Is a stochastic process being continuous a.e. completely determined by its law? 
*

*Given a measure space $(X, \mathcal F, \mu)$, it is always possible
to construct a measurable mapping defined from some other measure
space and with $X$ as its codomain, such that the measure induced by
the mapping is the measure $\mu$ on the codomain.
Suppose we further require the measurable mapping to satisfy that
given some subset $A$ of $X$ ($A$ is not necessarily measurable,
i.e. may not be in $\mathcal F$), the mapping maps into $A$ a.e..
Then the existence of such a mapping requires that there exists a $B
    \in \mathcal F$ such that $B \subseteq A$ and $\mu(X - B) = 0$ (correct me if I am wrong). I
was wondering if that necessary condition is also sufficient for
existence of such a mapping? 
Is it correct that existence of such a $B$ is completely determined
by the measure space $(X, \mathcal F, \mu)$? Among all the
measurable mappings that can induce $\mu$, either all of them map
into $A$ a.e., or neither of them will map into $A$ a.e.? It is
impossible that some of them map into $A$ a.e. and some don't?

*As an example and also source of my above question, let's look at
construction of a Brownian motion:
Given finite dimensional distributions of a Brownian motion, there
exists a unique probability measure $P$ on the sample path space
$\mathbb{R}^T$ where $T:=[0, \infty)$ with its Borel sigma algebra,
by Kolmogorov extension theorem.
A Brownian motion also requires its sample path to be continuous
a.e.. Is existence of a Brownian motion equivalent to that the set
$A$ of continuous functions from $T$ to $\mathbb{R}$ contains a
subset which has probability $1$ under $P$? (Note $A$ is not
measurable wrt the Borel sigma algebra over $\mathbb{R}^T$.)
Is it possible that there exists a stochastic process with $P$ being
its law, but not continuous a.e.?
Thanks and regards!
 A: *

*This is a sufficient condition, but it is not necessary.  A necessary and sufficient condition for the existence of such a mapping is that $\mu(S)=0$ for all $S \in\cal F$ with $S\cap A=\emptyset$.  Also, given a mapping that induces $\mu$, and given that this condition holds, it is not predetermined whether or not the mapping is a.s. into $A$.  It may be that some mappings are a.s. into $A$ and some are not.

*You can construct processes which have the same finite-dimensional distribution as Brownian motion but which are a.s. continuous nowhere.  For example, let $(B_t)_{t\ge 0}$ be the usual Brownian motion, which is a.s. continuous everywhere, and let $U$ be uniformly distributed on $[0,1]$.  Then
pick a dense set $M$ of measure $0$ and a badly behaved function $f$ and set
$$B'_t:=\left\{\begin{array}{ll} B_t+f(t),&\qquad t+U\in M,\\
B_t,&\qquad t+U\notin M.\end{array}\right.$$
For any choice $(t_1,\dots,t_k)$ of a finite number of times, $U$ will a.s. not be in any of $M-t_1$, $\dots$, $M-t_k$, so the finite-dimensional distributions of $B_t$ and $B'_t$ are identical.  However, if $f$ has been chosen so that $f \chi_M$ is discontinuous everywhere on ${\Bbb R}$, $B'_t$ will a.s. be discontinuous everywhere on ${\Bbb R}_{\ge 0}$.
The moral is that you cannot construct Brownian motion directly by using the Kolmogorov extension theorem on the product space over all real times.  It's possible to construct it by using the Kolmogorov extension theorem on the product space over a countable dense set of times and then extending the motion by continuity.
Re your additional questions:
(1), (4): Your condition is stronger.  It implies that all mappings inducing $\mu$ are a.s. into $A$.  My condition says that a $\mu$-inducing mapping a.s. into $A$ exists, not that all $\mu$-inducing mappings are a.s. into $A$.
(2)  Correct.  A.s. sample path continuity is not determined by finite-dimensional distributions, or, equivalently, by the measure on the product $\sigma$-algebra.
(3)  Applying the Kolmogorov extension theorem to ${\Bbb R}^{[0,\infty)}$ gives an underdetermined random variable, since, in the product space over all nonnegative real times, there are uncountably many times, but any event in the product $\sigma$-algebra can depend on only countably many times.
