# Bounding in Glivenko-Cantelli theorem

## Problem

Let $$X_1, X_2, \cdots, X_n$$ be iid random variables. The cdf and empirical cdf are $$F(t)=P[X\leq t]$$ and $$\hat{F}_n(t)=\frac{1}{n} \sum_{i=1}^n 1(X_i\leq t)$$. The Glivenko-Cantelli theorem says,

$$\sup_{t\in\mathbb{R}}\vert \hat{F}_n(t)-F(t)\vert \rightarrow 0\ a.s.$$

Then prove that with probability at least $$(1-\delta)$$ $$\sup_{t\in\mathbb{R}}\vert \hat{F}_n(t)-F(t)\vert\leq 2\sqrt{\frac{2\log(n+1)}{n}} +\sqrt{\frac{1}{2n}\log \frac{1}{\delta}}$$

## What I Have Done

Denote $$\Delta s(t) := \hat{F}_n(t)-F(t)$$, then by union bound, we have $$P[\sup_{t\in\mathbb{R}}\vert \Delta s(t)\vert - \mathbb{E}[\vert \Delta s(t)\vert]\geq \alpha]\leq P[\sup_{t\in\mathbb{R}}\Delta s(t)-\mathbb{E}[\Delta s(t)]\geq \alpha]+ P[\sup_{t\in\mathbb{R}}\mathbb{E}[\Delta s(t)]-\Delta s(t)\geq \alpha]$$

The two terms on LHS could be easily bounded using MacDiarmid inequality with $$\exp(-2n\alpha^2)$$, then we have $$P[\sup_{t\in\mathbb{R}}\vert \Delta s(t)\vert - \mathbb{E}[\vert \Delta s(t)\vert]\geq \alpha] \leq 2\exp(-2n\alpha^2)$$

However, if we set RHS with $$\delta$$, then we will end up getting $$\sqrt{\frac{1}{2n}\log \frac{2}{\delta}}$$. I am not sure how to get rid of this and get the required form?

Could anyone help me, thank you in advance!