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

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?

$$\sum_{n=1}^{\infty} \left| f_n - f \right|$$
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.
• $f_n,f \in L_m(X)$ is nowhere used, so that is not necessary? – Arindam Apr 29 '17 at 0:58
• @Arindam, Conditions on $(f_n)$ and $f$ themselves are not necessary. Another way of seeing this is to observe that both the assumption on $|f_n - f|$ and the conclusion does not change if we add any measurable function $g$ to both $f_n$ and $f$. – Sangchul Lee Apr 29 '17 at 1:02