Let $X=(X_t)_{t\in\mathbb{R}}$ be a stationary process, i.e. the shifted process $X^s=(X_t^s)_{t\in\mathbb{R}}=(X_{t+s})_{t\in\mathbb{R}}$ has the same distribution as $X=X^0$, and assume that $X$ has right continuous paths with left limits (cadlag). The process $X$ has the mixing property, if \begin{equation*} \lim_{t\rightarrow\infty} \mathbb{P}\{ X\in A, X^t \in B \} = \mathbb{P}\{ X\in A \} \mathbb{P}\{X\in B\} \end{equation*} Here $A$ and $B$ are in the $\sigma$-algebra over the set of cadlag functions generated by the finite dimensional projections (standard model).
In the literature i've read, where the authors proofed some sufficient conditions for the mixing property, they always assumed, that the above convergence is equivalent to \begin{equation*} \lim_{t\rightarrow\infty} \mathbb{E}\left[ \exp\left( i\sum_{k=1}^m a_k X_{t_k} + i\sum_{k=m+1}^{M} a_kX_{t_k}^t \right) \right] = \mathbb{E}\left[ \exp\left( i\sum_{k=1}^m a_kX_{t_k} \right) \right] \mathbb{E}\left[ \exp\left( i\sum_{k=m+1}^M a_kX_{t_k} \right) \right] \end{equation*} for $a_1,\dots,a_m,a_{m+1},\dots,a_M \in \mathbb{R}$ and $t_1,\dots,t_m,t_{m+1},\dots,t_M \in \mathbb{R}$, that is, the above mixing property convergence is true for all finite dimensional distributions and in the weak sense (?).
if i interpreted all the above in the right way, why is that so? why does the second convergence of the characteristic functions imply the first convergence for all $A$, $B$ ? does the weak convergence of the marginal distributions imply the stronger convergence for all $A$, $B$?