## Tuesday, July 14, 2020

### Regularity and rotational invariance

Suppose that we have some sort of (not merely real-valued) probability assignment P to the Lebesgue measurable subsets of the unit circle Ω.

Theorem: Suppose that the probability values are rotationally invariant (P(A)=P(ρA) for any rotation ρ) and satisfy the two axioms:

1. If A and B are disjoint, A and C are disjoint, and P(B)=P(C), then P(A ∪ B)=P(A ∪ C)

2. P(Ω − A)=P(Ω − B) if and only if P(A)=P(B).

Then P(A)=P(∅) for every singleton A.

In other words, we cannot have regularity (non-empty sets having different probability from empty sets) if we have the additivity-type condition (1), the complement condition (2) and rotational invariance.

Proof: Fix an irrational number r and let B be the set of points at angles in degrees r, 2r, 3r, .... Let x0 be the point at angle 0. Then B and C = B ∪ {x0} are rotationally equivalent (you get the former from the latter by rotating by r degrees). So, P(B)=P(C). Let A = Ω − C. Then A and B are disjoint as are A and C. Hence, P(A ∪ B)=P(A ∪ C). But A ∪ C = Ω. So, P(A ∪ B)=P(Ω) by axiom 1. But A ∪ B = Ω − {x0}. So, P({x0}) = P(∅) by axiom 2. But all singletons are rotationally equivalent so they all have the same measure.

This result is a variant of the results here.