In order to demonstrate that the theory generated by the Kolmogorov axioms corresponds with
classical probability, some elementary consequences are typically derived. • Since P is finitely additive, we have P(A) + P(A^c) = P(A\cup A^c)= P(\Omega) = 1, so P(A^c) = 1-P(A). • In particular, it follows that P(\emptyset) = 0. The empty set is interpreted as the event that "no outcome occurs", which is impossible. • Similarly, if A \subseteq B, then P(B) = P(A \cup (B\setminus A)) = P(A) + P(B\setminus A) \ge P(A). In other words, P is
monotone. • Since \emptyset \subseteq E \subseteq \Omega for any event E, it follows that 0 \le P(E) \le 1. By dividing A \cup B into the disjoint sets A \setminus (A \cap B) , B \setminus (A \cap B) and A \cap B, one arrives at a probabilistic version of the inclusion-exclusion principleP(A \cup B) = P(A) + P(B) - P(A \cap B).In the case where \Omega is finite, the two identities are equivalent. In order to actually do calculations when \Omega is an infinite set, it is sometimes useful to generalize from a finite sample space. For example, if \Omega consists of all infinite sequences of tosses of a fair coin, it is not obvious how to compute the probability of any particular set of sequences (i.e. an event). If the event is "every flip is heads", then it is intuitive that the probability can be computed as:P(\text{infinite sequence of heads}) = \lim_{n \to \infty} P(\text{sequence of n heads}) = \lim_{n \to \infty} 2^{-n} = 0.In order to make this rigorous, one has to prove that P is
continuous, in the following sense. If A_j,\,\, j = 1, 2, \ldots is a sequence of events increasing (or decreasing) to another event A, then\lim_{n \to \infty} P(A_n) = P(A). == Simple example: Coin toss ==