Hello everybody,


**DISCLAIMER**: I'm not a mathematician, but a computer scientist, so I hope the question is not trivial (or perhaps I hope so, in order to get a definitive answer). Anyway it's not a homework, as actually I'm working on some problem where the question arises.


Let $X$ be a $0$-dimensional Polish space (which in particular means we have a countable basis $\mathcal{C}$ of basic clopen sets).


Let $\mathcal{M}(X)$ be the set of *probability* measures over $X$. These are uniquely determined by assigning compatible values in $X$ to the basic clopen sets, i.e. there is a bijection between $\mathcal{M}(X)$ and a subset of the space $\mathcal{C}\rightarrow [0,1]$.


Let $(X,\Sigma^{B})$ be the Borel $\sigma$-algebra on $X$.


Let $(X, \Sigma^{\mu})$ be the Lebesgue $\sigma$-algebra on $X$ of all $\mu$-measurable sets.


Let $(X, \Sigma)$ be the $\sigma$-algebra, where $\Sigma=\bigcap\_{\mu\in \mathcal{M}(X)} \Sigma^{\mu}$. This is indeed a $\sigma$-algebra.


**QUESTION 1**
Is it actually possible that $\Sigma \not = \Sigma^{B}$? I think they should not be necessarily the same, but that's just an intuition. Obviously $\Sigma^{B}\subseteq \Sigma$, since $\forall \mu. \Sigma^{B}\subseteq \Sigma^{\mu}$.


**QUESTION 2**


Let $\phi: X \rightarrow [0,1]$ be a $\Sigma^{B}$-measurable function (i.e. inverse images of basic open sets are Borel).


Let $\mathcal{M}(X)$ be endowed with a topology. You can assume that this is also a $0$-dimensional polish space.


Let us define the function $g:\mathcal{M}(X)\rightarrow [0,1]$ as follows:


$g(\mu) = \displaystyle\int\_{X} \phi \ d \ \mu$ 


is $g$ (Borel)-measurable?


**QUESTION 3**
Same as question $2$, but taking $g$ just $\Sigma$-measurable (i.e. inverse images of basic open sets are in $\Sigma$). This is of course an interesting question only depending on the "outcome" of Question 1.


I kind of strongly suspect that the answer to Question 1 is YES. After all I'm plugging Borel measurable functions with integral operations, and that should be safe. 
But i'm less sure about Question 2, in particular perhaps $g$ is not Borel measurable in general, but (something-else)-measurable. I'm not sure.


Thank you in advance for any answer!


bye


matteo 


**PS**: Perhaps I should add some detail about the topology I have in mind for $\mathcal{M}(X)$. This basically is defined taking as basic open sets, the sets $S\_{C,\lambda}$ of probability measures $\mu$, such that $\mu(C)>\lambda$, for $C\in\mathcal{C}$ and $\lambda$ rational in $[0,1]$, or something like that. I haven't worked out the details so far, but that would be my guess.



measurability of integrated functions
Question 1.
Your $\Sigma$ is known as the $\sigma$-algebra of *universally measurable sets*. And, yes, there are universally measurable sets that are not Borel sets.


Question 2.
You say "a topology" ... there is a commonly-used one, the *narrow* topology, obtained by declaring that $\mu \mapsto \int \phi\,d\mu$ is continuous for any continuous $\phi$. (I believe this coincides with the one in your PS, since you are taking only this zero-dimensional case.) With that topology, your answer is "yes".


