I am reading the book Brownian Motion by Yuval Peres and Peter Morters which can be found [here][1].
Tail algebra for a brownian motion $\{B(t): t\geq 0\}$ is defined as $\mathcal{T} = \bigcap_{t\geq 0} \mathcal{G}(t)$ where $\mathcal{G}(t) = \sigma(B(s): s \geq t)$. The zero-one law for tail events says that:
Let $x \in \mathbb{R}^d$ and suppose $A \in \mathcal{T}$ is a tail event. Then $P_x(A) \in \{0,1\}$.
The book says that the proof follows from time inversion and Bluementhals 0-1 law (Which has the same conclusion except $A \in \mathcal{F}^+(0) \left( = \bigcap_{t> 0} \mathcal{F}^0(s) \text{ where } \mathcal{F}^0(s) = \sigma(B(t): 0 \leq t \leq s\right)$). I don't understand how it follows from time inversion. Here is what I think they mean but I am not sure:
$$X(t) = tB(\frac{1}{t}) \text{ if $t \neq 0$ and } 0 \text{ otherwise } $$ be the time inverted brownian motion. From this I think the following is true (But not sure):
$$\sigma(B(s): s \geq t) = \sigma(X(s): 0 \leq s \leq 1/t)$$ And thus $A \in \mathcal{T} \implies A \in \bigcap_{t > 0} \sigma(X(s): 0 \leq s \leq t) $ Now the bluementhals law applies to show that $P_x(A) \in \{0,1\}$ (as X and B have the same law).
Is my approach correct or do I need to do something more? Thanks in advance. [1]: https://people.bath.ac.uk/maspm/book.pdf
This follows immediately from Bluementhal's $0-1$ law applied to the BM $(X(s))$ since $\sigma(X(s): 0 \leq s \leq 1/t)=\sigma(B(s): s \geq t)$ which implies $\bigcap_{t\geq 0} \mathcal{G}(t)=\bigcap_{t > 0} \sigma(X(s): 0 \leq s \leq t) $.