Measurability of marginal distributions of a random measurable function For a probability space $(\Omega, \mathcal F, \mathsf P)$, let $X \colon \Omega \times [0,1] \to \mathbf R \colon (\omega, t) \mapsto X(\omega,t)$ be a random Borel function (i.e. an $(\mathcal F\times \mathcal B(\mathbf R))$-measurable map, see 4a1 here), and let $\mathcal M(\mathbf R)$ be the space of probability measures on $\mathbf R$ with the topology of weak convergence. For $t\in[0,1]$, denote by $P_t$ the distribution of $X(\cdot,t)$. 
Is the mapping $\Pi \colon [0,1] \to \mathcal M(\mathbf R) \colon t \mapsto P_t$ necessarily Borel (or Lebesgue) measurable?
If not, is the claim true at least for a random semicontinuous function $X$ (see 4c here)?
Thank you.
 A: I claim that $\Pi$ is Borel measurable.  Define the family
$$
   \mathscr{A} := \left\{\text{Borel measurable $E\subseteq\mathcal{M}(\mathbb{R})$}: \text{$\Pi^{-1}(E)$ is Borel measurable}\right\} \;.
$$
It is enough to verify that


*

*$\mathscr{A}$ is a $\sigma$-algebra, and

*$\mathscr{A}$ contains a countable sub-family $\mathscr{C}$ that generates the weak topology on $\mathcal{M}(\mathbb{R})$.


That $\mathscr{A}$ is a $\sigma$-algebra is straightforward ($\Pi^{-1}$ commutes with complementation, infinite unions and infinite intersections).
Since $\mathbb{R}$ is Polish (i.e., separable and completely metrizable), so is the space $\mathcal{M}(\mathbb{R})$ of Borel probability measures on $\mathbb{R}$ with the weak topology.  In particular, $\mathcal{M}(\mathbb{R})$ is metric and separable, and hence second-countable.  Therefore, any base of the weak topology on $\mathcal{M}(\mathbb{R})$ has a countable sub-family that is itself a base.  Moreover, every generating family for the weak topology on $\mathcal{M}(\mathbb{R})$ has a countable generating sub-family.
Now, as the family $\mathscr{C}$ in point (2), we use a countable generating sub-family of sets of the form
$$
   U(\pi,f,\varepsilon) := \left\{
      \mu\in\mathcal{M}(\mathbb{R}):
      \big|\mu(f)-\pi(f)\big|<\varepsilon
   \right\} \;.
$$
for probability measures $\pi\in\mathcal{M}(\mathbb{R})$, bounded continuous functions $f\in C_b(\mathbb{R})$ and real numbers $\varepsilon>0$.
It remains to see that $U(\pi,f,\varepsilon)\in\mathscr{A}$, or equivalently,
\begin{align}
   \Pi^{-1}\left(U(\pi,f,\varepsilon)\right) &=
      \left\{
         t\in[0,1]: \big|P_t(f)-\pi(f)\big|<\varepsilon
      \right\} \\
   &=
      \left\{
         t\in[0,1]:
            \int f\left(X(\omega,t)\right)\,P(\mathrm{d}\omega)
               \in(\pi(f)-\varepsilon, \pi(f)+\varepsilon)
      \right\}
\end{align}
is Borel measurable.  But this is an instance of the more general standard fact (a prerequisite of the Fubini-Tonelli Theorem) that for a bounded measurable function $g:X\times Y\to\mathbb{R}$ and a probability measure $\mu$ on $Y$, the function
$$
   x \mapsto \int g(x,y)\,\mu(\mathrm{d}y)
$$
is measurable.
