I have some questions to the proof of the recurrence of the Brownian motion where we consider the stopping time $\tau_a=\text{inf}\{t\geq 0:B_t=a\}$.
In the proof we consider a shift-operator $\theta_t$. (I give the definition at the bottom.) Together with the stopping time we get these equalities:
$\theta_{\tau_{-a}}^{-1}\{\tau_a<\infty\}= \{\omega \in \Omega_{\tau_{-a}}:\tau_a(\theta_{\tau_{-a}(\omega)}(\omega))<\infty\}=\{\tau_{-a}<\infty\}\cap\{\tau_a(\tau_{-a})<\infty\}$
where $\Omega_{\tau_{-a}}= \{{ \tau_{-a}}<\infty\}$.
By the definition of $\Omega_{\tau_{-a}}$ I think that the second equality uses that $\tau_a(\theta_{\tau_{-a}(\omega)}(\omega))=\tau_a(\tau_{-a})$, right? If yes, why does this hold and if not, why is the second equality satisfied?
And why does the first equality hold?
Definition of the shift-operators of a Markov process $(X_t,\mathcal{F}_t,P_x)$:
It is a family of maps $\theta_t:\Omega \rightarrow \Omega$ such that
- $\theta_0=Id_{\Omega}$
- $\theta_t \theta_s = \theta_{t+s}$
- $ X_s(\theta_t) = X_{s+t}$