Given $(X, \mathbb A, m)$ a measurable space and $\{E_i\}_{i \in \mathbb N} \subset \mathbb A$ a sequence such that $\sum_{i \in \mathbb N} m (E_i) < + \infty$.
I want to show that $m$-almost every $x \in X$ is contained in at most finitely many of the $E_i$.
So, I rewrote the statement and know that I have to show that $\{i \in \mathbb N: x \in E_i\}$ is finite for $m$-almost every $x \in X$.
But does this not follow from the simple fact that $\sum_{i \in \mathbb N} m (E_i) < + \infty$ ?
Define $f(x) := \sum_{i=1}^\infty 1_{E_i}(x)$. Now we have by assumptation (and monotone convergence theorem) $$\int f(x) \, d \mu(x) = \sum_{n=1}^\infty \mu(E_i) < \infty.$$ Thus $f(x) < \infty$ for almost all $x \in X$. This only happens if for each $x \in X$ the sum is finite, i.e. $x$ is only contained in finite many of the $E_i$.