Let $p\_1, ... ,p\_n$ be chosen independently from the uniform distribution on the unit torus $[0,1]^2$.


I want to prove a theorem of the form: "With high probability, every circle of radius $r$ contains approximately $\pi r^2 n$ points".


The problem is that there are infinitely many circles, so proving concentration for a single circle and then union bounding won't work. How do I do this?



Union bound over infinitely many events
$\newcommand\C{\mathscr C}\newcommand\ep{\varepsilon}$Let $\C$ denote the set of all disks of a radius $r\in(0,\infty)$ contained in the unit square. Using a rectangular grid of centers of disks in $\C$, we can cover $\C$ by $N=O(r^2/\ep^2)$ balls of radius $\ep\in(0,\infty)$, where the balls are considered with respect to the distance between disks equal the Lebesgue measure of the symmetric difference between the disks and the constant in $O(\cdot)$ is universal.


Hence, [Talagrand's Theorem 1.1 for empirical measures](https://www.jstor.org/stable/2244494?seq=1) (used here with $v=2$) implies the following:
$$P\Big(\sup\_{C\in\C}\Big|\sum\_1^n 1(p\_j\in C)-\pi r^2 n\Big|\ge z\sqrt n\Big)
\le K(r)z^3e^{-2z^2} \tag{1}$$
for all real $z>0$, where $K(r)\in(0,\infty)$ depends only on $r$ (in a way that can be found from the proof).


Obviously, a condition of the form $z\ge c$ with a universal constant $c\in(0,\infty)$ is missing here; cf. e.g. [Azencott and Vayatis, p. 564](https://www.sciencedirect.com/science/article/abs/pii/S0764444201018602).


