$\int_X |f_n - f| \,dm \leq \frac{1}{n^2}$ for all $n \geq 1$ $\implies$ $f_n \rightarrow f$ a.e.

Question

Let $(X, M, m)$ be an arbitrary measure space. Let $f_n, f \in L^1_m(X)$. Assume that $$\int_X |f_n - f| \, dm \leq \frac{1}{n^2} \text{ for all }n \geq 1. $$ Then I want to show that $f_n \rightarrow f$ a.e. on $X$.

I thought since we have norm convergence, we get a subsequence $f_{n_k}$ of $f_n$ converging to $f$ a.e. Does the condition $\int_X |f_n - f| \,dm \leq \frac{1}{n^2}$ for all $n \geq 1$ imply something stronger than norm convergence?

Answer 1

Since

$$ \sum_{n=1}^{\infty} \left| f_n - f \right| $$

is integrable, this sum is finite a.e. This implies that the series converges a.e., hence we have $|f_n - f| \to 0$ a.e.