Let $\mu$ measure of lebesgue stiltjes, ie $\mu([a,b>)=f(b)-f(a)$ where $f$ is nondecreasing function. The set E=[m,n] where exist $m<r<n$ such that $f$ is constant $c_1$ in [m,r] and $f$ is constant $c_2$ in $<r,n]$ then E is one atom.
I wonder if all the atoms will be that way , ie will contain at most one point of discontinuity. I appreciate any reference to it .
You can readily check that any monotonically increasing function has only countably many jump discontinuities.
Thus, the measure induced by a monotonically increasing function has at most countably many "point masses".
For example, the measure induced by $f(x) = \left \lfloor{x}\right \rfloor$ has point masses at the integers.
You can check, say, that in this case, $$\int_{(0,\infty)} 2^{-x} \, d \mu(f(x)) = 1.$$ In your example, the atom occurs precisely at $r$, so $r$ essentially is the only point in the interval where the measure changes. You can show by the definition of the measure induced that $\{r\}$ has non-zero measure on it's own, but $E \setminus \{r\}$ has a different measure.
In the example of the floor function, the sets containing subsets of the integers are the only sets with non-zero measure.