Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $T>0$
- $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration of $\mathcal A$
- $W$ be an $\mathcal F$-Brownian motion o $(\Omega,\mathcal A,\operatorname P)$
Assume we have defined the Itō integral for $\mathcal F$-predictable $\Phi:\Omega\times[0,T]\to\mathbb R$ with $$\operatorname E\left[\int_0^T\left|\Phi_t\right|^2\:{\rm d}t\right]<\infty\;.\tag1$$ Let $\Phi\cdot W$ denote the Itō integral process. For an $\mathcal F$-stopping time $\tau:\Omega\to[0,T]\cup\left\{\infty\right\}$, let $$[0,\tau]:=\left\{(\omega,t)\in\Omega\times[0,T]:t\le\tau(\omega)\right\}$$ and $$\Phi^{(\tau)}:=1_{[0,\:\tau]}\Phi\;.$$ The identity $$\Phi^{(\tau)}\cdot W=(\Phi\cdot W)^{\tau}\;\;\;\text{almost surely}\tag2$$ is a standard result found in many books.
However, all authors assume that $\operatorname P[\tau\le T]=1$, but I absolutely don't see any stage of the proof of $(2)$ where we need this restriction. So, does $(2)$ fail to hold, if $\operatorname P[\tau\le T]\ne1$?
That doesn't make sense to me. If $(2)$ holds in the case $\operatorname P[\tau\le T]=1$, then it should hold for general $\tau$ too just by replacing $\tau$ with $\tau\wedge T$ ...