Is the following an example of a non-measurable event? Goodmorning everyone.
I started reading DeGroot and Schervish's 'Probability and Statistics' (4th editon) and wondered,  is there an uncountable union of events, that is not an event?
This question has been answered on this site.
The topic of question is "What is an uncountable union of events?" (sorry ba I can't report the code or the link of the question that has already been answered).
The responses were very illuminating,
however, re-reading the example of an uncountable union of events that is not an event, I'm referring to Arthur's answer, I wonder if the example is adequate. This confuses me.
I report the example referred to in the aforementioned answer and then I ask my question.
Example:
Assuming the Axiom of Choice (which is a very reasonable and common thing to do, but not universal), you can construct unions of events which, if allowed to be events themselves, will have a problematic probability of ocurring. Basically, it can't be 0 and it can't be positive.
To see it in action, let's say your experiment is to pick a point uniformly at random on a circle. Then any point is an event. Using the AoC, we can construct (or more correctly, we can prove the existence of) a set of points A0 on the circle with a special property: Rotating the set along the circle by any rational angle α∈(0∘,360∘) results in a new set of points Aα. None of these Aα have any points in common with any of the others, but together they cover the entire circle.
So, if we were to assign some probability p to picking a point in A0, then by rotational symmetry the same probability should apply to any of the Aα. And since they are all pairwise disjoint, and they together cover the circle, the sum of all those p's should be 1. Thus we have
∑α∈[0,360)∩Qp=1
But if p is 0, the sum is 0, and if p is positive, then the sum is infinite. So it is impossible to assign a probability to this union A0, and therefore we are better off not calling it an event.
Question
Why can't I consider the set of rational points to have zero measure? If I imagine the circle composed of rational and irrational points, why not assign a measure of nothing on the first (rational) and measure equal to 1 on the second (irrational)? so, having admitted the above, am I wrong if I consider the circle as a union of rational and irrational points?
Thanks for any answer or clarification.
Francesco.
 A: Let us divide the answer in two parts:

*

*Consider the space $[0,1]$ with the Lebesgue $\sigma$-algebra and the Lebesgue measure.
We know that there are subsets of $[0,1]$ that are not Lebesgue. One of such subsets is the Vitali set. Let $V$ be the Vitali set. Then $V$ is uncountable and it is not an event. On the other hand, every single point in $V$ is an event (is measurable). So $V$ is an example of an uncountable union of events that is not an event.

(For more details on Vitali set, see, for instance : https://en.wikipedia.org/wiki/Vitali_set)


*Arthur answer's in What is an uncountable union of events? includes the "construction" of a non-measurable subset in the circle, very similar to the Vitali subset of $[0,1]$ .

If you want to go deeper and understand the proof of the existence of Vitali set, I suggest you do it, with the usual Vitali set (subset of $[0,1]$). After that, you can read Arthur's answer and see how the same argument is adapetd to the circle.
Remark 1: Regarding your questions at the end.
"Why can't I consider the set of rational points to have zero measure?"
Answer: the set rational points has zero measure
"If I imagine the circle composed of rational and irrational points, why not assign a measure of nothing on the first (rational) and measure equal to 1 on the second (irrational)?"
Answer: Assuming the circle has measure 1, the set of irrational points has measure 1.
"Am I wrong if I consider the circle as a union of rational and irrational points?"
Answer: No, you are not wrong. But Arthur's answer is not about this.
Remark 2: Explaining the "Vitali set" in the circle.
You don't get just one point to perform rational rotations.
Given any angle $\theta \in [0^\circ, 360^\circ)$, let us write $R_\theta$ to indicate the rotation by $\theta$.
You first consider the equivalence relation between points in the circle:
$x \sim y$ iff there a rational angle $\alpha$ such that the $R_\alpha(x)=y$. It is easy to see that such relation is actually an equivalence relation.
Then (using the Axiom of Choice) we choose one and just one element (one point) from each equivalence class and with them we have a set $A$. Note that $A$ has two interesting properties:

*

*For any two rational angles $\alpha, \beta$, if $\alpha \neq \beta$, then $R_\alpha(A) \cap R_\beta(A) =\emptyset$.

*If $C$ is the whole circle,
$$C = \bigcup_{\alpha \textrm{ rational angle in } [0^\circ, 360^\circ)} R_\alpha(A)$$
In fact, for 1: Let $\beta'$ be the (rational) angle in $[0^\circ, 360^\circ)$ congruent to $-\beta$ and let $\gamma$ be the (rational) angle in $[0^\circ, 360^\circ)$ congruent to $\alpha-\beta$.
If there is $p\in R_\alpha(A) \cap R_\beta(A)$, then
there is $q=R_{\beta'}(p)$ and  $q \in R_\gamma(A) \cap A$. But it means that $q \in A$ and there is $q_1 \in A$ such that $q=R_\gamma(q_1)$. Since $\alpha \neq\beta$, we have $\gamma \neq 0$ and $q \neq q_1$. But $q \sim q_1$. However, it is not possible, because in $A$ there is only one representative of each equivalence class.
For 2: Just note that every point of the circle is in one equivalence class. It means that, for every $p$ in the circle, there is $q\in A$ and a rational angle $\alpha \in [0^\circ, 360^\circ)$, such that $p=R_\alpha(q)$.
Since the set of rational angle is countable, the measure of the whole circle is finite, and for any rational angle $\alpha \in  [0^\circ, 360^\circ)$, $R_\alpha(A)$ would have the same measure of $A$, we conclude that $A$ is not measurable (it is not an event).
Remark 3: @Francesco, to answer your questions in the comments,
let me give more details regarding the "Vitali set" in the circle.
Let us pick just one point $p$ in the circle. By applying rotations by a rational angle, we produce a countable set of points (why countable? because there are only a countable "quantity" of rational angles). This countable set of points is the equivalence class of $p$.  In this equivalence class you must pick exactly one point to represent the equivalence class.
Is it possible that this equivalence class includes all points in the circle?  The answer is NO, because the circle is an uncountable set of points. So, there are others equivalence classes, all of them countable sets.  In fact, it is known that taking the union of any countable family of countable sets produces a countable set as result. Since the circle is uncountable, we can conclude that there is an uncountable "quantity" of such equivalence classes.
A simple example that one such equivalence does not include the whole circle is:  let us start with the point at $0^\circ$. Any rotation by a rational angle wil produce a point located at a rational angle. So, for instance, the point located at angle ${\sqrt{2}\,}^\circ$ can not be produced from our initial point by a rotation by a rational angle.
Another point that you need to keep in mind is that for any set $S$ and any equivalence relation on $S$, the equivalence classes are disjoint  and the union of all equivalence classes is the $S$ (that is, every point of $S$ is one equivalence class).
