The derivative of a discrete distribution function is zero a.e.? Suppose $\{q_n\}$ is an arbitrary enumeration of all rationals, and $$F(x)=\sum_{n=1}^{\infty}\frac1{2^n}\mathbf 1_{\{q_n\le x\}}$$  Then $F$ is a discrete distribution function with countable jump discontinuities (that are in $\mathbb Q$), but is continuous on $\mathbb {R\setminus Q}$, as shown in About the continuity of the function $f(x) = \sum\limits_k2^{-k}\mathbf 1_{q_k \leq x}$.  
My question is: Is $F'(x)=0$ for every $x\notin \mathbb Q$?  If so, how do we prove it?

Not necessarily. Assume for example that $$\pi<q_{2n}<\pi+5^{-n}$$ for every $n$ (note that enumerations with this property do exist since the subsequence $(q_{2n+1})$ is free), then $$F(\pi+5^{-n})>F(\pi)+4^{-n}$$ hence $$\frac{F(\pi+5^{-n})-F(\pi)}{5^{-n}}>\left(\frac54\right)^n$$ which implies $$\lim_{x\to0,x>0}\frac{F(\pi+x)-F(\pi)}x=+\infty$$
