Say that a preordering $\le$ on a set of subsets of some space preserves strict inclusion provided that $A\lt B$ whenever $A\subset B$ (where $A\lt B$ iff $A\le B$ and $B \not\le A$).

Let the space be the unit circle $T$. Then it's easy to see that there is no preorder on all countable subsets of $T$ that preserves strict inclusion and that satisfies the condition that $A \le B$ iff $r[A] \le B$ for every rotation $r$. (Just let $A = \{ e^{2\pi i n u} : n \in \mathbb N \}$ for an irrational number $u$, and let $r$ be a rotation by $-2\pi u$.)

Is there a preordering of all Borel-measurable subsets of $T$ that (a) preserves strict inclusion, (b) satisfies the condition that $A \le B$ iff $r[A] \le r[B]$ for every rotation $r$ and (c) is total, i.e., all Borel subsets are comparable? (Without (c), we could let $\le$ be $\subseteq$.)

(One can think of a total preorder as a very general generalization of measure or probability, so the question is whether there is any measure-like thing that preserves strict inclusion, i.e., is "regular" in Bayesian terminology.)

**Update:** The answer is negative if we require $r[A] \le r[B]$ for all isometries $r$ of the circle (i.e., we include reflections). That's because (b)-for-isometries and (c) in that case imply that $A \le B$ iff $r[A] \le B$ for every isometry $r$, and we already saw that in that case we have a problem.

To see that $A \le B$ iff $r[A] \le B$, note that all rotations are compositions of reflections, and so all we need to show is that if $A \le B$, then $\sigma[A] \le B$ for any reflection $\sigma$. If $\sigma$ is a reflection and $A \le B$, suppose for a contradiction that we don't have $\sigma[A] \le B$. Then by (c) we have $\sigma[A] > B$. By the analogue of (b), we have $A = \sigma^2[A] > \sigma[B]$. Since $B \ge A$, we have $B > \sigma[B]$. Thus, by (c) $\sigma[B] > \sigma^2[B]$ and so $\sigma[B] > B$, and we have a contradiction.

But I still don't know what happens if instead of all isometries we just have all rotations.