I am having trouble with the following problem
Let $(X, \mathcal{F}, \mu)$ a measure space where $\mu (X)<\infty.$ Let $f,f_n:X \to \mathbb{C}$ be measurable. Set $A_n=\{ |f_n-f|\geq a_n\}$ where $a_n>0$ and $a_n \to 0$. Show that if $\sum_n \mu (A_n)<\infty,$ then $f_n\xrightarrow{a.e.} f.$
I've been trying a lot this problem. For example, I tried to show that $\mu (\{f_n \nrightarrow f\})<\varepsilon$ for all $\varepsilon>0$ using facts as $\mu(A_n) \to 0$ (because the series is convergent) and even asumming that $(a_n)$ could be taken strictly decrasing. In my "closer" attempt I showed that every $x \in \{f_n \nrightarrow f\}$ is contained in infinitely many of the sets $A_n$. But at the end, it doesn't worked.
At every attempt I made, I though "I'm very very close to the solution"... but something failed.
Could you please help me solving this problem?
First observe that the set where $f_n$ does not converge to $f$ is measurable and it can be written as $A:=\{f_n\not\to f\}=\bigcap_{k=1}^{\infty}\bigcup_{n\ge k}A_n.$
Now note that showing $f_n\to f$ almost everywhere is equivalent to showing that $A$ has measure $0.$ To do this, we first observe that $$\mu(A)\le \mu(\bigcup_{n\ge k}A_n)\le \sum_{n\ge k}A_n.$$
Since $\sum_{n=1}^{\infty}\mu(A_n)$ is finite, by choosing $k$ large, we can make $\sum_{n=k}^{\infty}\mu(A_n)$ arbitrarily small. It follows that $\mu(A)=0.$