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 $\Sigma$?



Is every convex subset of a Borel-linearly ordered space measurable?
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](http://www.ams.org/journals/tran/1988-310-01/S0002-9947-1988-0965754-3/) (Trans. AMS 310 (1988), 293-302, [MR 0965754](http://www.ams.org/mathscinet-getitem?mr=965754)), 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](http://www.ams.org/mathscinet-getitem?mr=673790), [text at ScienceDirect](http://www.sciencedirect.com/science/article/pii/S0049237X09705087), 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](https://mathoverflow.net/questions/82642/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.


