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Let $X$ be a Polish space with the probability measure $P$ and the Borel sigma-algebra. Suppose that $X$ is also a group such that $(x,y)\mapsto xy^{-1}$ is Borel measurable and the probability $P$ is left and right quasiinvariant. Let $P_x$ denote the probability measure $P_x(A)=P(xA)$ for every $x$ in $X$. Obviously, for each $x$ in $X$, the Radon-Nikodym derivative $dP_x/dP$ is Borel measurable.

I am trying to show that there is a measurable function $\Phi:X \times X\rightarrow[0,\infty )$ such that for every $x$ in $X$, $\Phi(x,y)=(dP_x/dP)(y)$ for a.e. $y$ (notice that $\Phi$ needs to be measurable with respect to the Borel sigma-algebra of the product space).

I can show that there is a measurable function $\Phi$ such that for $P$ - almost every $x$ in $X$, $\Phi(x,y)=(dP_x/dP)(y)$ for a.e. $y$, by taking the derivative $dm/dP\times P$, where $m=(P\times P)\circ S$ and $S:X \times X\rightarrow X\times X$ is the function $S(x,y)=(x,x^{-1}y)$. But I need the equality for every $x$ in $X$.

Any suggestions?

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Are you aware that your hypotheses imply that $X$ is a locally compact second countable group and that $P$ is in the Haar class by a theorem of Weil and Mackey? (Essentially the point is that you can embed $X$ into the unitary group of $L^2(X)$ and it inherits a locally compact group topology from the unitary group of $L^2(X)$.) If you're willing to accept this, your question becomes obvious. However, you might as well be trying to prove that result, so I'm asking for a clarification on your ultimate goal. – t.b. Aug 17 '11 at 20:22
I am familiar with that theorem. In fact, I am trying to show that this embedding exactly is Borel measurable (as part of the proof of this theorem). In order to do so, I am trying to show that such a function Φ exists. – Arnold Aug 17 '11 at 23:52
Okay, I'll write something later today or tomorrow if nobody else does. However, the proof is somewhat technical and I'd probably refer you to Fisher—Witte Morris—Whyte anyway, so I can as well point you to that reference now. What you ask is a special case of their Proposition 2.22. – t.b. Aug 18 '11 at 8:49
Thank you, I believe that lemma 2.7 is the one I need. – Arnold Aug 18 '11 at 13:50
Yes, exactly, that's the main point. But note that it is not quite obvious that your $\Phi$ actually gives a Borel map $X \to \mathbf{F}(X,[0,\infty))$ (in their notation) doing what you want; you can't just shove $\Phi$ into lemma 2.6, apply 2.7 and get the thing you're looking for entirely for free! You need an argument similar to the one in 2.22 but you already have the main ingredients. – t.b. Aug 18 '11 at 14:02

Let me start by mentioning that the kind of things you are trying to understand are folklore in the worst sense. I can't resist quoting R.W. Thomason (in a completely different context, but equally applicable here): "Most of these facts are well-known in outline, although many people exhibit some confusion and fuzziness on the details when pressed."

Some good references:

  • Appendix B and Chapter 2 of R.J. Zimmer, Ergodic theory and semisimple groups, Birkhäuser 1984. MR776417

  • David Fisher, Dave Witte Morris, and Kevin Whyte, Nonergodic actions, cocycles and superrigidity, New York Journal of Mathematics, Volume 10 (2004) 249–269, MR2114789 and the references therein.

  • For an excellent introduction to the bare minimum on descriptive set theory and selection theorems, I recommend Chapter III of Arveson, An invitation to $C^{\ast}$-algebras, Springer GTM 39, 1979, MR512360.

  • By far the best introduction to descriptive set theory and the circle of ideas you're considering is A.S. Kechris, Classical descriptive set theory, Springer GTM 156, 1995. MR1321597.

  • Chapter V of Varadarajan's book The geometry of quantum theory, second edition, Springer 1969, MR805158. In particular the Mackey-Weil theorem is proved in section 6 of that chapter.

If you can manage to read Mackey's original texts, you will certainly only profit, but I found them to be on the extremely indigestible side if you pardon my blunt assessment.

Edit: (in view of Mark's comment below) If your ultimate goal is to learn about the "Mackey machine" and induced representations, you can follow Mark's recommendation and look at Folland's Abstract Harmonic Analysis, Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1995. MR1397028.

Another nice text on that topic is Barut–Rączka, Theory of group representations and applications. Second edition. World Scientific Publishing Co., Singapore, 1986. MR0889252.

So, let's look at the math, finally:

I'd like to prove a stronger statement than the one you ask about:

Let $G$ be a standard Borel group and let $X$ be a standard Borel space equipped with a Borel $G$-action and a quasi-invariant probability measure $\mu$. There is a Borel measurable map $\rho: G \times X \to [0,\infty)$ such that $$\int_{X} \rho(g,x)f(x)\,d\mu(x) = \int_{X} f(gx)\,d\mu(x)$$ for all $g \in G$ and all $f \in L^1(X,\mu)$.

You'll find this statement and some applications as Lemma 1.1.1 of Appendix D on page 84 of my thesis (where it is formulated for $G$ Polish but that isn't used here). It is extracted from Fisher–Witte Morris–Whyte's Proposition 2.22.

First of all recall that the space $\mathscr{F}(X)$ of $\mu$-equivalence classes of Borel-measurable functions $X \to [0,\infty)$ is Polish with respect to the topology of convergence in measure. In fact the metric $$d_{\mathscr{F}}(f,g) = \min{\{\varepsilon \geq 0\,:\,\mu(\{|f(s) - g(s)| \gt \varepsilon\})\leq \varepsilon\}}$$ is particularly convenient.

Translating Fisher–Witte Morris–Whyte's Lemma 2.7 into our situation we get:

Lemma. Given a Borel measurable function $f: G \to \mathscr{F}(X)$ there exists a Borel measurable function $\varphi: G \times X \to [0,\infty)$ such that for all $g \in G$ we have $$\varphi(g,x) = f(g)(x) \quad \text{ for almost every } x \in X.$$

Note that his implies the exponential law $\mathscr{F}(G \times X) = \mathscr{F}(G,\mathscr{F}(X))$.

The proof of the lemma is relatively simple: partition $\mathscr{F}(X)$ into countably many disjoint Borel sets $\{D_{i}^{(n)}\}_{i=1}^{\infty}$ of diameter $2^{-n}$, for each $i$ pick $\varphi^{(n)}_i \in D_{i}^{(n)}$ and put $\varphi^{(n)}(g,x) = \varphi^{(n)}_i(x)$ if $f(g) \in D_{i}^{(n)}$. Then it is easy to see that for fixed $g$ the function $\varphi^{(n)}(g, \cdot)$ is Borel and converges a.e. to a limit $\varphi(g,\cdot)$. For details see loc. cit.

Given this, it remains to show that the map $r: G \to \mathscr{F}(X)$ given by $g \mapsto \frac{d(g\mu)}{d\mu}$ is Borel measurable. To see this, choose a countable separating and generating set $\{A_n\}_{n=1}^{\infty}$ of $\Sigma$ and notice that $$ \mathscr{R} = \bigcap_{n=1}^{\infty} \left\{(g,f) \in G \times \mathscr{F}(X)\,:\,\mu(gA_n) = \int_{A_n} f(x)\,d\mu(x)\right\} $$ is the graph of $r$ because the fiber of $\mathscr{R}$ over $g$ is the Radon-Nikodym derivative $\frac{d(g\mu)}{d\mu}$ of the Borel automorphism $x \mapsto gx$ of $(X,\Sigma)$. Hence, it suffices to show that $\mathscr{R}$ is Borel by the Borel graph theorem (e.g. Kechris, Theorem 14.12 on page 88). We're done as soon as we show that $g\mapsto\mu(gA_n)$ and $f \mapsto \int_{A_n} f\,d\mu$ are Borel on $G$ and $\mathscr{F}$, respectively. For the latter map this is an application of the monotone convergence theorem, see Fisher–Witte Morris–Whyte, Lemma 2.17, and for the former we write $gA_n \subset G \times X$ as $\psi(\{g\} \times A_n) \cap (\operatorname{pr}_G)^{-1}(\{g\})$ where $\psi: G \times X \to G \times X$ is the Borel automorphism $\psi(g,x) = (g,gx)$, hence it maps Borel sets to Borel sets and measurability of $g \mapsto \mu(gA_n)$ now follows e.g. from Kechris's theorem 17.25 on page 113.

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A nice and friendly text which treats this particular subject is Folland's book on abstract harmonic analysis. – Mark Aug 18 '11 at 14:59
@Mark: Yes, Folland's book is very nice but it doesn't contain any cocycle reduction theorems nor does it contain the Mackey-Weil theorem or the necessary machinery for developing it, but thanks for pointing it out. – t.b. Aug 18 '11 at 15:20
@Mark: I'm asking because I was a bit puzzled by your comment: What exactly do you mean by "this particular subject"? Is my interpretation "induced representations" and "Mackey machine" accurate or did you have something more specific in mind? I looked again into Folland and I failed to find anything particularly relevant to the present question. – t.b. Aug 19 '11 at 0:00
@Theo: I gotta 'fess up: an anonymous user did the bulk of the editing, actually. All I did was delete all the  's s/he had to insert just so the software could let him/her edit... – J. M. Aug 19 '11 at 11:10
@J.M. I should've thanked for accepting the edit I suggested a bit earlier from a public machine, then (didn't think I'd be back that soon) :) – t.b. Aug 19 '11 at 11:12

Just to add to Theo's excellent answer, I will quote in full Exercise 6.10.72 from page 63 of V. Bogachev's Measure Theory (volume 2). A long hint is also given there.

Let $(X,{\cal A},\mu)$ be a probability space, ${\cal A}$ a countably generated $\sigma$-algebra, $(T,{\cal B})$ a measurable space, and let $\mu_t$, where $t\in T$, be a family of bounded measures on $\cal A$ absolutely continuous with respect to $\mu$ such that for every $A\in {\cal A}$, the function $t\mapsto \mu_t(A)$ is measurable with respect to $\cal B$. Prove that one can find an ${\cal A}\otimes{\cal B}$-measurable function $f$ on $X\times T$ such that for every $t\in T$, the function $x\mapsto f(x,t)$ is the Radon-Nikodym density of the measure $\mu_t$ with respect to $\mu$.

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Oh, great! Thanks for that reference and the kind words! – t.b. Aug 20 '11 at 15:05
No problem. I always learn a lot from your answers. – Byron Schmuland Aug 20 '11 at 15:17
Byron, thanks again, hearing this from someone like you means a lot to me. – t.b. Aug 21 '11 at 0:03

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