I'm working on the proof of the following theorem given by Durrett's "Probability."
Theorem 7.2.7 If $A \in \mathcal{T}$, then either $P_x(A) \equiv 0$ or $P_x(A) \equiv 1$.
Here, $\mathcal{T} = \cap_{t\geq 0}\mathcal{F}_t'$, where $\mathcal{F}_t^\prime = \sigma(B_s: s \geq t)$. In words, $\mathcal{F}_t^\prime$ is the future at time $t$ and $\mathcal{T}$ is the tail $\sigma$-field of Brownian motion.
In the proof, Durrett says that for $A \in \mathcal{T}$ we can write $1_A = 1_D \circ \theta_1$ since $A \in \mathcal{F}_1^\prime$. Here, $D \in \mathcal{F}_0^+$, and when $\omega$ is a Brownian path, i.e. $B_t(\omega) = \omega(t)$, $\theta_s$, the shift transformation is defined as $(\theta_s \omega)(t)= \omega (s+t)$.
This is where I don't quite understand.
While I understand that, intuitively, $A \in \mathcal{F}_1^\prime$ means information about $t\geq 1$ is given. So, it is reasonable to think that $1_A = 1_D \circ \theta_1$ holds. But, how can we justify this conclusion rigorously?
Any help is very much appreciated!
I think I've come up with a solution to this problem. Here's the solution.
Let $\mathcal{P}$ be the collection of sets of form $A = \cap_{i=1}^n \{\omega(t_i) \in A_i\}$, where $1\geq t_1 < t_2 < \cdots < t_n$ and $A_i$ are Borel sets in $\mathbb{R}$. Then, $\mathcal{P}$ is a $\pi$-system.
For $A\in \mathcal{P}$, $$ 1_A(\omega) = \prod_{i=1}^n 1_{A_i}(\omega(t_i)) = \prod_{i=1}^n 1_{A_i}(\omega(t_i-1+1)) = \prod_{i=1}^n 1_{A_i}[(\theta_1\omega)(t_i-1)] $$
By defining $D = \cap_{i=1}^n\{\omega(t_i-1)\in A_i\}$, we can write $1_A (\omega)= 1_D \circ \theta_1(\omega)$.
Now that the desired conclusion is verified for $A \in \mathcal{P}$, we can complete the proof by the $\pi$-$\lambda$ theorem.
Is there any flaw in my reasoning?
In general, a $\mathbb R$-valued (or $\{0,1\}$-valued, in your case) random variable $Z$ which is measurable with regards to the $\sigma$-algebra $\sigma(Y)$ where $Y$ is a random variable, can be rewritten $Z = f\circ Y$ where $f$ is measurable from the range of $Y$.
https://proofwiki.org/wiki/Factorization_Lemma/Real-Valued_Function
That's a useful "abstract nonsense" lemma to have in your pocket.
In your case, $\mathcal{F}_1^\prime = \sigma(B_s: s \geq 1) = \sigma(\theta_1)$, by definition of $\theta_1$.