Consider a function $F : \mathbb{R} \rightarrow [0,1]$ with the following $3$ properties:
- $F$ is monotonically increasing,
- $F$ is right-continuous and
- $\lim_{x \rightarrow -\infty} F(x) = 0$ and $\lim_{x \rightarrow \infty} F(x) = 1.$
Such a function $F$ is called a cumulative distribution function. Every $F$ defines a probability distribution $P_F$ on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ via $$ P_F((-\infty,t]) :=F(t) $$ On the other hand, every probability measure $P$ on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ defines a cumulative distribution function $F_P(t)$ via $$ F_P(t) := P((-\infty,t]).$$ So we can see, that there is a bijection between all probability distributions on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ and all cumulative distribution functions.
Now let $(q_n)_{n \in \mathbb{N}}$ be an enumeration of all rational numbers. Then we can define a probability measure $P_{\mathbb{Q}}$ on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ by demanding $P_{\mathbb{Q}}(q_n) = \frac{1}{2^n}$ and $P_{\mathbb{A}} (x) = 0$ if $A \subseteq \mathbb{R} \setminus \mathbb{Q}$.
My question is now: how can the corresponding cumulative distribution function $F_{P_{\mathbb{Q}}}$ be described? The question is puzzling to me; I cannot possibly imagine, that such a function can be (say) right-continuous.
I am happy to hear any thoughts on this!
Kind regards, Joker