[proofplan] We prove $\mathbb{P}(A) \in \{0,1\}$ for every tail event $A$ using a martingale argument. The conditional expectation $M_n = \mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_n]$ is a bounded martingale that converges a.s. to $\mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_\infty] = \mathbb{1}_A$. Independence forces $M_n = \mathbb{P}(A)$ a.s. for each $n$, so $\mathbb{P}(A) = \mathbb{1}_A$ a.s., which is only possible if $\mathbb{P}(A) \in \{0,1\}$. [/proofplan] [step:Construct the martingale $M_n = \mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_n]$ and identify its limit] Let $A \in \mathcal{T}$ and define $\mathcal{G}_n = \sigma(X_k : k \leq n)$ and $\mathcal{G}_\infty = \sigma(\mathcal{G}_n : n \geq 0)$. Then $M_n = \mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_n]$ is a martingale. By the $L^1$ martingale convergence theorem (applicable since $\mathbb{1}_A$ is bounded): \begin{align*} M_n \to \mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_\infty] \quad \text{a.s.} \end{align*} [/step] [step:Use independence to show $M_n = \mathbb{P}(A)$ a.s. for each $n$] Since $A \in \mathcal{T} \subset \sigma(X_k : k \geq n+1)$ and $\mathcal{G}_n$ is independent of $\sigma(X_k : k \geq n+1)$, the independence property of conditional expectation gives \begin{align*} M_n = \mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_n] = \mathbb{P}(A) \quad \text{a.s. for each } n. \end{align*} [guided] Why does independence give $\mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_n] = \mathbb{P}(A)$? The event $A$ belongs to $\sigma(X_k : k \geq n+1)$, which is independent of $\mathcal{G}_n = \sigma(X_k : k \leq n)$. When a random variable $Z$ is independent of a $\sigma$-algebra $\mathcal{H}$, the conditional expectation $\mathbb{E}[Z \mid \mathcal{H}]$ reduces to $\mathbb{E}[Z]$ (since $Z$ carries no information about $\mathcal{H}$, conditioning does not change the expectation). Applying this with $Z = \mathbb{1}_A$ and $\mathcal{H} = \mathcal{G}_n$ gives $M_n = \mathbb{E}[\mathbb{1}_A] = \mathbb{P}(A)$ a.s. [/guided] [/step] [step:Conclude $\mathbb{P}(A) \in \{0,1\}$ by comparing the two expressions for the limit] Since $A \in \mathcal{T} \subset \mathcal{G}_\infty$, the random variable $\mathbb{1}_A$ is $\mathcal{G}_\infty$-measurable, so $\mathbb{E}[\mathbb{1}_A \mid \mathcal{G}_\infty] = \mathbb{1}_A$ a.s. Taking limits: the a.s. limit of $M_n$ is both $\mathbb{P}(A)$ (a constant) and $\mathbb{1}_A$ (a $\{0,1\}$-valued random variable). Therefore $\mathbb{P}(A) = \mathbb{1}_A$ a.s., which forces $\mathbb{P}(A) \in \{0, 1\}$. [/step]