Let $X$ be a metrizable topological space and $\mathscr B_X$ the Borel $\sigma$-algebra on it. Let $\Delta X$ denote the set of probability measures on $(X,\mathscr B_X)$, and let $\mathscr B_{\Delta X}$ denote the Borel $\sigma$-algebra generated by the weak-star topology on $\Delta X$ (which is defined as the weakest topology on $\Delta X$ that makes the map $\mathbb P\mapsto\int_X f(x)\,\mathrm d\mathbb P(x)$ continuous for every bounded and continuous function $f:X\to\mathbb R$). In what follows, let $\mathscr B_{\mathbb R}$ denote the usual Borel $\sigma$-algebra on the real line.
Suppose that $(\Omega,\mathscr F)$ is a measurable space and, for each $\omega\in\Omega$, $\mathbb P_{\omega}\in\Delta X$. That is, the space $\Omega$ is supposed to “index” a set of Borel probability measures on $X$ in a measurable manner.
CONJECTURE: The map $\omega\mapsto\mathbb P_{\omega}$ is $\mathscr F/\mathscr B_{\Delta X}$-measurable if and only if the map $\omega\mapsto\mathbb P_{\omega}(A)$ is $\mathscr F/\mathscr B_{\mathbb R}$-measurable for each fixed $A\in\mathscr B_X$.
While I have not checked the details, the “only if” direction does not seem to be difficult to me. The trick is to establish that the map $\mathbb P\mapsto\mathbb P(A)$ is $\mathscr B_{\Delta X}/\mathscr B_{\mathbb R}$-measurable for any given $A\in\mathscr B_X$—see this thread for more details. As this other thread reveals, metrizability of $X$ is essential.
I have trouble proving the “if” direction; in fact, my hunch tells me that it may not even be true without separability of the metric topology on $X$ (which would imply the separability of the weak-star topology on $\Delta X$ as well). Yet, constructing a counterexample has equally eluded me.
I would be very grateful for any hint or insight regarding (i) the proof of the “if” direction; or (ii) a counterexample showing that separability is indispensable.
UPDATE: The “if” direction fails whenever $\mathscr B_{\Delta X}$ strictly includes the $\sigma$-algebra generated by sets of the form $\{\mathbb P\in\Delta X\,|\,\mathbb P(A)\in E\}$, where $A\in\mathscr B_X$ and $E\in\mathscr B_{\mathbb R}$. This is because if one takes $\Omega\equiv\Delta X$ with $\omega\mapsto\mathbb P_{\omega}$ being the identity map on $\Delta X$ and $\mathscr F$ to be the generated $\sigma$-algebra just described, then $\omega\mapsto\mathbb P_{\omega}(A)$ (i.e., $\mathbb P\mapsto\mathbb P(A)$) is $\mathscr F/\mathscr B_{\mathbb R}$-measurable for each $A\in\mathscr B_X$ but the identity map on $\Delta X$ is not $\mathscr F/\mathscr B_{\Delta X}$-measurable. I would still very much appreciate an example that explicitly shows the possibility $\mathscr F\subset\mathscr B_{\Delta X}$ (strictly).
