The infinity version of Blumenthal's 0-1 law Blumenthal's 0-1 law states that on the space of continuous maps with domain $[0, \infty)$ with the appropriate (Wiener) measure making the coordinate maps Brownian motions starting at $x$, any event depending only on the germ $\sigma$-algebra $\mathcal{F}_0^+$ of infinitessimal events is of trivial measure.  Supposedly, the same is true of tail events.  But everybody I've seen and textbooks all seem only to make an argument based on looking at the transformed Brownian $tB(1/t)$.  Okay, I see where the argument is going.  Problem is, nobody said Blumenthal's 0-1 law holds for abstract Brownian motions.  I've only seen the infinitessimal Blumenthal proven for the case where the Brownian motion is as above, with that space, that measure, and the coordinate maps as the stochastic process.  But the $tB(1/t)$ are no longer coordinate maps on the Wiener space.  Instead, we have kept the old measure, and changed the RVs to something else to invert the process.  This leads to an earlier question of mine (https://math.stackexchange.com/questions/383497/facts-about-brownian-motions-and-markov-chains-on-concrete-constructions), which would probably be answered by any answer to this question.  How do you generalize facts about this concrete Brownian to general Brownians? (Stochastic processes on any space with independent Gaussian increments which are continuous a.s. in the weak sense that although the continuity points may not form a measurable set, they contain a measurable set that is of probability 1.)
I feel like I must be asking something that is not a mathematical grammatically complete sentence or something.  Every time I ask anybody this family of questions, I never hear anything that makes sense to me.  So I must be totally lost in some really bad way.  It seems to me like everybody should be asking "Where is the justification for the change of representation of this BM while still using things only applying to the original construction." but the fact that not even 1 textbook addresses this question leads me to believe I have a gross misunderstanding somewhere.  Please help!
 A: I think what you're missing here is Levy's characterisation of Brownian motion.
If $X_t$ is a continuous local martingale and $X^2_t - t$ is also a local martingale then $X_t$ is a Brownian motion.
I'll explain a bit more.
Let $(\Omega,\mathcal F_t, P)$ be some filtered space with  $X_\circ:\Omega\times\mathbb R\to\mathbb R$ 
and let $\left(C^0,\mathscr F_t,\mathbb P\right)$ be Wiener space with "concrete" Brownian motion $\tilde X_\circ$.
If $X_t$ satisfies the martingale characterization above then it is possible to construct a measure on $\Omega\times C^0$ with marginals $P$ and $\mathbb P$ such that $X_t$ and $\tilde X_t$ agree almost surely.
Levy's characterization only shows that if $X_t$ satisfies the assumptions then the finite dimensional distributions agree. (It's a long time since I did all this.) The coupling (the measure on the product space) is done somewhere else and I can't remember where.
The construction of the coupling is rather trivial because the function $X_t$ 
is a member of $C^0$ almost surely. So we can just fix $\tilde X_t = X_t$ almost surely. So formally we define $\mathbb Q$ on $\mathcal F_t\times\mathscr F_t$ such that $$\mathbb Q(A\times B) = P\{\omega\in A: X_t(\omega)\in B\}.$$
So I only need to check that this defines a proper measure, but as the measures agree on finite dimensional distributions $\mathbb Q$ is defined and consistent on the algebra defined by restricting $B$ to cylinder sets. 
Therefore by Carathéodory's extension theorem there is a unique probability  measure $\mathbb Q$ satisfying the above and the maginals of $\mathbb Q$ are $P$ and $\mathbb P$ respectively.
What this means is that if I do a little trick, like consider $tB(1/t)$, we get a Wiener space for $tB(1/t)$ to live in automatically.  So if we use a theorem that holds for Wiener space then I've proved something that holds for a process $X_t$ that with probability one agrees with $tB(1/t)$ for every $t$. 
This is something that's assumed when we talk about Brownian motion, so it's not often mentioned explicitly, which may be your difficulty.
