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.
> 
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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.



Measurability of integrals with respect to different measures
If $Y$ is a metric space, then the arguments in the answers to [The borel $\sigma-$algebra of the set of probability measures](https://mathoverflow.net/q/167823/4832) 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.


