Consider any measure $\mu\in\Delta(\mathbb{R})$ and let $F_{\mu}\in\mathbb{R}^{[0,1]}$ be the distribution function induced by $\mu$.
We know that if $F_{\mu}$ is continuous, then $\text{supp}(\mu)$, the support of the measure $\mu$, is a perfect set in $\mathbb{R}$.
Question: Under what conditions it is possible to say that $\text{supp}(\mu)$ is connected?
A cantor distribution is continuous, has a perfect support, but not a connected one. Nevertheless, it is not absolutely continuous. Is absolutely continuity (w.r.t. to the Lebesgue measure) sufficient to ensure a connected support?
Since $\text{supp}(\mu)$ is closed, it is disconnected iff there is an entire interval in its complement which is between two points of $\text{supp}(\mu)$. An interval in its complement is just an interval on which $F_\mu$ is constant, and being between two points of $\text{supp}(\mu)$ just means that $F_\mu$ changes both before and after that interval, or equivalently that the constant value on the interval is not $0$ or $1$. So $\text{supp}(\mu)$ is connected iff there does not exist any (nondegenerate) interval $I\subset\mathbb{R}$ such that $F_\mu$ is constant on $I$ with value different from $0$ and $1$.
Another equivalent condition would be that if $a=\inf\{t:F_\mu(t)>0\}$ and $b=\sup\{t:F_\mu(t)<1\}$, then $F_\mu$ is strictly increasing on the interval $[a,b]$, since if it were not strictly increasing it would have to be constant on some subinterval.