Cumulative Distribution Function that is Discontinuous at All Rationals I'm completely lost about how to construct this function.  I write out the definition to be that it must be some function $F: \mathbb{R} \rightarrow \mathbb{R}$ that satisfies:
(1) $F$ is non decreasing
(2) lim$_{x \rightarrow \infty} F(x) = 1$, lim$_{x \rightarrow -\infty} F(x) = 0$
(3) $F$ is right continuous
(4) $F$ is continuous at all irrationals and discontinuous at all rationals
I need something that has "large jumps" to the left and "small jumps" to the right of every rational to make it right continuous but not left continuous at those points...but then how can it be both left and right continuous at every irrational?  I'm horribly confused.  What I thought about so far (with the help of other stackexchange questions) is a function like this:
$\sum_{i=1}^{\infty} 2^{-i} \mathbb{I}(x-q_i > 0)$, where $\{ q_i \}_{i \in \mathbb{N}}$ are the set of rational numbers and $\mathbb{I}$ is the indicator function.
Can someone help me prove the properties, if this function satisfies them?
 A: You just need to come up with a distribution that puts positive mass on every rational.
Let $(r_n)$ denote an enumeration of the rationals. Consider $(a_n)$ a sequence of positive reals such that $\sum_{n=1}^\infty a_n  = 1$. Let $X$ be a random variable with distribution $\sum_{n=1}^\infty a_n \delta_{r_n}$.
Then $F_X$, the cdf of $X$, is discontinuous at every rational. To be more specific,  if $r_{n_0}\in \mathbb Q$, then $$F_X(r_{n_0})-\lim \limits_{x\to r_{n_0}^-}F_X(x) = a_{n_0}$$

The following results on cdfs are well-known.
$F_X$ is cadlag (it is right-continuous and has left-sided limits at every point). Besides, $$\lim \limits_{x\to x_0^-}F_X(x) = P(X<x_0)$$
Here's a proof of the last equality: since the left-sided limit exists, we have $$\lim \limits_{x\to x_0^-}F_X(x) = \lim\limits_{n\to \infty}F_X(x_0-\frac 1n)= \lim\limits_{n\to \infty}P(X\leq x_0-\frac 1n)$$
The sequence of events $(X\leq x_0-\frac 1n)$ is increasing, so $$\lim\limits_{n\to \infty}P(X\leq x_0-\frac 1n) = P(\bigcup_n (X\leq x_0-\frac 1n))=P(X<x_0)$$
Thus $\lim\limits_{x\to x_0^-}F_X(x) = F_X(x)-P(X=x_0)$. 
$F$ is continuous at $x_0$ if and only if $\lim\limits_{x\to x_0^-}F_X(x) = F_X(x)$, i.e. $P(X=x_0)=0$. 
In your case, if $x_0$ is irrational, $P(X=x_0)=0$, hence $F_X$ is continuous at $x_0$.
A: If $\mu_{\#}$ is the counting measure on $\mathbb{R}$, choose $f=\sum_ia_i\chi_{\{r{i}\}}$, where $\sum_i a_i=1$ and $\{r_i\}_i$ an enumeration of $\mathbb{Q}$.
Note that $F(x)=\int_{(-\infty,x]}f(s)\mu_{\#}(ds)=\sum_i a_i \chi_{[r_i,\infty)}(x).$
Clearly
$$\lim_{t\to x^+}F(t)=F(x)$$
and
$$\lim_{t\to x^-}F(t)+\mu_{\#}(\{x\})f(x)=F(x)$$
i.e. 
$$F\,\mbox{is continuous at x}\Leftrightarrow f(x)=0\Leftrightarrow x\in \mathbb{R}\setminus \mathbb{Q}.$$
