I am trying to show that$f(s,\omega):=1_{[0,T \wedge \tau(\omega)))(s)}, 0 \le s \le T < \infty$ is progressively measurable.
To do this, I take the usual approximating sequence of stopping times $\tau_k \downarrow \tau$, and try to show that $1_{[0,T \wedge \tau_k (\omega))}(s)$ is progressively measurable. In this case, $T \wedge \tau_k(\omega)$ will take finitely many values $s_1 < \cdots < s_k$. From here, how can I show that this is progressively measurable? I know that I can decompose $1_{[0,T \wedge \tau_k (\omega))}(s)=\sum_j 1_{\{T \wedge \tau_k = T \wedge s_j\}}1_{[0,T \wedge s_j)}$. So we have $\{(s,\omega):1_{[0,T \wedge \tau_k(\omega))}(s)=1\}=\cup_j [0,T \wedge s_j) \times \{T \wedge \tau_k = T \wedge s_j\}$. But I cannot show that $\{(s,\omega):1_{[0,T \wedge \tau_k(\omega))}(s)=1\cap ([0,t] \times \Omega) \in \mathcal{B}[0,t] \otimes \mathscr{F}_t$ for all $t \le T$.
STEP1. Let $\omega \in \Omega$, $s \in {\mathbb R}_{+}$ and $\varepsilon>0$.
(1) If $s \in [0,T \wedge \tau(\omega))$, then we set $\delta:=T \wedge \tau(\omega)-s>0$. Then for any $u \geq 0$ with $0<u-s<\delta$, $$ \left|{\bf 1}_{[0,T \wedge \tau(\omega))}(s)-{\bf 1}_{[0,T \wedge \tau(\omega))}(u)\right|=\left|1-1\right|=0<\varepsilon $$ (Indeed, $u-s<\delta=T \wedge \tau(\omega)-s \Rightarrow u<T \wedge \tau(\omega)$).
(2) If $s \in [T \wedge \tau(\omega),\infty)$, then we choose $\delta>0$ arbitrarily. Then for any $u \geq 0$ with $0<u-s<\delta$, $$ \left|{\bf 1}_{[0,T \wedge \tau(\omega))}(s)-{\bf 1}_{[0,T \wedge \tau(\omega))}(u)\right|=\left|0-0\right|=0<\varepsilon $$ (Indeed, $0<u-s \Rightarrow T \wedge \tau(\omega) \leq s<u$).
Thus a function $s \mapsto {\bf 1}_{[0,T \wedge \tau(\omega))}(s)$ is right-continuous on ${\mathbb R}_{+}$ by (1) and (2).
STEP2. Fix $t \geq 0$.
(3) By STEP1, we first obtain $$ {\bf 1}_{[0,T \wedge \tau(\omega))}(s)= \lim_{n \uparrow \infty}\sum_{k=0}^{2^{n}-1}{\bf 1}_{[0,T \wedge \tau(\omega))}(\frac{k+1}{2^{n}}t){\bf 1}_{(\frac{k}{2^{n}}t,\frac{k+1}{2^{n}}t]}(s), \quad (\omega,s) \in \Omega \times [0,t]. $$
(4) A map $(\omega,s) \mapsto {\bf 1}_{(\frac{k}{2^{n}}t,\frac{k+1}{2^{n}}t]}(s)$ is ${\mathcal F}_{t} \otimes {\mathcal B}([0,t])$-measurable since a map $s \mapsto {\bf 1}_{(\frac{k}{2^{n}}t,\frac{k+1}{2^{n}}t]}(s)$ is ${\mathcal B}([0,t])$-measurable. Indeed, for any $E \in {\mathcal B}({\mathbb R}^{1})$, $$ \left\{(\omega,s) \in \Omega \times [0,t]; {\bf 1}_{(\frac{k}{2^{n}}t,\frac{k+1}{2^{n}}t]}(s) \in E\right\} =\Omega \times \left\{s \in [0,t]; {\bf 1}_{(\frac{k}{2^{n}}t,\frac{k+1}{2^{n}}t]}(s) \in E\right\}. $$
(5) We first obtain $\{\frac{k+1}{2^{n}}t<T \wedge \tau\} \in {\mathcal F}_{t}$ since $T \wedge \tau$ is a stopping time noting ${\mathcal F}_{\frac{k+1}{2^{n}}t} \subset {\mathcal F}_{t}$ (cf. page 23, Ikeda-Watanabe (1989)). Thus a map $(\omega,s) \mapsto {\bf 1}_{[0,T \wedge \tau(\omega))}(\frac{k+1}{2^{n}}t)$ is ${\mathcal F}_{t} \otimes {\mathcal B}([0,t])$-measurable since a map $\omega \mapsto {\bf 1}_{[0,T \wedge \tau(\omega))}(\frac{k+1}{2^{n}}t)={\bf 1}_{\{\frac{k+1}{2^{n}}t<T \wedge \tau\}}(\omega)$ is ${\mathcal F}_{t}$-measurable. Indeed, for any $E \in {\mathcal B}({\mathbb R}^{1})$, $$ \left\{(\omega,s) \in \Omega \times [0,t]; {\bf 1}_{[0,T \wedge \tau(\omega))}(\frac{k+1}{2^{n}}t) \in E\right\} =\left\{\omega \in \Omega; {\bf 1}_{[0,T \wedge \tau(\omega))}(\frac{k+1}{2^{n}}t) \in E\right\} \times [0,t]. $$ Thus a map $(\omega,s) \mapsto {\bf 1}_{[0,T \wedge \tau(\omega))}(s)$ is ${\mathcal F}_{t} \otimes {\mathcal B}([0,t])$-measurable by (3),(4) and (5) (cf. page 30, 31, Williams (1991)). Hence, the map $(\omega,s) \mapsto {\bf 1}_{[0,T \wedge \tau(\omega))}(s)$ is progressively measurable.