Consider a distribution function $F:\mathbb{R}\rightarrow [0,1]$ definining the positive, finite measures $\mu_F$ determined by $$ \mu_F((a,b])\equiv F(b)-F(a) $$ for each $a,b\in \mathbb{R}$ with $b> a$.
Suppose $F$ is such that $\mu_F((1,3])=0$.
Can $F$ still be a continuous distribution (i.e., a continuous function) in the sense outlined here?
(Note, I'm asking whether $F$ can be a continuous distribution. I'm not referring to the concept of absolute continuity)
We can make lots of examples by taking mixtures of compactly supported distributions where the supports are some positive distance apart.
For example, let $f_n = 2 \mathbf 1_{[n, n + 1/2]}$ so $f_n$ is the density of a uniform RV on $[n, n + 1/2]$ (I could have done this with translated beta distributions too, among many others). Then if I take $$ f =\sum_{n=0}^\infty f_n 2^{-n-1} $$ (where I'm using $2^{-n-1}$ as the mixing probability, but any other distribution over $\mathbb N$ could be used) I have a continuous density over $\mathbb R$. The corresponding CDF will be flat on $(n+1/2, n+1)$ for every $n\in\mathbb N$ since the density there is zero so the integration adds nothing. But this is a perfectly valid distribution and is continuous.
Here's what this looks like (just the beginning at least). Not only does the CDF have some flat regions but it has a countable infinity of them.
