Let $(X_t)_{t\geq0}$ and $(Y_t)_{t \geq0}$ denote two continuous stochastic processes on a probability space $(\Omega, \mathcal{F}, P)$ and let $C$ denote space of continuous functions from $[0,\infty)$ to $\mathbb{R}$.
Let $\pi_x\colon C \rightarrow \mathbb{R}$ given by $\pi_x(f)=f(x)$ and equip $C$ with the sigma algebra $\mathcal{E} = \sigma(\{\pi_x : x\geq 0\})$.
Consider the now the $\mathcal{F}$–$\mathcal{E} $-measurable mappings $\mathbb{X}, \mathbb{Y} \colon \Omega \rightarrow C$ given by
$$
\omega \mapsto X_{(\cdot)}(\omega) \quad \text{and} \quad \omega \mapsto Y_{(\cdot)}(\omega).
$$
I want to prove that if $(X_t)_{t\geq0}$ and $(Y_t)_{t \geq0}$ have the same marginal distributions, then $P_\mathbb{X}=P_\mathbb{Y}$.
My progress: Supposing that $(X_t)_{t\geq0}$ and $(Y_t)_{t \geq0}$ have the same marginal distributions, then for $t_1,\dots, t_n, A_1,\dots,A_n$
$$
P(X_{t_1}\in A_{1},\dots, X_{t_n} \in A_{n})=P(Y_{t_1}\in A_{1},\dots, Y_{t_n} \in A_{n})
$$
by definition. So working from this, we get that
$$
\bigcap_{k=1}^n \{X_{t_k} \in A_k\} = \bigcap_{k=1}^n \{\pi_{t_k} \circ\mathbb{X} \in A_k\} = \bigcap_{k=1}^n \{\mathbb{X} \in \pi_{t_k}^{-1}(A_k)\} = \{\mathbb{X} \in \bigcap_{k=1}^n \pi_{t_k}^{-1}(A_k) \}
$$
but this is where I become stuck. We want to show $P(\mathbb{X} \in B) =P(\mathbb{Y} \in B) $ where $B \in \mathcal{B}$, an intersection stable generator set of $\mathcal{E}$. So the question is if
$$
\{\bigcap_{k=1}^n \pi_{t_k}^{-1}(A_k) \colon n \in \mathbb{N}, t_1,\dots,t_n \in \mathbb{R}, A_1,\dots, A_n \in \mathcal{B}(\mathbb{R}) \}
$$
is an intersection stable generating set of $\mathcal{E}$.
Can anyone help me? Am I on the right track?
Also my textbook claims this is only true if the processes are continuous but I can't seem to find anywhere where I would use this continuity…
Best Answer
From the hypothesis and Dynkin's $\pi-\lambda$ theorem [1], you can deduce that the two processes have the same distributions when restricted to rational times. Then continuity gives the desired conclusion.
Without continuity, the following is a well known counterexample: Let $\{X_t\}$ be standard Brownian motion, let $U$ be uniformly distributed in $[0,1]$ and let $\{Y_t\}$ be obtained from $\{X_t\}$ by setting $Y_U=X_U+1$ and $Y_t=X_t$ for all $t \ne U$.
[1] https://en.wikipedia.org/wiki/Dynkin_system