite set. Here $X(\Omega) = \{X(\omega) \in \mathbb{R}: \omega \in \Omega\}$.

Def 2 (CDF based): A random variable $X:\Omega\rightarrow\mathbb{R}$ is "discrete" if there is a finite or countably infinite set $A\subseteq\mathbb{R}$ such that $P[X\in A]=1$.

Example under definition 1 (image based): Here are two examples of a random variable $X:\Omega\rightarrow\mathbb{R}$ that satisfies $X(\Omega) = \mathbb{R}$ but $P[X=0]=1$. So its CDF is $$ F_X(x) = \left\{\begin{array}{cc} 0 & \mbox{if $x<0$} \\ 1 & \mbox{if $x\geq 0$} \end{array}\right.$$ This random variable does not satisfy Definition 1, but its CDF is constant on a neighborhood of every point at which it is continuous.

Example 1: Define $(\Omega,\mathcal{F}, P)$ by $\Omega=\mathbb{R}$, $\mathcal{F}=2^{\mathbb{R}}$ (the set of all subsets of $\mathbb{R}$), $P:2^{\mathbb{R}}\rightarrow[0,1]$ is $$ P(A) = \left\{\begin{array}{cc} 1 & \mbox{ if $0 \in A$} \\ 0 & \mbox{ else} \end{array}\right.$$ It is easy to check this is a valid probability measure (it satisfies the 3 axioms of probability). Then define $X:\Omega\rightarrow\mathbb{R}$ by $X(\omega)=\omega$. We have $P[X=0]=1$ and $X(\Omega)=\mathbb{R}$.

Example 2: Use $(\Omega, \mathcal{F}, P)=([0,1], \mathcal{B}([0,1]), \lambda)$ where $\mathcal{B}([0,1])$ is the standard Borel sigma algebra on $[0,1]$ and $\lambda$ is the standard Borel measure. Let $C$ be any uncountably infinite Borel measurable subset of $[0,1]$ that has measure 0, such as a Cantor set. There is a theorem that says: Between any two uncountably infinite Borel measurable subsets of $\mathbb{R}$, there is a Borel measurable bijection (and its inverse is also Borel measurable). So let $\phi:C\rightarrow\mathbb{R}$ be a Borel measurable bijection. Define the Borel measurable function $X:\Omega\rightarrow\mathbb{R}$ by $$ X(\omega) = \left\{\begin{array}{cc} \phi(\omega) & \mbox{ if $\omega \in C$} \\ 0 & \mbox{ if $\omega \notin C$} \end{array}\right.$$ Then $X(\Omega)=\mathbb{R}$ and $P[X=0]\geq P[[0,1]\setminus C] = 1$.

Answer 2

This gives a proof sketch in the case you prefer "Definition 2 (CDF based)" from my other answer.

Claim: If random variable $X$ has a CDF $F_X:\mathbb{R}\rightarrow\mathbb{R}$ that is constant on a neighborhood of every point at which it is continuous, then there is a finite or countably infinite set $A\subseteq \mathbb{R}$ such that $P[X \in A]=1$ (so $X$ satisfies Definition 2).

Proof: Let $A\subseteq\mathbb{R}$ be the set of all points at which the CDF is discontinuous. The CDF is nondecreasing so all points in $A$ must be "jump discontinuities" with $P[X=x]>0$ for all $x \in A$. It follows that $A$ is a finite or countably infinite set (since every uncountably infinite set of positive numbers has a countably infinite subset of numbers that sum to $\infty$, whereas we know the sum of the probability masses sums to no more than 1). Note that $$ P[X \in A] = \sum_{x\in A} P[X=x]$$ If we can show $P[X \in A]=1$ then we are done.

Define $C = \mathbb{R}\setminus A$ as the set of points at which $F_X$ is continuous. Since every point in $C$ has a neighborhood in $C$ over which $F_X$ is constant, it can be shown that we can write $C$ as a union of a finite or countably infinite number of disjoint open intervals over which $F_X$ is constant: $$ C = \cup_{i \in D} I_i$$ where $D$ is a nonempty finite or countably infinite set; $I_i$ are open intervals that are disjoint over all $i \in D$; and $F_X$ is constant over each interval $I_i$.

Define $G:\mathbb{R}\rightarrow\mathbb{R}$ by removing all jump discontinuities of $F_X$: $$ G(x) = F_X(x) - \sum_{y \in A} P[X=y]1_{\{y\leq x\}} \quad \forall x \in \mathbb{R}$$ It can be shown that $G$ is continuous. Further, for each $i \in D$, since there are no points in $A$ in the interval $I_i$, and the function $F_X$ is constant over interval $I_i$, then $G$ is constant over the interval $I_i$. So $G$ is continuous everywhere, and $G$ is constant over all $I_i$ intervals. It follows that the cardinality of the image of $G$ is at most countably infinite: $$|G(\mathbb{R})| \leq |D| + |A|$$

Also $$ \lim_{x\rightarrow-\infty} G(x)=0$$ $$\lim_{x\rightarrow\infty}G(x)=1-P[X\in A]$$ If we can show that $G$ is a constant function then it must be $0=1-P[X\in A]$ and we will be done.

Now if $G$ is not a constant function then, since it is continuous, the intermediate value theorem ensures its image is uncountably infinite, a contradicton. So $G$ is a constant function. $\Box$