TITLE: Formally, why does a logical contradiction have probability zero?
QUESTION [14 upvotes]: In terms of formal probability theory, why does an event representing a logical contradiction (such as $A \wedge \neg A$) always have probability zero?
I understand intuitively why this is the case, but I don't see anything in the axioms of probability that directly explains why $A \wedge \neg A$ must have probability zero (or, equivalently, why a logical tautology such as $A \vee \neg A$ must have probability 1).

REPLY [5 votes]: Your axiom list states that 


Additivity: $P(E_1 \cup E_2)=P(E_1)+P(E_2)$, where  $E_1$ and $E_2$ are mutually exclusive.


Note that $A \wedge \neg A$ is mutually exlusive with itself.
Hence $$P(A \wedge\neg A)=P\left((A \wedge \neg A)\cup(A \wedge \neg A)\right) = P(A \wedge\neg A) + P(A \wedge\neg A) $$