Measurability of a map involving probability measures

Question

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.

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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).

Answer 1

Case I: There exists a real-measurable cardinal $\kappa$. Then a discrete space $X$ of cardinality $\kappa$ is a counterexample to the conjecture.

Denote by $\mathcal{A}$ the $\sigma$-algebra of all subsets of $X$. Let $\mu \colon \mathcal{A} \to [0,1]$ be a probability measure that witnesses the real-measurability of $\kappa$. The following lemma is probably well-known, although I don't have a reference on hand.

Lemma. For any countable set $\mathcal{C}\subseteq \mathcal{A}$ there exists another witnessing measure $\nu \colon \mathcal{A} \to [0,1]$, $\nu\neq\mu$, such that $\mu$ and $\nu$ agree on $\mathcal{C}$.

By the lemma, the Borel subset $\{\mu\}$ of $\Delta X$ is not in the $\sigma$-algebra generated by the sets of the form $\{\mathbb{P}\in\Delta X \mid \mathbb{P}(A)\in E\}$.

Case II: There are no real-measurable cardinals. Then the "if" direction of the conjecture holds.

In this case every finite $\sigma$-additive Borel measure on a metric space $X$ is $\tau$-additive, hence $\Delta X$ with the weak-star topology is metrizable.