I'm trying to prove the next:
Let $\{B(t)\}_{t\geq 0}$ be a standard Brownian motion on the line, and $T$ be a stopping time with $E(T)<\infty.$ Define $T_{1}=T$ and $T_{n}=T(B_{n})+T_{n-1}$ where $T(B_{n})$ is the same stopping time, but associated with the Brownian motion $\{B_{n}(t)\}_{t\geq 0}$ given by $$B_{n}(t)=B(t+T_{n-1})-B(T_{n-1}).$$
Then $T_{n}$ is stopping time with respect to the original Brownian motion filtration (natural filtration) for each $n\in\mathbb{N}.$
I was thinking to use induction over $n$ but I'm stuck. As first step,I'd like to prove this for $n=2,$ i.e. $T_{2}=T(B_{2})+T_{1}$ is stopping time respect to natural filtration of $\{B(t)\}_{t\geq 0}.$
So, for each $t\in\mathbb{R}$ We have that
$$\{T_{2}>t\}=\{T(B_{2})+T_{1}>t\}=\bigcup_{r\in\mathbb{Q}}\{T(B_{2})>r\}\cap\{T_{1}>t-r\}$$
Hence $T_{1}=T$ then $\{T_{1}>t-r\}\in\mathcal{F}_{t-r}\subset\mathcal{F}_{t},$ where $\mathcal{F}_{t}$ is the natural filtration of original Brownian motion. But I'm stuck proving that $\{T(B_{2})>r\}$ belongs to natural filtration of original Brownian motion.I know $B_{n}$ is Brownian motion because of the strong Markov property, but I can't see how is the natural filtration of the Brownian motion $B_2(t).$
Any kind of help is thanked in advanced.