Let
- $T>0$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\in[0,\:T]}$ be a complete right-continuous filtration on $(\Omega,\mathcal A)$
- $(X_t)_{t\in[0,\:T]}$ be a $\overline{\mathbb R}$-valued $\mathcal F$-adapted almost surely right-continuous stochastic process on $(\Omega,\mathcal A,\operatorname P)$ with $$X_s\le X_t\;\;\;\text{for all }0\le s\le t\le T\tag 1$$ and $$\operatorname P[X_T<\infty]=1\tag 2$$
Let $$\tau_n:=\inf\left\{t\in[0,T]:X_t\ge n\right\}\wedge T$$ for $n\in\mathbb N$. It's clear that $\tau_n$ is an $\mathcal F$-stopping time with $\tau_n\le\tau_{n+1}$ for all $n\in\mathbb N$. I want to show that $$\operatorname P\left[\tau_n\xrightarrow{n\to\infty}T\right]=1\;.\tag 3$$
I think we only need to prove that "almost surely $\left\{t\in[0,T]:X_t\ge n\right\}$ will become the empty set" ($\inf\emptyset:=\infty$) as $n\to\infty$. However, I got problems to write this down rigorously.
If $N:=\left\{X_T=\infty\right\}$, then it's clear that $N$ is a $\operatorname P$-null set. Since $$\left\{A_t=\infty\right\}\subseteq N\;\;\;\text{for all }t\in[0,T]\;,\tag 4$$ we obtain $$1_{\left\{A_t\:\ge\:n\right\}}\xrightarrow{n\to\infty}1_{\left\{A_t\:\ge\:\infty\right\}}\le 1_N\tag 5$$ and hence $$\operatorname P\left[1_{\left\{A_t\:\ge\:n\right\}}\xrightarrow{n\to\infty}0\right]=1\;,\tag 6$$ but I still don't know why this implies $(3)$ even when it's intuitively clear. I guess I'm thinking to complicated.