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Let $(X,\Sigma)$ be a standard measurable space, and let $\,\preceq\,$ be a total order on $X$ with the property that $\,\{(x,y) \in X \times X: x \preceq y\} \in \Sigma \otimes \Sigma$.

Let $A \subset X$ be a set with the property that for all $x,z \in A$ and $y \in X$ with $x \preceq y \preceq z$, we have $y \in A$.

Is it necessarily the case that $A \in \Sigma$? If not, is $A$ necessarily universally measurable with respect to $\Sigma$?

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    $\begingroup$ A thought: is it possible for a Borel order to have a set isomorphic to $\omega_1$? It seems unlikely. If not, then take a cofinal well-ordered subset $C$ of $A$, which must then be countable. Fix some $x_0 \in A$ and then you have $A \cap [x,\infty) = \bigcup_{y \in C} [x,y]$ which is Borel. Applying the same argument to the reverse order, $A \cap (-\infty, x]$ is also Borel. $\endgroup$
    – Nate Eldredge
    Commented Sep 20, 2015 at 17:24

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I believe the answer is yes. (Please check this answer carefully, as this is rather outside my field. There may well be a much easier solution.)

In the paper "Borel Orderings" by Harrington, Marker and Shelah (Trans. AMS 310 (1988), 293-302, MR 0965754), the authors consider Borel (partial) orderings. A Borel ordering is said to be thin if there does not exist a perfect set of pairwise incomparable elements. Of course, in a total order there are no incomparable elements, so every Borel total order is thin. Corollary 3.2 states that a thin Borel order does not contain an $\omega_1$-chain. (The original reference given for this result is a 1982 paper of Harrington and Shelah, "Counting equivalence classes for co-$\kappa$-Souslin relations", MR 673790, text at ScienceDirect, for which I lack the appropriate subscription.)

Given this, suppose $\preceq$ is a Borel total order on the Polish space $X$ and $A \subset X$ is convex. Using Zorn's lemma, choose a subset $C \subset A$ which is well-ordered by $\preceq$ and cofinal in $A$ (see Well-ordered cofinal subsets). If $C$ is uncountable then it contains a subset with the order type of $\omega_1$, which is forbidden by the Harrington–Shelah result. So $C$ is countable. Fix some $x_0 \in A$; then we have $A \cap [x_0,\infty) := \{x \in A: x \succeq x_0\} = \bigcup_{y \in C} [x_0,y]$. (The $\subseteq$ direction is because $C$ is cofinal, and the $\supseteq$ direction is because $A$ is convex.) This is a countable union, and each closed interval $[x_0,y]$ is Borel. So $A \cap [x_0, \infty)$ is Borel. By applying the same argument to the reverse ordering, we also get that $A \cap (-\infty, x_0]$ is Borel. The union of these two sets is $A$, so $A$ is Borel.

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  • $\begingroup$ Bravo! ${{{{}}}}$ $\endgroup$
    – Joel David Hamkins
    Commented Sep 21, 2015 at 19:42
  • $\begingroup$ @NateEldredge: Yes, thank you very much for this. I was actually about to post a comment myself that I have read somewhere that every Borel-totally ordered space $X$ admits no copies of $\omega_1$, since there exists a countable ordinal $\xi$ such that $X$ can be embedded into $2^\xi$. [I don't know whether this embedding is just an order embedding or has additional measurable-structure-respecting properties.] So then, by precisely the argument in your comment and in your answer, we have the desired result. $\endgroup$
    – Julian Newman
    Commented Sep 21, 2015 at 20:06
  • $\begingroup$ @JulianNewman: In fact, maybe we can give a more direct proof by embedding $X$ into $2^\xi$, which I think would let us construct the desired countable cofinal set more explicitly. $\endgroup$
    – Nate Eldredge
    Commented Sep 21, 2015 at 20:14
  • $\begingroup$ @NateEldredge: Yes, I have had precisely this thought - in particular, the "well-ordered cofinal subsets" argument probably requires at least $\omega_1$-dependent choice, whereas perhaps the embedding of $X$ into $2^\xi$ doesn't need anything like this. $\endgroup$
    – Julian Newman
    Commented Sep 22, 2015 at 0:00

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