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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.

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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
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See en.wikipedia.org/wiki/Universally_measurable_set for the basics about universally measurable sets. –  Bill Johnson Feb 22 '11 at 20:28
    
Hi George and Bil! thanks a lot for the answers! @George: just to be sure (since you also gave an answer below), $\Sigma$ - the universal $\sigma$-algebra$ - on an uncountable Polish space does not coincide with $\Sigma^{B}$, right? Just wanted to be sure about the meaning "YES". Anyway I borrowed Kechris "Classical Descriptive Set Theory", so tomorrow i'll search for it. –  Matteo Mio Feb 22 '11 at 21:03
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@Matteo: I didn't give an answer below, but yes, $\Sigma$ does not coincide with $\Sigma^B$. Here's an example of an element of $\Sigma\setminus\Sigma^B$ for the real numbers; planetmath.org/?op=getobj&from=objects&id=11351 (or, if you prefer, in the irrational real numbers, which is zero dimensional). –  George Lowther Feb 22 '11 at 21:22
    
@George: Uh, embarrassing!! I miss-read Gerald below with George and just assumed it was the same entity :) I must be a bit tired. Thanks a lot for the answer! –  Matteo Mio Feb 22 '11 at 21:31
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2 Answers

up vote 7 down vote accepted

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".

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Hi gain Gerald :) Beside point $1$, which you answered already, could you please expand a little bit on points 2-3? Unlike for "universal $\sigma$-algebra", wikipedia does not seem to cover "narrow topology". Could expand and/or you give me some reference on this? (I already have Kechris' "Classical Descriptive Set Theory") just from the post I'm not sure I understood what the opens are in the narrow topology. And, if you have time, could you expand a little bit on your intuition that "my topology" coincides with the narrow one for $0$-dimensional POlish spaces? thanks in adanve. –  Matteo Mio Feb 22 '11 at 21:07
    
Yes, "narrow" is also called "weak" or "weak*", which can be confusing. If $\phi \colon X \to \mathbb R$ is a continuous function and $U$ is an open set in $\mathbb R$, then $\{\mu \colon \int \phi\, d\mu \in U\}$ is a basic open set for this topology. –  Gerald Edgar Feb 23 '11 at 4:07
    
Once you have $\mu \mapsto \int \phi\,d\mu$ continuous for every bounded continuous $\phi$, you easily get $\mu \mapsto \int \phi\,d\mu$ Borel for every bounded Borel $\phi$. This is because the class of bounded Borel functions is the smallest class containing the bounded continuous functions and closed under limits of pointwise-convergent sequences. –  Gerald Edgar Feb 23 '11 at 4:12
    
Note that this topology does make $\mathcal{M}(X)$ a polish space, but not a $0$-dimensional one. Indeed, if $x\neq y\in X$ you can produce continuous curves $t\mapsto t \delta_x +(1-t)\delta_y$ where $\delta$ is the Dirac mass. –  Benoît Kloeckner Mar 11 '11 at 15:31
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I know in computable analysis, which is closely related to the descriptive set theoretic questions you are asking, that the Lévy–Prokhorov metric on the weak topology is useful. (I think the narrow topology mentioned by Gerald Edgar is the same. Also it is called the weak* topology in functional analysis. See http://en.wikipedia.org/wiki/Convergence_of_measures.)

If your space $X$ is a separable metric space, then $\mathcal{M}(X)$ is also a separable metric space under the Lévy–Prokhorov metric. Further if $X$ is a computable metric space, then $\mathcal{M}(X)$ is a computable metric space under the Lévy–Prokhorov metric. (I'm not sure if such computable concerns matter to you, but a proof that it is computable can be found in this paper: Computability of probability measures and Martin-Löf randomness over metric spaces, along with other useful facts.)

Update: Also I should mention that there are (at least) two definitions of a computable measure $\mu$ on $X$. One is that you can, given a finite union $U$ of open basis sets $B_1,\ldots B_k$, compute all rationals $\lambda$ such that $\mu(U)>\lambda$. The other is that it is a computable point in the Lévy–Prokhorov metric on $\mathcal{M}(X)$. The first definition seems similar to what you are looking for, but without the computability concerns. Again, the paper I mentioned above shows the equivalence of these two definitions. (But then again, there may be more standard references that don't dive into the computable stuff.)

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