Is every convex subset of a Borel-linearly ordered space measurable? 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 $\Sigma$?
 A: 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.
