Proof in Oksendal Book

Question

I'm stack into a proof of Oksendal's Book. Page 182, equation (9.3.7).
Let $\mathcal{H}$ be the set of all real $\mathcal{M}_{\infty}$-measurable functions (sigma algebra generated by $X_t$, $t \geq 0$). For $t \geq 0$ we define the shift operator $\theta_t: \mathcal{H} \rightarrow \mathcal{H}$ as follows:
$ \theta_t \eta = g_1(X_{t_1+t}) \dots g_k(X_{t_k+t}), $ where $\eta=g_1(X_{t_1}) \dots g_k(X_{t_k})$ ($g_i$ Borel measurable, $t_i \geq 0$).
Let $\alpha \leq T$ be a stopping time ($T$ is a stopping time). We have that
$ \theta_{\alpha} \Big(\int_{0}^{T}F^u(Y_r)dr\Big) = \int_{\alpha}^{T}F^u(Y_r)dr. $ Using the approximation proposed in the book I don't understand why the integral is from $\alpha$ to $T$ and not $T +\alpha$.

Answer 1

$T$ in the situation you refer to is not just any old stopping time, but a terminal time, meaning that on the event $\{\alpha<T\}$ you have $T=\alpha+\theta_{\alpha}T$. (This terminal time property it true of all exit times like $T=\tau_D$.)