This work when the involved measure space $(X,\mathcal A,\mu)$ is $\sigma$-finite. Let $\left(A_n\right)_{n\geqslant 1}$ be a non-decreasing sequence of measurable subsets of finite measure such that $\bigcup_{n\geqslant 1}A_n=X$. Notice that for all fixed $n$, the following convergence holds: $ \lim_{R\to +\infty}\mu\left(A_n\cap\left\{x\in X\mid \left\lvert f_n(x)\right\rvert\gt 2^{-n} R\right\}\right)=0 $ hence using the definition of the limit with $\varepsilon=2^{-n}$, we see that we can choose $R_n\gt 0$ such that $ \mu\left(A_n\cap \left\{x\in X\mid \left\lvert f_n(x)\right\rvert\gt 2^{-n} R_n\right\}\right)\lt 2^{-n}. $ Choose $c_n=1/R_n$. Then for all $n$, the following inequality holds $ \mu\left(A_n\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right)\lt 2^{-n}. $ Fix a $N\geqslant 1$. Since $A_N\subset A_n$ for $n\geqslant N$, it follows that for such $n$'s, $ \mu\left(A_N\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right)\lt 2^{-n}. $ hence exploiting the convergence of the series $\sum_n \mu\left(A_N\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right) $, we derive that there exists $E_N\subset A_N$ of measure $0$ such that for all $x\in A_N\setminus E_N$, there exists $m(x)$ such that for all $n\geqslant m(x)$, $c_n\left\lvert f_n(x)\right\rvert\leqslant 2^{-n}$. This proves that for all $x\in A_N\setminus E_N$, the series $\sum_n c_n\left\lvert f_n(x)\right\rvert$ converges. Finally, let $E:=\bigcup_{N\geqslant 1}E_N$. For all $x\in X\setminus E$, there exists a $N$ such that $x\in A_N\setminus E_N$ and $E$ has measure zero.