In a book I'm working through there is a proof that $$\int_0^{\tau(\omega)\land t}f(\omega,s)dB_s(\omega) = \int_0^t f(\omega,s)1\{s\le \tau(\omega)\}dB_s(\omega)$$ The proof begins by claiming that $\tau_n = 2^{-n}(\lfloor2^n\tau\rfloor+1)$ is a stopping time converging to $\tau$ with $\tau \ge k2^{-n}$ iff $\tau_n \ge (k+1)2^{-n}$, that $\int_0^{\tau_n\land t}fdB = \int_0^t 1_{(0,\tau_n]}(s)f(s)dB_s$ and that $\int_0^t 1_{(0,\tau_n]}(s)f(s)dB_s \to \int_0^t 1_{(0,\tau]}(s)f(s)dB_s$ in $\mathcal{L}^2$. Then it says that using path continuity on $\int_0^{\tau_n\land t}fdB$ completes the proof.
I have two questions about this. First, I'm unsure about the claim that $\tau_n$ is a stopping time and was wondering how you could prove that it is true. Second is that for the last step, I am unsure of what is meant by applying path continuity, so any explanation of what's going on in that step would be greatly appreciated! Thanks for any help.
Hi for the 1st question observe that $\forall n\geq0, \tau_n\leq \tau$ by definition of $\tau_n$.
It is sufficient to prove that $\{\tau_n\leq t\}\in \mathcal{F}_t$, but $\{\tau_n\leq \tau\}$ so $\{\tau_n\leq t\}\subset \{\tau\leq t\}$ and as the last set is in $\mathcal{F}_t$ ($\tau$ is a stopping time) this in turn shows that $\{\tau_n\leq t\}\in \mathcal{F}_t$ .
For the second question the first $L^2$ argument shows that :
$\int_0^{\tau_n(\omega)\land t}f(\omega,s)dB_s(\omega)\to \int_0^t f(\omega,s)1\{s\le \tau(\omega)\}dB_s(\omega)$
The path continuity argument (of the stochastic integral seen as a process) argument shows that :
$\int_0^{\tau_n(\omega)\land t}f(\omega,s)dB_s(\omega)\to \int_0^{\tau(\omega)\land t}f(\omega,s)dB_s(\omega)$
which allows to conclude that both quantities are the same.
Best regards