This is a problem from Cohn's Measure Theory book:
Suppose that $(X, \mathscr A, \mu)$ is a measure space and that $f$ and $f_1, f_2, \ldots $ belong to $\mathscr L ^1 (X, \mathscr A, \mu , \mathbb R)$. Show that if $\{ f_n \}$ converges to $f$ in mean so fast that $\sum_n \lVert f_n -f \rVert _1 < +\infty$ then $\{ f_n \}$ converges to $f$ almost everywhere.
Here's my attempt which might be helpful in proving something later on:
Let us take $\varepsilon >0$ and $m \in \mathbb N$. Then $\mu \left( \{ x \in X : |f_n (x) -f(x) | > \varepsilon \}\right)\le \frac{1}{\varepsilon} \lVert f_n - f \rVert _1$ for any $n \in \mathbb N$.
Thus, for any $N \in \mathbb N$, we have $\sum_{n=N}^{\infty} \mu \left( \{ x \in X : |f_n (x) -f(x) | > \varepsilon \}\right)\le \sum_{n=N}^{\infty} \frac{1}{\varepsilon} \lVert f_n - f \rVert _1$.
Since $\sum_n \lVert f_n -f \rVert _1 < +\infty$, the RHS of the aforementioned inequality can be made arbitrarily small, hence we can find $N_m \in \mathbb N$ such that for $N\ge N_m$, we have $\sum_{n=N}^{\infty} \mu \left( \{ x \in X : |f_n (x) -f(x) | > \varepsilon \}\right)\le \frac{\varepsilon}{2^m}$.
I need to find set $A$ of measure zero such that $\{ x \in X : \{ f_n (x) \} \text{ does not converge to } f(x)\}$ is contained in $A$. It feels like I need to make an $\varepsilon/2^m$ argument but I am unable to see how.
Any hints to complete the problem will be appreciated.