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 $f:Y\to\mathbb R$.

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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 $x\in X$.

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**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?

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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.

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Disclaimer: I also asked this question on [Mathematics StackExchange](https://math.stackexchange.com/questions/1748864/measurability-of-integrals-with-respect-to-different-measures). I repeat the question here in the hope that it can attract more attention and feedback.