Let $X$ be a measurable space, in which $C$ is a $\sigma$-ring of measurable sets, with $\mu$ as the measure. Let $\{A_n\}_{n=1}^\infty$ be a sequence of sets in $C$. Suppose that $\sum_{n=1}^\infty \mu(A_n) < +\infty$. Show that $\lim_{n\rightarrow \infty} \chi_{A_n} = 0$ almost everywhere on $X$.
Here, $\chi_{A_n}$ is the characteristic function for $A_n$.
Idea: So, it's immediate that you get $\mu(A_n) \rightarrow 0$ as $n \rightarrow \infty$. This implies that $$\lim_{n\rightarrow \infty } \int_X \chi_{A_n} d\mu = 0$$.
From here, I think you can use Fatou's lemma somehow? That gets you this:
$$\int_X \liminf \chi_{A_n} d\mu \leq \liminf \int_X \chi_{A_n} d\mu = 0$$
using the fact that convergence of $\mu(A_n)$ implies we have convergence of the $\liminf$ to the same limit. But now I'm stuck. I was thinking you could somehow do something with the limsup, but I'm not sure. And if you had convergence of $\chi_{A_n}$, then you could just use DCT I think.
Hint: Read the proof of the first Borel-Cantelli Lemma.