Why can we rewrite an event $\{\sup_{n\geq N}W_{n}>t\}=\bigcup\limits_{n\geq N}\{W_{n}>t, \sup_{N\geq k\geq n}W_{k}\leq t\}$?

Question

I would like to understand, why $\{\sup_{n\geq N}W_{n}>t\}=\bigcup\limits_{n\geq N}\{W_{n}>t, \sup_{N\geq k\geq n}W_{k}\leq t\}$ true is? $W_{n}$ is a sequence of random variables, so that $W_{n}$ converges almost surely to $W$. Thank you for your help!

Answer 1

The second sup should read $N\leq k<n$, else this wouln’t make sense.

$\sup_{n\geq N} W_n>t$ if and only if $W_n>t$ for some $n\geq N$. Thus $$ \{\sup_{n\geq N} W_n > t\} = \bigcup_{n\geq N} \{ W_n > t \} $$

Now call $E_n:=\{W_n>t\}$. Then if $\omega \in E_n$, but for some $k<n$ we have $W_k(\omega)>t$, then (and exactly then) is $\omega$ already in $E_k$. Thus $E_{N+1}\setminus E_N = \{W_{N+1}>t, W_N \leq t\}$, $E_{N+2}\setminus (E_{N+1}\cup E_N) = \{W_{N+2} >0, W_{N+1}\leq t, W_N\leq t\}$ and generally $$ E_n \setminus \bigcup_{N\leq k<n} E_k = \{W_n>t, W_N,\ldots,W_{n-1}\leq n\} = \{W_N>t, \sup_{N\leq k< n} W_k \leq t\} $$

So we have $$ \{\sup_{n\geq N} W_n > t\} = \bigcup_{n\geq N} E_n = \bigcup_{n\geq N} \left(E_n \setminus \bigcup_{N\leq k<n} E_k\right) = \bigcup_{n\geq N} \{W_N>t, \sup_{N\leq k< n} W_k \leq t\} $$ where this last union is a union of pairwise disjoint sets.