Understanding dense discontinuities

Question

I am reading Durrett's text on probability and stumbled upon the dense discontinuities, which is example 1.2.6. This example says that we first enumerate the rationals $q_1,q_2,...$. We consider some variable $\alpha_i > 0$ such taht $\sum_{i=1}^{\infty} \alpha_i = 1$ and so we have the distribution function $$F(x) = \sum_{i=1}^{\infty}\alpha_i 1_{[q_i,\infty)}(x)$$ where $1_{[\theta,\infty)}(x) = 1$ if $x \in [\theta,\infty)$ and $0$ otherwise. I am having a hard time wrapping my qhead around this function. Why is the text calling it dense discontinuities, I don't see it from the formula. Also, what is the main importance of this function or why is it particularly interesting. Finally, how do we know what discrete probability measure is this function related to. Thanks for your help.

Answer 1

In Durrett's Probability: Theory and Examples, the paragraph above Example 1.2.6 (i.e. this example) reads

In Section 1.6, we will see the Bernoulli, Poisson, and geometric distributions. The next example shows that the distribution function associated with a discrete probability measure can be quite wild.

From this, I think it is reasonable to assume that Durrett doesn't consider this example to be particularly useful for anything, only that it is interesting in it's pathology. It is an example of how badly a function can misbehave, yet still be tractable with the tools that are going to be developed later on.

To understand the example a little better, let's start simple. Consider the distribution function $$ F(x) = 1_{[0,\infty)}(x) = \begin{cases} 0 & \text{if $x<0$, and} \\ 1 & \text{if $x\geq 0$.} \end{cases} $$ Observe that this function is nondecreasing and takes values between 0 and 1 (inclusive), hence defines a distribution function as claimed. Moreover, it has a single discontinuity at $0$.

Generalizing just a little bit, suppose that $a_1 + a_2 = 1$, and that $x_1<x_2\in\mathbb{R}$. Then we can define a distribution function by setting $$ F(x) = a_11_{[x_1,\infty)}(x) + a_21_{[x_2,\infty)}(x) = \begin{cases} 0 & \text{if $x < x_1$,} \\ a_1 & \text{if $x_1 \le x < x_2$, and} \\ a_1 + a_2 = 1 & \text{if $x\ge x_2$.} \end{cases} $$ Again, this defines a distribution function. In this case, the set of discontinuities is $\{x_1,x_2\}$. Even more generally, if $a_1 + \dotsb + a_n = 1$ for some finite set of values $\{a_n\}$. Then the function $$ F(x) = \sum_{k=1}^{n} a_k 1_{[x_k,\infty)}(x) $$ defines a distribution function with discontinuities at $x_k$. This can be shown via a fairly straightforward induction argument.

All of the above is really just prolog which is meant to give some intuition for the more interesting cases. For example, suppose that $\{a_k\}_{k\in\mathbb{N}}, \{x_k\}_{k\in\mathbb{N}} \subseteq \mathbb{R}$ with $a_k > 0$ for each $k$ and $$ \sum_{k=1}^{\infty} a_k = 1. $$ Then the function $$ F(x) = \sum_{k=1}^{\infty} a_k 1_{[x_k,\infty)}(x) $$ is nondecreasing and satisfies $$ \lim_{x\to -\infty} F(x) = 0 \qquad\text{and}\qquad \lim_{x\to \infty} F(x) = 1, $$ and therefore defines a distribution function. On the other hand, it can be shown that $F$ is discontinuous at each $x_k$ (we can't use the induction above, but it can be done via a fairly simple $\varepsilon$—$\delta$ argument), which is a fairly interesting property—the distribution function has infinitely many discontinuities.

Finally, we get to Durrett's example, which is a special case of the example above, but chosen in such a way that the distribution function is discontinuous at every rational number. Since the rational numbers are dense in $\mathbb{R}$, he has constructed a distribution function that has a dense set of discontinuities. If you are used to a world where most of your functions are smooth (not to mention continuous), a pathological example like this can be somewhat surprising.