Formally, why does a logical contradiction have probability zero? 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).
We see that in probability, we represent the event $A$ as a set of elements in our sample space, and $\neg A$ as the complement of $A$ in our sample space. Thus, in probabilistic terms,

$$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$

by the definition of the complement. And by the Kolmogorov Axioms, we see that

$$P(\emptyset) = 0.$$
A logical contradiction can never be true. &nbsp; Something that surely will never happen (an impossibility) has probability measure of zero.

A tautology will always be true. &nbsp; Something that surely will happen (a certainty) has a probability measure of one.

That is all.