I am trying to understand a piece of this proof (of a version of Helly's theorem). I understood everything except the passage in which the author says that $F(x)$ is right continuous. How can I prove that? Is it so clear from the contruction he made? Also, is it true that $F$ is continuous on rationals? And why? Thank you.
Since we know that $F$ is increasing, to prove that $F$ is right continuous, it is sufficient to show that for any $x \in \mathbb{R}$ and for any $\epsilon > 0$ there exists $ y > x $ such that $F(y) < F(x) + \epsilon$. So take $x\in \mathbb{R}$ and $\epsilon >0$. Then by definition we can find $y \in \mathbb{Q}$ for which $x < y$ and $F(y) < F(x) + \epsilon$. Therefore $F$ is right continuous. For the second question $F$ is not necesarrily continuous at rationals. For instance let $\mu$ be a Borel probability measure on $\mathbb{R}$ supported on $\{0\}$, and $\mu_n = \mu$ for all $n \in \mathbb{N}$. In this case $F = 0$ on $(-\infty, 0)$ and $F=1$ on $[0, \infty)$. Thus $F$ is not continuous at $0$.