Measurability of integrals with respect to different measures Let $Y$ be a locally compact Hausdorff topological space (further assumptions like metrizability, separability, etc., may be added if necessary) and let $\mathscr Y$ denote the Borel $\sigma$-algebra on it. Let $\Delta (Y)$ be the set of probability measures on $(Y,\mathscr Y)$ and endow it with the weak-$\star$ topology. This is, by definition, the weakest topology on $\Delta(Y)$ such that a net $(\mathbb P_{\alpha})$ in $\Delta(Y)$ converges to $\mathbb P\in\Delta(Y)$ if and only if $$\int_{y\in Y}f(y)\,\mathrm d\mathbb P_{\alpha}(y)\to\int_{y\in Y}f(y)\,\mathrm d\mathbb P(y)$$ for any bounded continuous function $f:Y\to\mathbb R$.

Now consider another measurable space $(X,\mathscr X)$ and a map $$\mathbb P:X\to\Delta (Y)$$ that is measurable, where the relevant $\sigma$-algebra on $\Delta(Y)$ is the Borel $\sigma$-algebra derived from the weak-$\star$ topology. Take any bounded measurable function $f:Y\to\mathbb R$, so that $f$ is integrable with respect to the probability measure $\mathbb P_x$ for each $x\in X$.

My question is as follows: given the measurability of the map $\mathbb P:X\to\Delta (Y)$, is the following map from $X$ to $\mathbb R$: $$x\mapsto\int_{y\in Y}f(y)\,\mathrm d\mathbb P_x(y)$$ measurable?

So far, I managed to reduce the problem to the following claim:

For any $c\in\mathbb R$ and any $B\in\mathscr Y$, the set $$\{\mathbb P\in\Delta(Y)\,|\,\mathbb P(B)\geq c\}$$ is in the Borel $\sigma$-algebra on $\Delta(Y)$ generated by the weak-$\star$ topology.

Is this claim true? I was thinking about using Lusin’s theorem to show it is, but to no avail so far.

Disclaimer: I also asked this question on Mathematics StackExchange. I repeat the question here in the hope that it can attract more attention and feedback.
 A: If $Y$ is a metric space, then the arguments in the answers to The borel $\sigma-$algebra of the set of probability measures show that the map $I_f$ defined by $I_f(\mu) = \int f\,d\mu$ is measurable for any bounded measurable $f$.  Then the answer to your question is affirmative, since your map is just $I_f \circ \mathbb{P}$.  (The question there is about compact metric spaces, but compactness is not needed.)
If you drop metrizability, things get weird.  For instance, if you let $Y = \omega_1+1$ where $\omega_1$ is the first uncountable ordinal, then the weak-* topology on $\Delta(Y)$ is not Hausdorff; if $\mu_1$ is the Dieudonne measure and $\mu_2$ is the point mass at the endpoint $\omega_1$, then $\int f\,d\mu_1 = \int f\,d\mu_2 = f(\omega_1)$ for every continuous $f$, so $\mu_1, \mu_2$ are topologically indistinguishable.  Thus every Borel set in $\Delta(Y)$ contains both measures or neither.  So if we take $f = 1_{\{\omega_1\}}$, which is a bounded Borel function on $Y$, then the map $\mu \mapsto \int f\,d\mu$ is not measurable, since $\int f\,d\mu_1 = 0 \ne 1 = \int f\,d\mu_2$.  If you like, you may then take $X=Y$ and $\mathbb{P}$ the identity map.
