From what I understand, modern probability theory assumes the following axioms:

  1. $0 \le P(E) \le 1$.
  2. $P(S) = 1$.
  3. $P(\bigcup_{i=1}^\infty E_i) = \sum_{i=1}^\infty P(E_i)$ where $E_i \cap E_j=\emptyset$ for $1 \le i < j$.

Clearly, the inclusion–exclusion principle reduces to axiom 3 whenever events are mutually exclusive. Moreover, the inclusion–exclusion principle may be proven by induction, without any help from axiom 3 whatsoever. Why then is axiom 3 even an axiom at all?

  • 11
    $\begingroup$ Could you elaborate on how you prove inclusion--exclusion without axiom 3? $\endgroup$ Commented Jan 16, 2021 at 21:55
  • 5
    $\begingroup$ Specifically, one could define $P : \{x\} \to [0, 1]$ by $P(A) = 1$ for all $A \in 2^{\{x\}}$. Then $P$ satisfies axioms 1 and 2, but certainly not axiom 3; so axiom 3 is not redundant. $\endgroup$
    – LSpice
    Commented Jan 16, 2021 at 21:56
  • 1
    $\begingroup$ If you take the conjunction, then you only have one axiom! $\endgroup$
    – Asaf Karagila
    Commented Jan 16, 2021 at 21:58

1 Answer 1


Contra your claim, axioms $(1)$ and $(2)$ alone are extremely weak. For example, taking $S=[0,1]$ for concreteness, let $P_*(X)=1$ iff $0\in X$ or $1\not\in X$, and $P(X)=0$ otherwise. This $P_*$ is absolutely horrible: beyond merely failing to satisfy $(3)$, it's not even monotonic, since e.g. $P_*([{1\over 2}, 1))=1$ but $P_*([{1\over 2}, 1])=0$. For that matter, it also has $P_*(\emptyset)=1$.

  • $\begingroup$ I think you're missing the point. I'm not saying to forget the statement expressed by axiom 3 and build an entire theory of probability on axioms 1 and 2 alone. In that case you would surely derive nonsense. What I'm wondering is why we're calling "axiom" 3 an axiom when it appears to be a theorem. As I mentioned before, it is a simple case of the inclusion-exclusion principle (when dealing with disjoint sets), and the inclusion-exclusion principle is proven by induction without any assistance from axiom 3 at all. We need not assume anything is true if it can be proven. $\endgroup$ Commented Jan 18, 2021 at 6:17
  • $\begingroup$ @RyRytheFlyGuy "the inclusion-exclusion principle is proven by induction without any assistance from axiom 3 at all" What exactly is inclusion-exclusion proven from? As my answer shows, you cannot hope to prove inclusion-exclusion from (1) and (2) alone. So until you add to (1) and (2) whatever is necessary to get inclusion/exclusion, you can't do away with (3). $\endgroup$ Commented Jan 18, 2021 at 6:18
  • $\begingroup$ ahahaha nevermind. I just realized the proof by induction for the inclusion-exclusion principle actually does utilize axiom 3, so it is rightfully referred to as an axiom. $\endgroup$ Commented Jan 18, 2021 at 6:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .