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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?

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  • $\begingroup$ Is $\le$ the usual ordering of the reals, or did you mean $\le_R$? $\endgroup$ – Andrés E. Caicedo Oct 7 at 2:16
  • $\begingroup$ In case you meant $\le_R$: Consider a subset $\Omega$ of $\mathbb R$ of size $\aleph_1$ and ordered in type $\omega_1$. 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. (Cont.) $\endgroup$ – Andrés E. Caicedo Oct 7 at 2:18
  • $\begingroup$ (cont.) 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 $\Omega$. $\endgroup$ – Andrés E. Caicedo Oct 7 at 2:18
  • $\begingroup$ I meant $\le_R$. Thank you for your post, I will think about it. $\endgroup$ – Ray Oct 7 at 2:30
  • $\begingroup$ Hmm, how is it relevant to your question that $\Omega$ is a subset of $\mathbb{R}^n$? None of the properties you discuss seem to be related to any properties of the spaces $\mathbb{R}^n$ (except that it bounds the cardinality of $\Omega$). $\endgroup$ – Jochen Glueck Oct 7 at 6:07
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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 $\Omega$.

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