I started reading the book Brownian motion by Mörters and Peres where I don't quite understand the following example. For completeness I've provided the definitions for brownian motion and finite dimensional distributions used in the book.
Definition $1.1$$ . \quad$ A real-valued stochastic process $\{B(t): t \geqslant 0\}$ is called a (linear) Brownian motion with start in $x \in \mathbb{R}$ if the following holds:
$\bullet\hspace{0,2cm} B(0)=x,$
$\bullet\hspace{0,2cm}$the process has independent increments, i.e. for all times $0 \leqslant t_{1} \leqslant t_{2} \leqslant \ldots \leqslant t_{n}$ the increments $B\left(t_{n}\right)-B\left(t_{n-1}\right), B\left(t_{n-1}\right)-B\left(t_{n-2}\right), \ldots, B\left(t_{2}\right)-B\left(t_{1}\right)$ are independent random variables,
$\bullet\hspace{0,2cm}$ for all $t \geqslant 0$ and $h>0,$ the increments $B(t+h)-B(t)$ are normally distributed with expectation zero and variance $h$
$\bullet\hspace{0,2cm}$ almost surely, the function $t \mapsto B(t)$ is continuous.
We say that $\{B(t): t \geqslant 0\}$ is a standard Brownian motion if $x=0$.
By the finite-dimensional distributions of a stochastic process $\{B(t): t \geqslant 0\}$ we mean the laws of all the finite dimensional random vectors $$ \left(B\left(t_{1}\right), B\left(t_{2}\right), \ldots, B\left(t_{n}\right)\right), \text { for all } 0 \leqslant t_{1} \leqslant t_{2} \leqslant \ldots \leqslant t_{n} $$ To describe these joint laws it suffices to describe the joint law of $B(0)$ and the increments $$ \left(B\left(t_{1}\right)-B(0), B\left(t_{2}\right)-B\left(t_{1}\right), \ldots, B\left(t_{n}\right)-B\left(t_{n-1}\right)\right), \text { for all } 0 \leqslant t_{1} \leqslant t_{2} \leqslant \ldots \leqslant t_{n} $$
Example 1.2 Suppose that $\{B(t): t \geqslant 0\}$ is a Brownian motion and $U$ is an independent random variable, which is uniformly distributed on [0,1] . Then the process $\{\tilde{B}(t): t \geqslant 0\}$ defined by $$ \tilde{B}(t)=\left\{\begin{array}{ll} B(t) & \text { if } t \neq U \\ 0 & \text { if } t=U \end{array}\right. $$ has the same finite-dimensional distributions as a Brownian motion, but is discontinuous if $B(U) \neq 0,$ i.e. with probability one, and hence this process is not a Brownian motion.
Now my questions are
- Why does $\tilde{B}(t)$ have the same finite-dimensional distributions as a Brownian motion ?
- Why is $B(U)\neq 1$ with probability one ?
