Consider the measurable space $\left(\mathbf{C}\left[0,\infty\right), \mathcal{B}\left(\mathbf{C}\left[0,\infty\right)\right)\right)$ and the stochastic process $\left(X_t\right)_{t \in \left[0,\infty\right)}$ defined over this space, where for every $t \in \left[0,\infty\right)$, $X_t$ is the projection on the $t^\textrm{th}$ coordinate, i.e. $\forall f \in \mathbf{C}\left[0,\infty\right),\ X_t\left(f\right) = f\left(t\right)$. Denote by $\mathcal{T}$ the associated tail $\sigma$-algebra. For every $x \in \mathbb{R}$, let $P_x : \mathcal{B}\left(\mathbf{C}\left[0,\infty\right)\right) \rightarrow \left[0,1\right]$ be a probability measure that renders $\left(X_t\right)_{t \in \left[0,\infty\right)}$ a Brownian motion with start at $x$.
Let $A \in \mathcal{T}$. For any $x \in \mathbb{R}$, $P_x\left(A\right) \in \left\{0, 1\right\}$ (this can be derived using Blumenthal's $0$-$1$ law together with the time inversion invariance of Brownian motion). I'd like to show that for every $x, y \in \mathbb{R}$, $P_x\left(A\right) = P_y\left(A\right)$. In proving this fact, I'd like to avoid using the strong Markov property of Brownian motion, if possible (the simple Markov property is legit, though), but this is not a requirement, just an extra challenge.