Let
- $T>0$
- $I:=[0,T]$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\in I}$ be a filtration of $\mathcal A$, $$\mathcal R:=\bigcup_{F\in\mathcal F_0}F\times\left\{0\right\}\cup\bigcup_{0\le s<t\le T}\bigcup_{F\in\mathcal F_s}F\times(s,t]$$ and $$\mathcal P:=\sigma(\mathcal R)$$
- $\tau:\Omega\to I$ be an $\mathcal F$-stopping time
Are we able to show that $$X(\omega,t):=1_{\left\{\:t\:\le\:\tau(\omega)\:\right\}}\;\;\;\text{for }(\omega,t)\in\Omega\times I$$ is $\mathcal P$-measurable?
Since $$\left\{\omega\in\Omega:t\le\tau(\omega)\right\}=\Omega\setminus\left\{\omega\in\Omega:\tau(\omega)<t\right\}\in\mathcal F_t\;\;\;\text{for all }t\in I\;,\tag1$$ it's clear that $(X(\;\cdot\;,t))_{t\in I}$ is $\mathcal F$-adapted. The desired claim should be equivalent to $$[0,\tau]:=\left\{(\omega,t)\in\Omega\times I:t\le\tau(\omega)\right\}\in\mathcal P\;.\tag2$$ How can we prove that?
For simplicity of notation we consider only $T=1$. If we define a sequence of stopping times by
$$\tau_n := \frac{ \lfloor 2^n \tau \rfloor +1}{2^n}$$
then $\tau_n \downarrow \tau$ and $\{\tau_n \leq k 2^{-n}\} \in \mathcal{F}_{k2^{-n}}$. In particular,
$$[0,\tau] = \bigcap_{n \geq 1} [0,\tau_n]$$
and therefore it suffices to show that $[0,\tau_n]$ is predictable for all $n \geq 1$. To this end, note that
$$\tau_n(\omega)> (k-1)2^{-n} \implies \tau_n(\omega) \geq k2^{-n}$$
and so
$$[0,\tau_n] = \left( \{0\} \times \Omega \right) \cup \left( \bigcup_{k=1}^{2^n} ((k-1)2^{-n},k2^{-n}] \times \{\tau_n>(k-1)2^{-n}\} \right).$$
As
$$\{\tau_n > (k-1)2^{-n}\} = \{\tau_n \leq (k-1)2^{-n}\}^c \in \mathcal{F}_{(k-1)2^{-n}}$$
this proves that $[0,\tau_n]$ is predictable.