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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?

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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.

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  • $\begingroup$ $f_n,f \in L_m(X)$ is nowhere used, so that is not necessary? $\endgroup$ – Arindam Apr 29 '17 at 0:58
  • $\begingroup$ @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 $f$. $\endgroup$ – Sangchul Lee Apr 29 '17 at 1:02

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