# Why Not Fewer Axioms for Probability Theory?

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?

• Could you elaborate on how you prove inclusion--exclusion without axiom 3? Commented Jan 16, 2021 at 21:55
• 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. Commented Jan 16, 2021 at 21:56
• If you take the conjunction, then you only have one axiom! Commented Jan 16, 2021 at 21:58

## 1 Answer

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 $$P_*(\emptyset)=1$$.

• 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. Commented Jan 18, 2021 at 6:17
• @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). Commented Jan 18, 2021 at 6:18
• 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. Commented Jan 18, 2021 at 6:30