The statement in the title of my post is a basic "fact" from probability theory and for quite a long time I thought I had fully understood its meaning. However, today I stumbled across an issue which made me rethink. First of all, let me state a brief survey of what is commonly known (hopefully this is helpful for some of the readers of this post):
Suppose we are given two stochastic processes $X = (X_t)_{t \geq 0}$ and $Y = (Y_t)_{t \geq 0}$ on a common probability space ($\Omega,\mathcal{F},\mathbb{P})$, mapping to $\mathbb{R}$ (of course the state space may also be some arbitrary topological space $S$ equipped with a $\sigma$-algebra, but let's keep things simple).
1) Assume we do not know any regularity of the paths of $X$ and $Y$, hence when considering $X$ and $Y$ as path-valued random variables, we have no (reasonable) choice but to consider $\mathbb{R}^{[0,\infty]}$ as their state space (or, in general, $S^{[0,\infty]}$ if we consider arbitrary state spaces $S$). In most situations one will equip this path space with the $\sigma$-algebra generated by all canonical projections, i.e. with $\sigma(\pi_t|t \geq 0)$ and then it is well-known that the laws of $X$ and $Y$ coincide if and only if they have the same finite-dimensional distributions. This follows easily by the structure of the mentioned $\sigma$-algebra (essentially, this is also how the uniqueness-part of the Kolmogorov existence theorem can be dealt with). Everything is fine here.
2) Assume we have quite a lot of regularity of the paths of $X$ and $Y$, i.e. the paths belong to $C(\mathbb{R}_+,\mathbb{R})$. As in 1), this space will usually be equipped with the $\sigma$-algebra $\sigma(\pi_t|t \geq 0)$ (one reason is that this $\sigma$-algebra naturally appears as the Borel-$\sigma$-algebra of natural topologies on this space, such as the topology of local uniform convergence). Hence the equivalence of equality in law and equality in the fdd's stated in 1) remains true. Things are still good.
3) Now assume the paths of $X$ and $Y$ are known to be cadlag, i.e. they belong to the Skorokhod space - which offers quite a few reasonable topologies and - hence - Borel-$\sigma$-algebras. The most common one, often called $J_1$-topology (allowing to wiggle space and time a bit) behaves well in the sense that its Borel-$\sigma$-algebra is also the one generated by all projections as in 1) and 2).
BUT: If we choose the norm of local uniform convergence, then the Borel-$\sigma$-algebra is a different one. In fact it does not seem to be generated by cylinder sets and this leads me to the impression that in this situation we cannot conclude "the fdd's uniquely determine the law of the stochastic process", because the $\sigma$-algebra of the path space is, in this case, not such that the sets involved in the fdd's generate it.
Can somebody clarify this? Can it really happen that (for a quite reasonable topology and Borel $\sigma$-algebra) the fdd's are not enough to specify the law on the path space? Or would one typically still consider the $\sigma$-algebra $\sigma(\pi_t | t \geq 0)$ in such situations, even if this might not be the Borel-$\sigma$-algebra for the underlying topology?
I hope to get interesting and helpful comments and answers and would be happy if this long post was valuable as a brainteaser or a review for some of the readers :-)
I think, the simplest way to see this is simply asking about an event for which the probability cannot possibly be given by the probability of the finite dimensional distributions.
Assume we're in a topological setting. Let $U$ be open and consider the event that there exists a time $t\in [t_1,t_2]$ such that $X_t\in U$. Now, if $X$ is continuous and $U$ is open, this reduces to a question about the rational times in that interval, so this event is measurable on the cylinder borel algebra.
Let however $u\in U$ and $v\in U^c$ and let $(\tau_n)_{n\in \mathbb{N}}$ be any sequence of positive, independent random variables such that their distributions have no point-masses. Define now $X_t=u$ if $t\in \{\tau_n\}_{n\in \mathbb{N}}$ and $X_t=v$ else. Then, the probability that $X_t\in U$ for exactly $t$ is $0$, no matter which random times I've picked.
But CLEARLY, at least discarding concerns over measurability, if there is, indeed, such a well-defined random function, I can also clearly pick random times such that the probability that $\{\tau_n\}_{n\in \mathbb{N}}\cap [t_1,t_2]\neq \emptyset$ with probability 1 (for instance, pick distributions that are supported in exactly this interval). And, going further, I could get the probability that the two sets intersect non-trivially to be any number between $0$ and $1$.