# L1 convergence gives pointwise convergent subsequence

I have been reading Terry Tao's notes on Real Analysis and there's a part he just says, but does not really explain, so I am wondering if someone here would. The notes are http://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/ and my particular question is from Section 4, Corollary 3. It goes as follows,

Let $f_n \rightarrow f$ in $L^1$ then there exists a sub sequence $(f_{n_j}) \subset (f_n)$ such that $f_{n_j} \rightarrow f$ pointwise a.e. Moreover $(f_{n_j})$ converges almost uniformly to $f$.

The proof he gives is simply that since $||f_n-f||_1 \rightarrow 0$ as $n \rightarrow \infty$ we can pick a sub sequence such that $||f_{n_j}-f||_1<2^{-j}$ which is enough to show pointwise a.e and almost uniform convergence. But what allows you to pick such a sub sequence is it maybe some Cauchy property or is it some weird construction? Then how do you go from that to pointwise a.e and even almost uniform convergence. I am assuming that for almost uniform, you do something similar to Egorov's theorem without the assumption the domain of $f$ has finite measure. Also I am aware that if you get almost uniform, you immediately have pointwise a.e, but I'd like to see how to get to both. Thank you.

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Choose $N_{k}$ such that $N_{1} \le N_{2} \le N_{3} \le \cdots$ and such that $\|f_{m}-f_{n}\| < 1/2^{k}$ whenever $m, n \ge N_{k}$. This is possible because $\{ f_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence. Then $\{ f_{N_{k}}\}_{k=1}^{\infty}$ is a subsequence such that $\|f_{N_{l}}-f_{N_{m}}\| < 1/2^{k}$ whenever $l,m \ge k$. Then $$f_{N_{m}}=f_{N_{1}}+\sum_{l=1}^{m}f_{N_{l+1}}-f_{N_{l}}$$ converges pointwise a.e. absolutely because $$g_{m}=|f_{N_{1}}| + \sum_{l=1}^{m}|f_{N_{l+1}}-f_{N_{l}}|$$ converges pointwise a.e. to an extended real function $0 \le g \le \infty$ such that, by the monotone convergence theorem, $$\int g\,d\mu = \int |f_{N_{1}}|d\mu+\sum_{l=1}^{\infty}\int |f_{N_{l+1}}-f_{N_{l}}|d\mu = \|f_{N_{1}}\|+\sum_{l=1}^{\infty}\|f_{N_{l+1}}-f_{N_{l}}\| < \infty.$$ So $g < \infty$ a.e., which means that $\lim_{l}f_{N_{l}}$ converges pointwise a.e.. to an $L^{1}$ function.