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Let $(X,\mathcal{A},\mu)$ be a measure space and $E_1,E_2,...\in \mathcal{A}$ with $\sum_{i=1}^{\infty}{\mu(E_i)<\infty}$.

How to show that

$\sum_{i=1}^{\infty}{\mu(E_i)}=\mu(\bigcup\limits_{i=1}^{\infty} E_{i}) \Leftrightarrow \mu(E_i \cap E_j) =0$ for all $i \neq j$

For $\Leftarrow$:

Since all $E_i$ are measurable and $\mu(E_i \cap E_j) =0$, then with the definition of the measure space $\sum_{i=1}^{\infty}{\mu(E_i)}=\mu(\bigcup\limits_{i=1}^{\infty} E_{i})$.

I'm not sure if this way works.

For $\Rightarrow$ I thought about constructing a disjoint sequence to prove it, but how can it be done?

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Let $A=\cup_{i\neq j} (E_i\cap E_j)$. If $F_i=E_i\setminus A$ then you can verify that $\{F_i\}$ is disjoint. Also $\cup_i E_i\setminus A =\cup_i F_i$. If $\mu (E_i\cap E_j)=0$ for $i\neq j$ the $\mu(A)=0$ and countable additivity applied to $\{F_i\}$ gives $\sum \mu(E_i) =\mu (\cup E_i)$. Converse: let $i\neq j$. We have $$\mu (\cup E_i) \leq \mu (E_i\cup E_j) +\mu (\cup _{k\neq i,j} E_k)$$ $$ \leq \mu(E_i)+\mu(E_j)+\sum_{k\neq i,j} \mu(E_k)=\sum_k \mu(E_k)=\mu (\cup E_k).$$ This forces equality to hold throughout; in particular we must have $\mu (E_i\cup E_j)=\mu(E_i)+\mu (E_j)$ which is equivalent to $\mu (E_i\cap E_j)=0$.

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