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This question is related to another one that I asked two days ago.

Question. Does there exist a Borel subset $ M $ of $ \mathbb{R}^{2} $ with the following two properties?

  • The projection of $ M $ onto the first component is a non-Borel analytic subset of $ \mathbb{R} $.
  • Every vertical cross-section of $ M $ is finite, i.e., the set $ \{ y \in \mathbb{R} \mid (x,y) \in M \} $ is finite for every $ x \in \mathbb{R} $.

An affirmative answer to this question will provide a counterexample in the topic of measure theory, as explained in the linked post. It therefore holds some importance.

Thank you very much for your help!

Community
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Transcendental
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    I think the answer is no, and it could be proved with results from Borel equivalence relations.
    Consider $M$ as a standard Borel space, and the Borel equivalence relation $E$ on $M$ which is $(x,y) E (x', y')$ iff $x=x'$. Since every equivalence class of $E$ is finite, it has a transversal: a Borel set $A \subset M$ containing one element from each $E$-class. (See Proposition 1.4.4 in http://homepages.math.uic.edu/~kslutsky/papers/cber.pdf.) Then the projection $\pi : A \to \mathbb{R}$ is an injective Borel function, hence its image $\pi(A) = \pi(M)$ is Borel.     – Nate Eldredge Nov 02 '16 at 06:45
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    But there should be a more elementary argument. – Nate Eldredge Nov 02 '16 at 06:46
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    @Nate: Thank you for your comments, Nate. Yes, a more elementary argument might do away with the requirement that $ M $ be a standard Borel space. Hopefully, a descriptive set theorist will be able to weigh in. – Transcendental Nov 02 '16 at 09:56
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    Oh, that part's not a requirement: since $M$ is a Borel subset of $\mathbb{R}^2$ it is already a standard Borel space. I just mean to think of it as a space in its own right, since we don't really want to work with an equivalence relation on all of $\mathbb{R}^2$. – Nate Eldredge Nov 02 '16 at 12:51

3 Answers3

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No, no such set exists. This is a special case of the Lusin–Novikov theorem; see e.g. Kechris, Classical Descriptive Set Theory, Theorem 18.10.

In general, let $X,Y$ be standard Borel spaces, and suppose $M \subset X \times Y$ is Borel. (Here we are taking $X=Y=\mathbb{R}$.) For $x \in X$, let $M_x = \{y : (x,y) \in M\}$ be the section of $M$ at $x$. The Lusin–Novikov theorem asserts that if all but at most countably many of the $M_x$ are at most countable (i.e. $|\{x : |M_x| > \aleph_0\}| \le \aleph_0$) , then $M$ has a Borel uniformization: there is a Borel set $M^* \subset M$ such that $M^*$ intersects every nonempty section $M_x$ in exactly one point. In particular, the projection map $\pi : X \times Y \to X$, which is Borel, is injective when restricted to $M^*$; so as a consequence (see Kechris Corollary 15.2) we have that $\pi(M^*) = \pi(M)$ is Borel in $X$.

Any set $M$ satisfying your second condition certainly satisfies the hypothesis of Lusin–Novikov (since $\{x : |M_x| > \aleph_0\} = \emptyset$), so its projection is Borel, and thus it does not satisfy your first condition.

Nate Eldredge
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    Or you can even prove it bare-handedly by noticing there is a borel linear ordering of the space, define the uniformizing function by picking the least element under that – Jing Zhang Nov 02 '16 at 16:46
  • @JingZhang: So you're suggesting something like let $f(x) = \inf M_x$? I considered that but did not see how to show it's a Borel function. We have to use somewhere the fact that the sections are finite. For general $M$ this will not give a Borel function. – Nate Eldredge Nov 02 '16 at 16:53
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    @NateEldredge: Yeah the definition is $(x,y)\in f \Leftrightarrow (x,y)\in B \wedge \forall z<y (x,z)\not \in B$, which is just co-analytic. I believe Luzin-Novikov is needed to reduce $\forall z<y$ to a countable operation. (I thought of Feldman-Moore which states every countable Borel equivalence relation is the same as an orbit equivalence relation generated by some action of a countable group on the space, but I don't think this is any easier than LN). – Jing Zhang Nov 02 '16 at 18:04
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The projection is always Borel provided the set is Borel and each cross-section is at most countable. This is an old theorem of Luzin and/or Novikov, valid for sets in any product of two polish spaces, and the countability can be weakened to sigma-compactness (Novikov and/or Kunugui).

Vladimir Kanovei
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There is a pure recursion theoretical proof of the result. The idea is as follows: By Spector-Gandy theorem, a lightface Borel set $(x,y)$ is an r.e set over $L_{\omega_1^{CK}}[x,y]$. If there are at most countably many $y$'s corresponded to the $x$ for every $x$, then it can be reduced to $L_{\omega_1^{CK}}[x]$. Then the projection to the first section becomes a $\Pi^1_1$ statement.

喻 良
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