# measurability of integrated functions

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.

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.
All uncountable Polish spaces are Borel isomorphic. So the answer to your first question is yes, for any uncountable Polish space. $\Sigma$ is the universal sigma algebra, but this is quite standard. – George Lowther Feb 22 '11 at 20:18