I understand that there are many discontinuous modifications of Brownian Motion, one of which seen in my course was:
$$X_t = \begin{cases} B_t & t\neq u \\ 0 & t=u \\ \end{cases} $$
where $(B(t):t\in[0,\infty))$ is BM and $u$ is $\text{Unif}([0,1])$. However, I don't understand the claim that $(X(t))$ has the same finite dimensional marginals as $(B(t))$. Recall that same finite-dimensional marginals means that with probability 1, for any finite sets $\{t_j\}_1^k\subseteq T$, the collection of marginals of both stochastic processes equate in law with each other. Any help is appreciated.
Edit: I'm adding "with probability 1" to the beginning the definition of same finite-dimensional marginals for congruence with the answer below. Also, in the same light as the answer below, it's clear that if the BM was $(B(t): t\in [0,n] \cap \mathbb{N})$ for some $n$, then equality in f.d. marginals would no longer hold.
The reason is that for any finite set on which you're taking marginals, the uniform random variable will almost surely never be on one of those points - this is true for a simpler example too:
For $t \in [0,1]$, take $X_t := 0$, a uniform $[0,1]$ random variable $U$, and $Y_t = \chi_\{U\}$ to be 1 at the value chosen by $U$, 0 otherwise. Then for any finite set $\{t_i\}$, we have $$ \mathbb P(Y_{t_i} = 0 \textrm{ for all }i) = 1 $$ so that $\mathbb P(X_{t_i} = Y_{t_i} \textrm{ for all }i) = 1$, but that $\mathbb P(X_t = Y_t \textrm{ for all }t) = 0$ since $X_U = 0 \not = Y_U = 1$ (there's always exactly one $t$ for which the two processes differ)
Your example just has more complicated (but still identical) definitions of processes away from the uniformly selected point.