Let
- $T>0$
- $I:=(0,T]$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\in\overline I}$ be a filtration of $\mathcal A$
- $\tau_n:\Omega\to\overline I$ be an $\mathcal F$-stopping time with $$\tau_n\le\tau_{n+1}\tag1$$ for all $n\in\mathbb N_0$ and $$\tau_n\xrightarrow{n\to\infty}T\tag2$$
- $X_n:\Omega\to\mathbb R$ be $\mathcal F_{\tau_{n-1}}$-measurable for all $n\in\mathbb N$
How can we show that $$\Phi_t:=\sum_{n\in\mathbb N}1_{\left\{\:\tau_{n-1}\:<\:t\:\right\}}X_n\;\;\;\text{for }t\in\overline I\tag3$$ is well-defined?
Let $(\omega,t)\in\Omega\times\overline I$. To be precise, I don't see why the series $\Phi_t(\omega)$ is convergent. If $t<T$, then $$\varepsilon:=T-t>0$$ and hence $$\sup_{n\in\mathbb N}\tau_n(\omega)=T\tag4$$ by $(1)$ and $(2)$. Thus, $$t=T-\varepsilon=\tau_N(\omega)\le\tau_n(\omega)\;\;\;\text{for all }n\ge N\tag5$$ for some $N\in\mathbb N$ by definition of the supremum. Thus, $$\Phi_t(\omega)=\sum_{n=1}^N1_{\left\{\:\tau_{n-1}\:<\:t\:\right\}}(\omega)X_n(\omega)\tag6\;.$$ So, the only problematic case is $t=T$.