Continuing my foray into Brownian motion (apologies for the bombardment...), I'm trying to verify the details of a proof of Durrett of the following 0-1 property of the tail $\sigma$-algebra of Brownian motion, $\mathcal{T}$:
If $A \in \mathcal{T}$, then for all $x$ either $P_x(A) = 0$ or $P_x(A) = 1$.
So to set the stage, let $\mathcal{F}_t' = \sigma(B_s: s \geq t)$ so that $\mathcal{T} = \cap_{t \geq 0} \mathcal{F}_t'$. Also let $\mathcal{F}_0^+$ be $\cap_{t>0} \mathcal{F}_t^0$ where $\mathcal{F}_t^0 = \sigma(B_s: s \leq t)$. Finally set $X_t = t B(1/t)$. The proof of the above fact starts with the claim that $\mathcal{T}$ (for $B_s$) is the same as $\mathcal{F}_0^+$ for the time-inverted process $X_s$. Now it seems pretty clear to me that $$ \sigma(B_s: s \geq t) = \sigma(X_s: s \leq t) $$ But when I take intersections to get the tail $\sigma$-field, shouldn't the RHS be $\cap_{t \geq 0} \mathcal{F}_t^0$ instead of $\cap_{t > 0} \mathcal{F}_t^0$, which would reduce to the ordinary $\sigma$-field $\mathcal{F}_0^0$?
I know that $\mathcal{F}_0^+$ is essentially the same as $\mathcal{F}_0^0$ by Blumenthal's 0-1 law, but I'm still having trouble seeing exactly if and how they differ, and exactly how much it matters. I get a sense that tail algebra events are somehow "nicer" than ones in the germ field because whether tail events happen or not is independent of the starting point $x$. I'm just trying to piece together how this relates to the ordinary Brownian motion $B_s$ and the time-inverted one, $X_s$.