I'm trying to prove this property. Could you have a check on my attempt?
Let $\mu$ be a Borel probability measure on $\mathbb R$ and $F$ its c.d.f. Then $F$ is right-continuous and non-decreasing.
Theorem: If $\mu$ is atomless then $F$ is continuous.
My attempt: Assume the contrary that $F$ is not continuous at $a \in \mathbb R$. Then $F$ is not left-continuous at $a$. Then there is a sequence $(x_n)$ such that $x_n \neq a$, $x_n \nearrow a$, and $F(x_n) \not\to F(a)$. Because $F$ is non-decreasing, there is $\alpha < F(a)$ such that $F(x_n) \nearrow \alpha$.
By continuity of measure from below $$ \mu ((-\infty, a)) = \lim_n \mu ((-\infty, x_n]) = \lim_n F(x_n) =\alpha. $$
It follows that $$ \mu (\{a\}) = \mu ((-\infty, a]) - \mu ((-\infty, a)) = F(a)- \alpha >0. $$
Hence $a$ is an atom, which is a contradiction.
