Is the Axiom of Finite Additivity even an axiom?
$\newcommand{\P}[1]{\mathbb{P}\left(#1\right)}$ In Casella and Berger's Statistical Inference (2nd ed., p. 9), there is the Axiom of Finite Additivity. That is, if $\mathcal{B}$ is a $\sigma$-algebra of subsets of a sample space $S$ and $A, B\in \mathcal{B}$ are two disjoint events, then $$\P{A \cup B} = \P{A} +\P{B}\text{.}$$
Is this even an axiom, assuming we have the definition of a probability measure? Casella and Berger define a probability "function" (I prefer measure) as follows:
Definition 1.2.4 Given a sample space $S$ and an associated sigma algebra $\mathcal{B}$, a probability function is a function $P$ with domain $\mathcal{B}$ that satisfies:
- $P(A) \geq 0$
- $P(S) = 1$
- If $A_1, A_2, \dots \in \mathcal{B}$ are pairwise disjoint, then $P(\cup_{i=1}^{\infty}A_i) = \sum_{i=1}^{\infty}P(A_i)$.
It is clear that that $\P{\varnothing} = 0$ from this definition. In $\mathcal{B}$, we know $\varnothing$ exists, and $\P{S} = \P{S \cup \varnothing} = \P{S}+\P{\varnothing} = 1$, so that $1 + \P{\varnothing} = 1$, or $\P{\varnothing} = 0$.
An event is defined as a subset of $S$. Take two events $A$, $B \in \mathcal{B}$. We can write $A$ and $B$ as a disjoint union of two sets: $$\begin{align} A &= (A \cap B) \cup (A \cap B^{c}) \\ B &= (A \cap B) \cup (B \cap A^{c})\text{.}\end{align}$$ and $$A \cup B = (A \cap B) \cup (A \cap B^{c}) \cup (B \cap A^{c})\text{.}$$It follows that $$\begin{align} \P{A} &= \P{A \cap B} + \P{A \cap B^{c}} \\ \P{B} &= \P{A \cap B} + \P{B \cap A^{c}} \\ \P{A \cup B} &= \P{A \cap B} + \P{A \cap B^{c}} + \P{B \cap A^{c}}\text{,} \end{align}$$ so that $$\P{A \cup B} = \P{A}+\P{B}-\P{A \cap B}\text{.}$$
Suppose now that $A$ and $B$ are disjoint. Then $A \cap B = \varnothing$ and as proven before, $\P{\varnothing} = 0$, implying the Axiom of Finite Additivity.
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$\begingroup$You repeatedly used the Axiom of Finite Additivity in your proofs (starting with the proof of $P(\emptyset)=0$). To correctly show that finite additivity follows from countable additivity, just let $A_1=A$, $A_2=B$ and $A_i=\emptyset$ for $i>2$. Then the equality $$P(A\cup B)=P(\bigcup_i A_i)=P(A)+P(B)+``\infty\cdot P(\emptyset)\!"$$ can only hold if $P(\emptyset)=0$. However, nobody said that the given collection of axioms is minimal. For example the group axioms are also often listed in a redundant (=non-minimal) way.
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