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 $S_i\cap S_j=\emptyset$?
 A: Let $\{S_n: n\in \Bbb{N}\}$ of  sets with finite measure and with the property in the post.
$m(S_i \cap S_j)=m(S_i)+m(S_j)-2m(S_i \cap S_j)=m(S_i)+m(S_j)$
Now assume that holds for the first $N$ sets,i.e $m (\bigcup_{n=1}^NS_n)=\sum_{n=1}^Nm(S_n)$
Then $$m (\bigcup_{n=1}^{N+1}S_n)=m((\bigcup_{n=1}^{N}S_n \cap S_{N+1})$$ $$=m((\bigcup_{n=1}^{N}S_n)+m(S_{N+1})-2m(\bigcup_{n=1}^{N}(S_n\cap S_{N+1}))$$ $$=\sum_{n=1}^Nm(S_n) +m(S_{N+1})$$ because $m(\bigcup_{n=1}^{N}(S_n\cap S_{N+1})) \leq \sum_{n=1}^{N}m(S_n\cap S_{N+1})=0$
Now take the increasing sequence  $A_N=\bigcup_{n=1}^{N}S_n$
Then $\bigcup_{n=1}^{\infty}S_n=\bigcup_{N=1}^{\infty}A_N$ and $$m(\bigcup_{n=1}^{\infty}S_n)=m(\bigcup_{N=1}^{\infty}A_N)$$ $$=\lim_Nm(A_N)$$ $$=\lim_Nm(\bigcup_{n=1}^{N}S_n)$$ $$=\lim_N\sum_{n=1}^Nm(S_n)=\sum_{n=1}^{\infty}m(S_n)$$

some $S_k$ has infinite measure ,then $$+\infty=m(S_k) \leq m(\bigcup_{n=1}^{\infty}S_n)$$ $$\sum_{n=1}^{\infty}m(S_n)=+\infty$$

So we have the desired conclusion in either case.
A: 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.
