Consider the set of non-negative real numbers $\mathbb{R}_{\geq 0}$ as a measurable space with the Borel $\sigma$-algebra $\mathbb{B}$. Let $\delta_{0}(E) = \begin{cases} 1 & \text{if } x \in E \\ 0 & \text{otherwise}\end{cases}$ be the Dirac measure at $0$.
Then $(\mathbb{R}_{\geq 0}, \mathbb{B}, \delta_0)$ is a measure space. Given any measure $\mu$, the CDF $F_\mu$ of $\mu$ is defined as $F_\mu(x) = \mu((-\infty,x])$, so for the Dirac measure we get $F_{\delta_0}(x) = 1$ for all $x \geq 0$, in particular for $x = 0$. So the CDF of the Dirac measure violates my usual intuition that a CDF on the non-negative reals starts in $0$ and then increases monotonically, tending towards $1$, because the CDF of the Dirac measure is positive in $0$.
My question is: Is the CDF of the Dirac measure the only CDF $F$ such that $F(0) > 0$? In particular, given any number $0 < r < 1$, is there a CDF $F$ such that $F(0) = r$?
A CDF is a non-decreasing and right-continuous function $F:\mathbb R\to\mathbb R$ with: $$\lim_{x\to-\infty}F(x)=0\text{ and }\lim_{x\to+\infty}F(x)=1$$
So evidently for every $r\in[0,1]$ there are lots of CDF's with $F(0)=r$.
This remains true under the extra condition that $F(x)=0$ for negative $x$, except for $r=1$.
Exactly one CDF with $F(x)=0$ for negative $x$ also satisfies $F(0)=1$
Every CDF determines a probability measure and the CDF with $F(x)=0$ for $x<0$ and $F(0)=1$ determines the Dirac measure at $0$.