I understand that if for a real-valued random variable $X$ we have $X = O_p(1)$, then it means that for any $\epsilon>0$, there exists a positive real number $M>0$ such that $p(|X|>M)<\epsilon$. (Please correct me if I am wrong.) However, I read the following result which I have no idea what it meant.
Consider the metric space $(\mathcal P, \delta)$ of all probability distributions on $(\mathcal X, \mathcal B)$, with $\delta$ satisfying $$ \sqrt n\delta(\mathbb P_n, \mathbb P) = O_p(1)\; n \to \infty, (*) $$ where $\mathbb P_n$ is the empirical probability distribution of the random sample $(X_1, \dots, X_n)$. In particular, it is claimed that the Kolmogorov distance, $\delta(F,G):=\sup_x|F(x) -G(x)|$, satisfies this condition.
My question is what $(*)$ means. I guess that I should treat $\sqrt n\delta(\mathbb P_n, \mathbb P)$ as a random variable because for a different sample we would end up with a different $\mathbb P_n$. However, I still do not get this condition. Could anyone clarify this, please? How to show this for Kolmogorov distance? Thank you!