Proving $m(S_i\cap S_j)=0\;\forall i\ne j \Rightarrow m(\cup_iS_i)=\sum_im(S_i)$ Let $(E,\Sigma,m)$ be a measure space and let $S_1,S_2,\ldots\in\Sigma$ so that $m(S_i\cap S_j)=0\;\forall i\ne j$. 
How to prove that $m(\cup_iS_i)=\sum_im(S_i)$?
I know that since $m$ is a measure and therefore countably additive, then $m(\cup_iS_i)=\sum_im(S_i)$ would hold for disjoint $S_i$. But how to show the equality if I only know $m(S_i\cap S_j)=0$ and not that $S_i\cap S_j=\emptyset$?

Let $N= \cup_{i,j} (S_i \cap S_j)$ and note that $mN = 0$.
Let $S'_k = S_k \setminus N$. Note that $m(S_k') = m(S_k)$ and $m( \cup_k S_k ) = m (\cup_k S_k')$.
Since the $ S_k'$ are disjoint, we have the desired result.
