Embeddings of spaces of probability measures What is the relationship between the spaces $X_1\triangleq \mathscr{P}(C([0,1],\mathbb{R}))$ and $X_2\triangleq C([0,1],\mathscr{P}(\mathbb{R}))$; where $\mathscr{P}(\cdot)$ denotes the Borel probability measures on a space and it is equipped with the total-variation topology.  Specifically, I wonder, is $X_1$ continuously embeded in $X_2$ or the converse?
 A: This is an extended version of my comment above. It is not an answer, or at least: not a positive answer. (Perhaps it is sort of a negative answer?)

There is a (sort of natural) candidate for an embedding $\Phi$ from $X_2$ into $X_1$, defined as follows. Take a family of measures $\mu = (\mu_t, t \in [0,1])$ in $X_2$, and denote by $F_t(x)$ the corresponding distribution functions. Let $F_t^{-1}$ be the generalised inverse function of $F_t$, so that if $U$ is a random variable uniformly distributed on $[0, 1]$, then
$$ F_t^{-1}(U) $$
is a random variable with distribution $\mu_t$. Finally, define a stochastic process
$$ x_t = F_t^{-1}(U) ,$$
and set $\Phi(\mu)$ to be the law of the process $x_t$ (again with $U$ distributed uniformly over $[0,1]$).
The above construction is natural in the following sense: $x_t$ is the unique process with one-dimensional distributions (a.k.a. marginals) $\mu_t$ with the following monotonicity property: if $x_t(\omega_1) \leqslant x_t(\omega_2)$ for some $t$, then $x_t(\omega_1) \leqslant x_t(\omega_2)$ for all $t$.

The above 'embedding' looks nice, but it remains to prove that $\Phi(\mu)$ is indeed in $X_1$. In other words: $x_t$ has almost surely continuous paths. Equivalently: $t \mapsto F_t^{-1}(u)$ is continuous for almost every $u \in [0, 1]$. This need not be true, though!
Set $\mu_t = (1 - t) \delta_0 + t \delta_1$. Then $x_t = 0$ for $t < U$ and $x_t = 1$ for $t > U$, which means that the path of $x_t$ is almost surely discontinuous!

Clearly, $\mu$ constructed above is not a family of one-dimensional distributions of a continuous stochastic process, so in a sense there is no hope to fix the above construction. The only thing one can easily say is that if $t \mapsto \mu_t$ is a function of bounded variation (with respect to the total variation distance), then $x_t$ is almost surely a function of bounded variation:
$$ \mathbb E \sum_i |x_{t_i}-x_{t_{i-1}}| = \sum_i \|\mu_{t_i} - \mu_{t_{i-1}}\|_{TV} \leqslant \operatorname{TV}(t \mapsto \mu_t) , $$
and hence
$$ \mathbb E \operatorname{TV}(t \mapsto x_t) \leqslant \operatorname{TV}(t \mapsto \mu_t) . $$
