Definition: Let $\mathcal{F}_t$ be a filtration. A mapping $T:\Omega\rightarrow\mathbb{R}_{\geq 0}$ is called a stopping time, if $\{T\leq t\}\in\mathcal{F}_t$ for all $t\geq 0$. A stopping time $T:\Omega\rightarrow\mathbb{R}_{\geq 0}$ is called predictable, if the set $\{(\omega,t)\in\Omega\times\mathbb{R}_{\geq 0}|T(\omega)>t\}$ is an element of the predictable $\sigma$-algebra.
There is the following theorem:
If $(T_n)_{n\in\mathbb{N}}$ is a sequence of predictable stopping times and $\Omega=\bigcup_{m\in\mathbb{N}}\{\bigwedge_{n\in\mathbb{N}}T_n=T_m\}$, then $\bigwedge_{n\in\mathbb{N}}T_n$ is a predictable stopping time.
There is the following example, why $\bigwedge_{n\in\mathbb{N}}T_n$ is not necessarily a predictable stopping time, when $\Omega=\bigcup_{m\in\mathbb{N}}\{\bigwedge_{n\in\mathbb{N}}T_n=T_m\}$ is not fulfilled: Let $S$ be a stopping time, which is not predictable and define $T_n=S+\frac{1}{n}$ for $n\in\mathbb{N}_{> 0}$. Then $T_n$ is a predictable stopping time for each $n\in\mathbb{N}$ but $S=\bigwedge_{n\in\mathbb{N}} T_n$ is not predictable by assumption.
My problem is to see why $\bigcap_{n\in\mathbb{N}}\{(\omega,t)\in\Omega\times\mathbb{R}_{\geq 0}|T_n(\omega)>t\}=\{(\omega,t)\in\Omega\times\mathbb{R}_{\geq 0}|\bigwedge_{n\in\mathbb{N}}T_n(\omega)>t\}$ is true when $\Omega=\bigcup_{m\in\mathbb{N}}\{\bigwedge_{n\in\mathbb{N}}T_n=T_m\}$ holds and not true otherwise.