Assume that $(\tau_n)$,$(T_n)$ are finite strictly increasing stopping times and set $\tau=\sup_{n}\tau_n$,$T=\sup_{n}T_n$. Now consider a process $(W_t)_{t<\tau}$, and a Brownian motion $B$, which satisfies that $(W^{\tau_n},\tau_n)\stackrel{D}{=}(B^{T_n},T_n)$ . Moreover, assume that $(W^{\tau_n},\tau_n)\stackrel{wk}{\rightarrow}(B^{T},T)$. I claim that:
$W$ has left limit at $\tau$, and defining $W_\tau:=W_{\tau^-}$ gives that $W^\tau\sim B^T$.
Can anybody verify if my claim is true? If so, a proof or a reference would be appreciated. I feel like i have tried every trick in the book, but i have a hard time adapting the ideas to the case where $W$ is not defined on the entire $[0,\infty)$ interval.
I will assume that $W$ has continuous sample paths on $[0,\tau)$, and that $\Bbb P(\tau<\infty)=1$.
The process $\exp(B_t-t/2)$ is a positive martingale; consequently $X_t:=\exp(B_t-t/2)1_{\{t<T\}}$ is a non-negative supermartingale. It is clear from the hypotheses that the process $X$ has the same finite-dimensional distributions as the process $Y_t:=\exp(W_t-t/2)1_{\{t<\tau\}}$, which is therefore also a supermartingale. Moreover, the process $Y$ is right-continuous on $[0,\infty)$. But a right-continuous supermartingale possesses left limits on $(0,\infty)$ almost surely. It follows that the left limit $Y_{\tau-}=\lim_{t\uparrow \tau}Y_t$ exists almost surely, and $\Bbb P(Y_{\tau-}>0)=\Bbb P(X_{T-}>0)=1$. Consequently $W_{\tau-}=\lim_{t\uparrow \tau}[\log(Y_t)+t/2]=\log(Y_{\tau-})+\tau/2$ exists, almost surely. The rest is clear sailing.