In the book on Brownian Motion by Yuval Peres, he makes the following claim:
Suppose that $\{ B(t); \ t \geq 0 \}$ is a Brownian Motion and $U$ is an independent random variable, uniformly distributed on $[0, 1]$. Then the process $\{ \tilde{B} (t); \ t \geq 0 \}$ defined as, $$\tilde{B}(t) = \begin{cases} B(t) & \mbox{ if $t \neq U$} \\ 0 & \mbox{ if $t = U$} \end{cases}$$
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.
I'm having difficulty in proving that the finite dimensional joint distribution ( basically the joint distribution of $( \ \tilde{B}(t_1), \ \tilde{B}(t_2), \tilde{B}(t_3), \dots \tilde{B}(t_n) \ )$ ) is the same as the finite dimensional distribution of $B(t)$. Can you please give me some clues.
Since
$$\{B_t \neq \tilde{B}_t\} \subseteq \{U=t\}$$
for any $t \geq 0$ we have
$$\bigcup_{j=1}^n \{B_{t_j} \neq \tilde{B}_{t_j}\} \subseteq \bigcup_{j=1}^n \{U=t_j\},$$
and so
$$\mathbb{P}(\exists j \in \{1,\ldots,n\}: B_{t_j} \neq \tilde{B}_{t_j}) \leq \sum_{j=1}^n \mathbb{P}(U=t_j)=0;$$
this implies readily that $(B_t)_{t \geq 0}$ and $(\tilde{B}_t)_{t \geq 0}$ have the same finite dimensional distribution.
Since $(B_t)_{t \geq 0}$ has continuous sample paths, we know that $$\lim_{\substack{s \to U(\omega) \\ s \neq U(\omega)}} \tilde{B}_s(\omega) = \lim_{\substack{s \to U(\omega) \\ s \neq U(\omega)}} B_s(\omega) = B_U(\omega)$$
for almost all $\omega \in \Omega$. This shows that
$$\mathbb{P}(\{\omega; t \mapsto \tilde{B}_t(\omega) \, \, \text{is discontinuous at $t=U(\omega)$}\}) = \mathbb{P}(B_U \neq 0) = 1.$$
In particular, each sample path $t \mapsto \tilde{B}_t(\omega)$ has at least one discontinuity with probability 1.