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?

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 $$\mu (E_i\cap E_j)=0$$.