Convergence of integrals implies a.e. convergence for subsequence

Question

I have a sequence of (Lebesgue) integrable functions $f_n: \mathbb{R} \to \mathbb{R} $ s.t. $\int |f_n| \to0$ and I have to prove the existence of a subsequence $f_{{n}_{k}}$ that converges almost everywhere to some function $f$. First I tried using Fatou's Lemma, but it doesn't seem reasonable to assume the existence of a positive subsequence (or is it?), and then I tried proving that a subsequence converges in measure, but then again this doesn't imply almost everywhere convergence.

Answer 1

First prove the following.

If $\sum_{n=1}^\infty \int|g_n|$ converges, then $g_n\to 0$ a.e.

Details in spoiler:

Let $M_k:=\bigcap_{N\in\mathbb{N}}\bigcup_{n\ge N}[|g_n|\ge 1/k]$ and $M:=\bigcup_{k\in\mathbb{N}}M_k$. For each $k\in\mathbb{N}$, use the fact that $M_k\subseteq \bigcup_{n\ge N}[|g_n|\ge 1/k]$ for each $N\in\mathbb{N}$ and $$ \mu\left(\bigcup_{n\ge N}[|g_n|\ge 1/k]\right) \le \sum_{n=N}^\infty \mu[|g_n|\ge 1/k]\le \sum_{n=N}^\infty k\int|g_n| \longrightarrow 0\quad\text{as $N\to\infty$} $$ to obtain $\mu(M_k)=0$. Hence $\mu(M)=0$. If $x\in M^c$, then $x\in \bigcap_{k\in\mathbb{N}}M_k^c$ and hence for every $k\in\mathbb{N}$ there exists $N\in\mathbb{N}$ such that $n\ge N$ implies $|g_n(x)|<1/k$. Therefore $g_n\to 0$ a.e.

Then use the fact that $\int |f_n|\to 0$ to find a subsequence $(f_{n_k})$ such that $\int|f_{n_k}| < k^{-2}$.

Alternatively, if you already know that convergence in measure implies convergence almost everywhere for a subsequence (whose proof is similar to the argument above), then you can apply the fact that $\int|f_n|\to 0$ implies $f_n\to 0$ in measure.