I have a measure space $(X, \xi, \mu)$ and $(A_n)_{n \in \mathbb{N}} \in \xi^\mathbb{N}$ with $\mu(A_l \,\cap\,A_k) = 0$ for $l \neq k$. I do want to show:
$$\mu(\bigcup_{n \in \mathbb{N}} A_n) = \sum_{n \in \mathbb{N}}\mu(A_n)$$
but seem to get stuck. I start by defining $B_n := A_n \setminus \bigcup_{m < n}A_m$ and get:
$$\mu(\bigcup_{n \in \mathbb{N}} A_n) = \mu(\dot{\bigcup}\,B_n) = \sum_{n \in \mathbb{N}}\mu(B_n) = \sum_{n \in \mathbb{N}}\mu(A_n \setminus \bigcup_{m < n}A_m)$$
but fail to include the extra information $\mu(A_l \,\cap\,A_k) = 0$ for $l \neq k$ at this point. How do I proceed from here?