I would like some help to understand a step in a proof or/and a proof verification:
I've tried doing it by myself but I can't justify exactly the same as in the proof (is mine correct as well?)
Setup: For $t\in\left[\frac{k-1}{n},\frac{k}{n}\right]$ where $k$ is the unique integer such that $k-1\leq nt<k$, and $T_k$ are increasing $T_k\geq T_{k-1}$ random variables (stopping times here) of expectations $k$.
Here we have take $\delta\in(0,1)$ and $n>\delta/2$.
The step in the proof I dont understand:
$$\mathbb P\left(\exists t\in [0,1):\max\left(\left|\frac{T_k}{n}-t\right|,\left|\frac{T_{k-1}}{n}-t\right|\right)\geq\delta\right)\leq\mathbb P\left(\sup_{1\leq k\leq n}\frac{\max(T_k-(k-1),k-T_{k-1})}{n}\geq\delta\right)$$ $$\leq \mathbb P\left(\sup_{1\leq k\leq n}\frac{T_k-k}{n}\geq \delta/2\right)+\mathbb P\left(\sup_{1\leq k\leq n}\frac{k-1-T_{k-1}}{n}\geq\delta/2\right)$$
I've tried developing the maximum as $\frac{1}{2n}\left(|T_k-nt|+|T_{k-1}-nt|+|T_k-T_{k-1}|\right)$ and other reasonings but none lead me to the same result. The further I've gotten is:
My proof: $$\{\exists t\in [0,1]:\max\left(\left|\frac{T_k}{n}-t\right|,\left|\frac{T_{k-1}}{n}-t\right|\right)\geq\delta\}=\bigcup_{k=1}^n\{\exists t\in [(k-1)/n,k/n]: \max\left(\left|\frac{T_k}{n}-t\right|,\left|\frac{T_{k-1}}{n}-t\right|\right)\geq\delta\}=(\star)$$ Using the fact that $T_i$ is increasing and decomposing the absolute value, I obtain that $$\mathbb P(\star)\leq\mathbb P\left(\sup_{1\leq k\leq n}\max\left(\frac{T_k-(k-1)}{n},\frac{k-T_{k-1}}{n}\right)\geq\delta\right)$$ $$\leq \mathbb{P}\left(\sup_{1\leq k\leq n}\frac{T_k-(k-1)}{n}\text{ or }\sup_{1\leq k\leq n}\frac{k-T_{k-1}}{n}\geq\delta\right)$$ And using the union bound for measures: $$\leq \mathbb P\left(\sup_{1\leq k\leq n}\frac{T_{k-1}-(k-1)}{n}\geq \delta\right)+\mathbb P\left(\sup_{1\leq k\leq n}\frac{k-T_k}{n}\geq\delta\right)$$
The proof is by Peter Mörters and Yuval Peres on Brownian motions.
NB: I have seen a similar question asked here but the answer has not been approved and has me puzzled.
Thanks in advance for the proof verification or the help understanding the steps.
Yes, that is correct. In the end there is one extra step where the inequality $n >\delta/2$ is used.