Well definition of measure on measurable spaces

Question

I don't really understand the consistency of the following definition of measure :

Given a measurable space $(\Omega,\Sigma)$, a measure on $(\Omega,\Sigma)$ is a map $\mu : \Sigma \to [0,\infty]$ which satisfies

1. The empty set has zero measure, i.e. $\mu(\varnothing) = 0$
2. $\mu$ is countably-additive, it means that for any family $\{A_n\}_{n\in\mathbb N}$ of pairwise disjoint measurable sets, the measure of countable union is the sum of measures : $$\mu\left(\bigcup_{n=0}^\infty A_n\right) = \sum_{n=0}^\infty \mu(A_n) $$

But, the second condition assumes that the sum converges unconditionally ? Since if I want to reindex my family, I want the measure be the same...

However, $[0,\infty]$ with usual operations is a commutative monoid, hence we are able to define finite sum and we must show that any family of elements of $[0,\infty]$ is unconditional convergent with respect to the Alexandrov's topology on $[0,\infty]$ :/

I don't see how to proceed, in order to legitimate the above definition. Thanks for answer :)

Answer 1

Since the summands $\mu(A_n)$ are all positive, the summation $\sum_{n=0}^\infty \mu(A_n)$ automatically converges absolutely. Therefore, rearrangements give the same sum, and there are no problems.