finite additivity condition The definition of a probability measure $ P $ requires countable aditivity: $ P \left( \bigcup_{n = 1}^\infty A_n\right) = \sum_{n = 1}^\infty P (A_n)$ whenever $ A_1, A_2, \ldots $ is a sequence of disjoint sets in the relevant sigma algebra. I don't understand the proof of the countable finite aditivity property:
$ P \left( \bigcup_{n = 1}^N A_n\right) = \sum_{n = 1}^N P (A_n)$. If someone could explain that to me that would be terrific.
Thanks
 A: The definition of countable additivity says that $P ( \bigcup_{n = 1}^{\infty} A_{n}) = \sum_{n = 1}^{\infty} P ( A_{n} )$ whenever $ A_{1}, A_{2}, \dots $ is a sequence of disjoint sets in the relevant sigma algebra. 
Assuming countable addivity, let's prove finite additivity, which says that $P ( \bigcup_{n = 1}^{N} A_{n}) = \sum_{n = 1}^{N} P ( A_{n} )$ whenever $ A_{1}, A_{2}, \dots A_{N} $ is a sequence of disjoint sets in the relevant sigma algebra. Given $ A_{1}, A_{2}, \dots A_{N} $, we note that $$\bigcup_{n=1}^{N}A_{n} = A_{1} \cup A_{2} \cup \cdots \cup A_{N} = A_{1} \cup A_{2} \cup \cdots \cup A_{N} \cup \emptyset \cup \emptyset \cup \cdots.$$ 
The point of including all those empty sets in the union above is to obtain an infinite sequence of disjoint sets (since the empty set is disjoint from everything) so we can apply countable additivity. Doing so yields
$$
\begin{align*}
P (A_{1} \cup A_{2} \cup \cdots \cup A_{N}) &= P(A_{1} \cup A_{2} \cup \cdots \cup A_{N} \cup \emptyset \cup \emptyset \cup \cdots) \\
&= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}) + P(\emptyset) + P(\emptyset) + \cdots \\
&= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}) + 0 + 0 + \cdots \\
&= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}).
\end{align*}
$$
Since the sequence $A_{1}, A_{2}, \ldots, A_{N}$ was arbitary, we're done.
A: lets say we want to apply finite additivity to only $N$ events (such that $N$ is finite). In this case too we can construct an infinite union using  $A_n=∅$  for $n> N$ (and $n< \infty$) and apply countable union axiom on it. It will prove finite union.
