In DeGroot and Schervish's 'Probability and Statistics'(4th editon), they define the sample space as the union of all outcomes, and an event as a set of possible outcomes.
Definition 1.3.1 on page 5.
"Experiment and Event.
An experiment is any process, real or hypothetical, in which the possible outcomes can be identified ahead of time. An event is a well-defined set of possible outcomes of the experiment.
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They then list 3 conditions that an event must satisfy, but I'm confused with the 3rd condition.
Condition 3 on page 10:
"If $A_1$, $A_2$, . . . is a countable collection of events, then $\bigcup_{i=1}^{\infty} A_i$ is also an event.
In other words, if we choose to call each set of outcomes in some countable collection an event, we are required to call their union an event also. We do not require that the union of an arbitrary collection of events be an event. To be clear, let I be an arbitrary set that we use to index a general collection of events ${A_i : i \in I }$. The union of the events in this collection is the set of outcomes that are in at least one of the events in the collection. The notation for this union is $\bigcup_{i \in I} A_i$. We do not require that $\bigcup_{i \in I } A_i$ be an event unless I is countable.
Condition 3 refers to a countable collection of events. We can prove that the condition also applies to every finite collection of events.
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What confuses me is that, is there an uncountable union of events, that is not an event?
Best Answer
Yes. 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 $\mathscr A_0$ on the circle with a special property: Rotating the set along the circle by any rational angle $\alpha\in (0^\circ, 360^\circ)$ results in a new set of points $\mathscr A_\alpha$. None of these $\mathscr A_\alpha$ 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 $\mathscr A_0$, then by rotational symmetry the same probability should apply to any of the $\mathscr A_\alpha$. 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 $$ \sum_{\alpha\in [0, 360)\cap \Bbb Q}p = 1 $$ 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 $\mathscr A_0$, and therefore we are better off not calling it an event.