I am looking for an example:


Suppose $(\Omega, \mathcal{F}, \mathbb{P})$ is a probability space, where $\Omega\subseteq \mathbb{R}^n$ (for some $n\in \mathbb{N}$) is a totally ordered set with respect to the relation $\le\_R$ (which is not necessarily the lexicographical order). We assume the sigma algebra $\mathcal{F}$ contains all sets of the form $\{z\in \Omega : z <\_R x\}$ and $\{z\in \Omega : z\le\_R x\}$ for $x\in \Omega$. Is there a set $S\in \mathcal{F}$ with positive measure such that $\sup\_{x, y\in S} \mathbb{P}\{z\in S: x\le\_R z\le\_R y\} = 0$.


If $(S, \le\_R)$ has both countable coinitiality and countable cofinality, then this situation will not happen. Is there such an example in the general cases?



"Inner Regularity" of probability measure on totally ordered sets
Consider a subset $\Omega$ of $\mathbb R$ of size $\aleph\_1$ and ordered in type $\omega\_1$. (This uses the axiom of choice.) 


Let $\mathcal F$ be the $\sigma$-algebra generated by the initial segments of $\Omega$ under the well-ordering (so all sets in $\mathcal F$ are countable or co-countable), with the measure that assigns $0$ to the countable sets and $1$ to the cocountable ones. 


For any $x,y\in\Omega$ with $x\le\_R y$, the interval $\{z\in \Omega: x\le\_R z\le\_R y\}$ has measure 0, so you can take as $S$ any cocountable subset of $\Omega$.


