Revuz and Yor's Book “Continuous Martingales and Brownian Motion” - Chapter 1 - Exercise 1.11 (again) Context : This post is the first of a series of posts taking their origins from the exercises in the Revuz and Yor's Book "Continuous Martingales ans Brownian Motion". 
The reason for doing so is that the exercises of this book are hard sometimes very hard but still very interesting and that there is no definitive or authorative source for the solutions. I am not alone on this project but I am starting with the first exercise of the book (which is easy and for which I am posting an answer). 
Lastly I would like this post (and the next ones) to get a "community wiki" status but do not have clearance for doing so at the question level so I will only do this for the answer. 
Exercise : 
Let $B$ be the Standard Linear Brownian Motion where we consider only $t\in [0,1]$. Prove that the process $\tilde B$ defined by :
$$\tilde B_t =B_{1-t}-B_1$$
is another version of $B$, in other words, a standard BM on $[0,1)$.
Standard Linear Brownian Motion is the almost surely continuous version of the process defined by its finite dimensional laws via Kolmogorov consistency theorem and such version exists via the Censtov Kolmogorov theorem (cf. Theorem 1.9 page 19 of the book for existence).
 A: Here is the solution : 
The properties required by $\tilde B_t$ to be a version of Linear BM $B_t$ are the following on $[0,1)$


*

*"$\tilde B_0 = B_0$"
Proof :
$\tilde B_0 = B_{1-0}-B_1=B_1-B_1=0=B_0$

*"Increments of $\tilde B$ are normal centered with variance equal to the time difference"
Proof :
For $t>s$, $\tilde B_t -\tilde B_s  = B_{1-t}-B_1- B_{1-s}+B_1= B_{1-t}-B_{1-s}=-(B_{1-s}-B_{1-t} )$ which equals in law $B_{1-s}-B_{1-t}$ by the symmetry of this normal centered r.v.. 
As $B_t$  is a linear BM, its increments $(B_t-B_s)$ are normal centered with variance equal to $t-s$. So the law of $\tilde B_t -\tilde B_s$ is the same as $B_{1-s}-B_{1-t}$ i.e. normal, centered with variance $t-s$.

*"almost sure continuity of the trajectories"
Proof :
As trajectories of $B_t$ are a.s. continuous, so are those of $B_{1-t}$ on $[0,1]$ as the time change do not affect the continuity of the trajectories and so the trajectories of $\tilde B_t$ are a.s. continuous over $(0,1)$. The only point left to examine is over is the continuity at $t=0$. This is assured by the fact that the shift $-B_1$ coupled with the time change brings back all the trajectories in 0, i.e. for almost all $\omega \in \Omega,~Lim_{t→0^+} \tilde B_t (\omega)= Lim_{t→1^-}B_{t}(\omega) - B_1 (\omega) = 0$ by the a.s. continuity of the trajectories of the Standard Linear Brownian Motion Standard $B_t$.  
The points 1 to 3 being proved are enough to say that $\tilde B$ is a version of $B$, indeed they entail that both processes have the same finite dimensional laws of increments and start at 0, as this implies that  $\mathbb{P}(\tilde B_t \in A)=\mathbb{P}( B_t \in A), \forall t\in [0,1)$ and $A\in \mathcal {B}(\mathbb {R})$ this what was asked for.
Last remark : $\tilde B_t$ is not adapted to the natural filtration $\mathcal{F}_t$ of $B_t$ because  $\tilde B_t$ is $\mathcal{F}_1$-measurable and not $\mathcal{F}_t$-measurable, the natural filtration of $\tilde{B_t}$ is the one generated by $\tilde{ \mathcal{F}}_t= \mathcal{F}_t\vee \sigma(B_1)$
