# Abstract transverse measure theory

After reading Noncommutative Geometry book (see here) I came across the notion of the so called abstract transverse measure theory which is a generalization of standard measure theory well adapted to deal with singular examples like the space of leaves of the foliation. All the relevant information for me can be found on pages 78 and 79 of the above mentioned book. However this exposition is rather brief and I have several questions regarding this topic. I will divide them into more than one post in order to avoid asking too many questions in one post. So the context is as follows: $$X,Y$$ are standard Borel spaces and $$p:Y \to X$$ is a Borel map such that for each $$x \in X$$ the preimage $$p^{-1}(x)$$ is at most countable.

The first claim is as follows: one defines $$F(x)$$ to be the cardinality of $$p^{-1}(x)$$. In this way we obtain a function $$F:X \to \mathbb{N} \cup \{0\} \cup \{ \infty \}$$ and the claim is that

There exists a Borel bijection $$\psi: Y \to \{ (x,n): 0\leq n < F(x) \}$$

My guess is that $$\psi$$ should be defined as follows: let $$Y=\sqcup_{n=0}^{\infty}Y_n$$ be a partition of $$Y$$ with $$Y_n:=\{y \in Y: card(p^{-1}(p(y))=n\}$$ and $$\psi(y):=(p(y),n)$$ for $$y \in Y_n$$. However

Q1 How to prove that $$\psi$$ is Borel?

According to the discussion here the function $$F$$ need not no be measurable (and if we had defined $$F$$ as $$F(y)=card(p^{-1}(p(y))$$ still the same argument seems to work).

Now the second claim is the following: suppose that we have two pairs $$(Y_1,p_1), (Y_2,p_2)$$ as above and let $$F_1,F_2$$ be as the functions as above constructed with respect to $$p_1,$$ and $$p_2$$. The claim is that:

$$F_1=F_2 \iff$$ there exists Borel bijection $$\psi:Y_1 \to Y_2$$ such that $$p_2 \circ \psi=p_1$$.

One direction of the above claim is clear: if we have $$\psi$$ then $$F_1=F_2$$. However I don't see how to prove the converse: so I'd like to ask

Q2 How to prove that if $$F_1=F_2$$ then there exists a Borel bijection $$\psi:Y_1 \to Y_2$$ such that $$p_2 \circ \psi=p_1$$?

After stating that the author drops the assumption for $$X$$ being standard Borel space and regard the second condition, i.e. the existence of $$\psi$$ as the definition of ,,having the same functions $$F_i$$''. I don't see what is the point here:
-First of all there is no need for $$F_i$$ to be Borel so even if $$X$$ is a standard Borel space and $$p$$ is Borel map still $$F_i$$ may not be Borel so in some sense (from the point of view of $$F_i$$) there is no difference whether $$X$$ was standard Borel or not -Second thing: even if $$F_i$$ are not Borel functions they still make perfect sense even if $$X$$ is no more a standard Borel space: thus we can still ask whether $$F_1=F_2$$
-and on the top of that: why are we interested in this strange notion of ,,having the same functions'' i.e. $$F_1=F_2$$? If we are interested in pairs $$(Y,p)$$ where $$Y$$ is standard Borel and $$p$$ is a map $$p:Y \to X$$ where $$X$$ is fixed then the natural notion of equivalence is the existence of $$\psi$$ and ,,having the same funtions'' seems artificial to me.

So to summarize:

Q3 Why should we care whether $$F_1=F_2$$?

Your proposed $$\psi$$ can't be bijective, since two $$y,y'\in Y$$ with $$p(y)=p(y')$$ are sent to the same pair. But indeed, the sets $$Y_n$$ are measurable (and hence the functions $$F$$ are Borel). This follows from Exercise 18.15 in Kechris' Classical Descriptive Set Theory, which states exactly what you need that but for a subset $$P\subseteq X\times Y$$ with countable sections $$P_x$$ (you can take $$P:=(\mathrm{graph}(p))^{-1}$$).
Let $$A_n= \{x\in X : \mathrm{card}(p^{-1}(x)) = n\}$$ for $$n=0,\dots,\infty$$. Then 18.15 says that for each $$n$$, $$A_n$$ is Borel and there are Borel functions $$f_i^{(n)}:A_n\to Y$$ with disjoint graphs such that for $$x\in A_n$$, $$p^{-1}(x) = \{f_i^{(n)}(x) : i . This exercise follows from the Lusin-Novikov Theorem (18.10 in the same book) that states the projection of such a $$P$$ must be Borel in $$X$$.
You can then define $$\psi$$ by cases using the functions above. Moreover, it will satisfy $$\psi(y)=(p(y),n(y))$$ for some $$n:Y\to\mathbb{N}_0$$.
For Q2, note that Borel bijections between standard Borel spaces are indeed Borel isomorphisms, and hence they have Borel inverses. Let $$\psi_i:Y_i\to \{ (x,n): 0\leq n < F(x)\}$$, where $$F=F_1=F_2$$. Then $$\psi:=\psi_2^{-1}\circ \psi_1$$ is the required Borel bijection.
• Thank you for your answer. Regarding Q2, I don't see why it follows that $p_2 \circ \psi=p_1$ – truebaran Jul 25 '19 at 20:16
• You are welcome. The reason is the shape of the maps $\psi_i$ in coordinates, described in the previous paragraph. I'm on my phone right now, if you need further clarification I'll answer when I get to a keyboard. – Pedro Sánchez Terraf Jul 25 '19 at 20:35