How would I show that $m(\cup_{n \in \mathbb{N}}A_{n}) = \displaystyle\sum_{n \in \mathbb{N}}m(A_{n})$?

Question

Suppose we have that $A_{1}, A_{2}, A_{3}, ...$ is a countable collection of elements of an algebra $\mathcal{A}$. Suppose also that $m$ is a probability measure on $\mathcal{A}$ where $m(A_{i} \cap A_{j}) = 0$ for $i \neq j$. How would I show that $m(\cup_{n \in \mathbb{N}}A_{n}) = \displaystyle\sum_{n \in \mathbb{N}}m(A_{n})$?

Answer 1

A probability measure is defined on a $\sigma$-algebra $\mathcal{A}$, not just an algebra. Also, you don't have to show anything, it's part of the definition of a probability measure $m$ that $m\bigl(\cup_{n=1}^\infty A_n\bigr) = \sum_{n=1}^\infty m(A_n)$ for disjoint $A_1, A_2, \dots$.