After I posted this question, a couple of months ago, and got from MO-users several good hints, I think i'm ready, after some study, to ask another related question (or rather, to focus on the main point of my previous question, after I got aware of all the necessary background).

First let me describe the setting:


Given any Polish space $X$, we denote with $\mathcal{M}(X)$ the set of all Borel-probability measures on $X$. The set $\mathcal{M}(X)$ is endowed with the smallest topology such that the map $\mathcal{M}(f): \mathcal{M}(X)\rightarrow[0,1]$ defined as

$\mu \mapsto \displaystyle \int_{X} f \ d\mu$

is continuous, for every continuous $f:X\rightarrow[0,1]$, where $[0,1]$ has the usual topology. This topology is called the "weak topology" on $\mathcal{M}(X)$, and is itself a Polish space.

As Gerald Edgar pointed out in my previous question, it turns out that the map $\mathcal{M}(g)$ defined as above, is Borel-measurable for every $g:X\rightarrow [0,1]$ Borel measurable function, i.e.

$\Big(\mathcal{M}(g)\Big)^{-1}\big( (\lambda,1] \big)$

is a Borel set in $\mathcal{M}(X)$ for every $\lambda \in [0,1)$.


Given a Polish space $X$, a subset $A\subseteq X$ is called universally measurable if and only if is $\mu$-measurable, for every (completion of) $\mu\in\mathcal{M}(X)$. The set of universally measurable sets forms a $\sigma$-algebra. A function $f:X\rightarrow[0,1]$ is called universally measurable if the inverse images of Borel sets is universally measurable. (of course this is equivalent to "inverse image of every $(\lambda,1]$ sub-basic open.." )

In particular (I look at) the set of universally measurable functions $f$ as the largest set such that $\mathcal{M}(f)$ is well defined.

FACT 1: Universally measurable functions are closed under composition.

FACT 2: Every $\Sigma^{1}_{1}$ set is universally measurable.

FACT 3: ZFC + V=L $\vdash$ "there exists a $\Delta^{1}_{2}$ set which is not universally measurable".

FACT 4: ZFC+ $\Sigma_{n}^{1}$-Determinacy, implies that every $\Sigma^{1}_{n+1}$-set is universally measurable.

FACT 5: ZFC $\Sigma_{n}^{1}$-Determinacy implies that every function $f:X\rightarrow[0,1]$ having graph $X\times[0,1]$ which is a $\Pi_{n}^{1}$-set is universally measurable. This follows from 4, because the set $f^{-1}\big( (\lambda,1] \big)$ is $\Sigma^{1}_{n+1}$.


Q1) Prove that if $f:X\rightarrow[0,1]$ is universally measurable, so is $\mathcal{M}(f)$.

I don't have many ideas for proving directly this result. Please let me know if you have any suggestions.

Anyway, I'd be happy enough (and actually equally interested) to prove the following result:

Q2) Prove that if $f:X\rightarrow[0,1]$ is function having $\Pi^{1}_{n}$ graph, so is $\mathcal{M}(f)$.

I'm not sure if the above theorem is right, or perhaps need to be weakened for example by saying that if $f:X\rightarrow[0,1]$ has a $\Pi^{1}_{n}$ graph, then $\mathcal{M}(f)$ has a $\Pi^{1}$graph, for some other $m$, maybe $m=n+1$.

This route would prove that the result of Question 1 holds (under PD), whenever $f$ has a reasonable (i.e. projective) description.

Now I would very much appreciate any suggestion for developing these ideas for Question 2. In particular I don't know precisely how to reason about the graph of $\mathcal{M}(f)$ starting from the graph of $f$. Unfortunately I'm not a mathematician and I lack proper background to work freely on this problem, and this idea of attacking the problem using PD and by considering the complexity of the graph of $f$ is the only one I had so far.

THank you again for any suggestion!



  • $\begingroup$ I just realized it is possible to offer a bounty! Ahah very funny! 350 points to the best answer! :P $\endgroup$
    – user11618
    Commented May 7, 2011 at 12:10

1 Answer 1


After some time I found the solution to Question 1 in the excellent book:

"Stochastic Optimal Control: The Discrete-Time Case", by Dimitri P. Bertsekas and Steven E. Shreve, freely available at link

The answer is YES, see Corollary 7.46.1 page 177, Chapter 7.

For what concerns Question 2 I have not found a solution yet, but I guess that after reading properly this book, I'll be able to solve it by myself.

thank you


  • $\begingroup$ that was quite some time ago, but did you happen to find the answer to the second question of yours? Bertsekas and Shreve do not seem to contain it. $\endgroup$
    – SBF
    Commented Mar 24, 2014 at 10:17

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