Setting We work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\in[0,T]},P)$ with finite time horizon $T$. Assume we are given a result of the following form:
Theorem If a stochastic process $X$ satisfies a property (A) on $[0,T']$ for some $T'\leq T$, then there exists a stochastic process $Y$ on $[0,T']$ with $X\leq Y$ a.s. on $[0,T']$, i.e. $$P[X_t\leq Y_t, \forall t\in [0,T']]=P[\{\omega: X_t(\omega)\leq Y_t(\omega), \forall t\in [0,T']\}=1.$$
Question Let $\tau$ be a $[0,T]$-valued stopping time. Assume now $X$ satisfies property (A) on the stochastic interval $[[0,\tau]] := \{(\omega,t)\in\Omega\times[0,T]:0\leq t\leq \tau(\omega) \}$. How can we then apply the theorem? More precisely:
- Do we obtain that there exists a stochastic process $Y$ on $[[0,\tau]]$ with $X\leq Y$ a.s. on $[[0,\tau]]$?
- And if so, is this meant in the sense $$P[\{\omega: X_t(\omega)\leq Y_t(\omega), \forall (\omega,t) \text{ with } 0\leq t \leq \tau(\omega) ]?$$
- Or is it meant in the sense $X\leq Y, \enspace \lambda\otimes P$-a.s., where $\lambda$ is the Lebesgue measure on $[0,T]$? This would be equivalent to (at least I think so) $X_t(\omega)\leq Y_t(\omega)$ for a.e. $t\in[0,T]$ and a.e. $\omega\in\{t\leq\tau\}$.
It is difficult to answer this without knowing what exactly property (A) is; let's assume for simplicity that property (A) is the process being nonnegative. The usual approach is to localise: consider the stopped process $X_t^\tau= X_{\min(t , \tau)}$. $X_t^\tau$ by construction will satisfy property (A) on $[0,T]$. Then, by your theorem, there exists a process $Y_t$ which satisfies property (A) on $[0,T]$ and almost surely dominates $X_t^\tau$. Thus, we have \begin{align} \mathbb{P}(\{\omega : X_t^\tau(\omega) \leq Y_t(\omega) , \forall t \in [0,T]\}) =1 \, . \end{align} But this implies that \begin{align} \mathbb{P}(\{\omega : X_t(\omega) \leq Y_t(\omega) , \forall t \in [0,\tau(\omega)]\}) =1 \, . \end{align}