Lemma in Donsker's Theorem in Moerters/Peres In Lemma 5.24 (S.132) in Moerters/Peres "Brownian Motion" as a precursor for the theorem of Donsker Principle.
They prove a version of the Skorokhod embedding theorem. After having constructed a sequence of stopping times $0=T_0\leq T_1\leq ...$ there is in the end a series of inequalities that i haven't made much progress with.
They state for $nt\in[(k-1),k)$ and $n>2/\delta$ with $\delta>0$
$$\mathbb P(\exists t\in[0,1]:|T_k/n-t|\vee |T_{k-1}/n-t|\geq \delta)
\leq \mathbb P\Big(\sup_{1\leq k\leq n}\frac{(T_k-(k-1))\vee (k-T_{k-1})}{n}\geq \delta\Big)$$
Can somebody help me understand this inequality?
EDIT: I was able to identify the reason. If $A_k=\{\exists t\in[0,1]:|T_k/n-t|\vee |T_{k-1}/n-t|\geq \delta\}$, then
$$A_k\subset\bigcup_{i=1}^nA_i=\sup_{1\leq k\leq n}A_k.$$
Then, the proof proceeds by estimating further
$$...\leq \mathbb P\Big(\sup_{1\leq k\leq n}\frac{T_k-k}{n}\geq \delta/2\Big)+\mathbb P\Big(\sup_{1\leq k\leq n}\frac{(k-1)-T_{k-1}}{n}\geq \delta/2\Big).$$
I think this was due to the general property, as for $X$, $Y$ r.v.s
$$\mathbb P(\max(X,Y)\geq t)=\mathbb P(\{X\geq t\}\cup\{Y\geq t\})\\=\mathbb P(X\geq t)+\mathbb P(Y\geq t)-\mathbb P(\{X\geq t\}\cap\{Y\geq t\}).$$
However, I find the $\delta/2$ still confusing. Also, the absolute has vanished.
 A: Writing
$$
\left\{\exists t\in[0,1]:|T_{\lfloor nt\rfloor+1}/n-t|\vee |T_{\lfloor nt\rfloor-1}/n-t|\geqslant \delta\right\}
=\bigcup_{k=1}^n\left\{\exists t\in[(k-1)/n,k/n]:|T_k/n-t|\vee |T_{k-1}/n-t|\geqslant \delta\right\}
$$
we are reduced to show that for each $k\in\{1,\dots,n\}$, the inclusion
$$
\left\{\exists t\in[(k-1)/n,k/n]:|T_k/n-t|\vee |T_{k-1}/n-t|\geqslant \delta\right\}\subset \left\{ \frac{(T_k-(k-1))\vee (k-T_{k-1})}{n}\geqslant \delta\right\}.
$$
To do so, we can decompose $\left\{\exists t\in[(k-1)/n,k/n]:|T_k/n-t|\vee |T_{k-1}/n-t|\geqslant \delta\right\}$ as $A_1\cup A_2\cup A_3\cup A_4$, where
$$
A_1=\left\{\exists t\in[(k-1)/n,k/n]: T_k/n-t  \geqslant \delta\right\}\\
A_2=\left\{\exists t\in[(k-1)/n,k/n]: t-T_k/n  \geqslant \delta\right\}\\
A_3=\left\{\exists t\in[(k-1)/n,k/n]: T_{k-1}/n-t  \geqslant \delta\right\}\\
A_4=\left\{\exists t\in[(k-1)/n,k/n]: t-T_{k-1}/n  \geqslant \delta\right\}.
$$
Note that since $T_{k-1}\leqslant T_k$, we have $A_3\subset A_1$ and $A_2\subset A_4$ hence we are reduced to show that $A_1 \subset  \left\{ \frac{(T_k-(k-1))\vee (k-T_{k-1})}{n}\geqslant \delta\right\}$, $A_4 \subset  \left\{ \frac{(T_k-(k-1))\vee (k-T_{k-1})}{n}\geqslant \delta\right\}$ which follows the fact that $t\mapsto T_k/n-t$ and $t\mapsto t-T_{k-1}/n$ are respectively decreasing and increasing.
For the last inequality, we use the fact that $1/n<\delta/2$ hence $(k-T_{k-1})/n\leqslant (k-1)/n-T_{k-1}/n+\delta/2$
hence
$$
\max_{1\leqslant k\leqslant n}\frac{(T_k-(k-1))\vee (k-T_{k-1})}{n}
\leqslant \frac{\delta}2+\max_{1\leqslant k\leqslant n}\frac{(T_{k-1}-(k-1))\vee (k-T_{k-1})}{n}
$$
then conclude by the inequality mentioned in the opening post.
