Mutually Exclusive biconditional It was taught in one of my previous probability courses that when two events have no elements in common, they are said to be mutually exclusive. Hence, for mutually exclusive events $A$ and $B$, we say that $A\cap B=\emptyset$.
($\clubsuit$) Furthermore, if $A_1, A_2, A_3,...,$ is a finite or infinite sequence of mutually exclusive events of sample space $S$, then $P(A_1 \cup A_2 \cup A_3 \cup...) = P(A_1)+P(A_2)+P(A_3)+...$
My question is this: Can the statement at ($\clubsuit$) be applied in the opposite direction? That is, can it be said that if: $$P(A_1 \cup A_2 \cup A_3 \cup...) = P(A_1)+P(A_2)+P(A_3)+...$$
Then $A_1, A_2, A_3,...,$ is a finite or infinite sequence of mutually exclusive events of $S$?
Thank you.
 A: In probability theory there is a difference between an impossible event and a zero-probability event.
$A\cap B = \emptyset$ defines an impossible event -- it cannot happen, ever, no matter how large a sample we take.
$P(A\cap B) = 0$ is a zero-probability event. It's a weaker condition than for an impossible event. All impossible events are zero-probability events but not vice versa.
Huh? What is this thing "zero probability event?" It sounds like something that would never happen, so what is the difference?
For all practical purposes, you should never expect to observe a zero probability event... but you could!
Let $\Omega$ be our sample space, and $E \subset \Omega: E\neq \emptyset \;\rm{ and}\; P(E)=0$. Let's also define as a random variable $X(\omega):\Omega \to\Omega $ the identity function $X(\omega) \mapsto\omega$.
Since $E$ is a zero-probability (but not impossible) event then we can expect to see $E$ occur at most a finite number of times in an infinite sequence of realizations of $X$:
$$\lim_{n\to\infty}\sum_1^n \mathbb{1}_{E}(X_i)<\infty$$
As I said earlier, for all practical purposes, we can ignore zero-probability events as "effectively impossible" but mathematically, we need to clarify if an event is an empty set or not.
