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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.

Elvin Feinberg
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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.

Disintegrating By Parts
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    $\int|\lim f_{n_l}|=\int \lim|f_{n_l}|\leq \int\lim g_l=\int g<\infty$. – KC. Jul 11 '17 at 19:04
  • Uniqueness: Denote the $L^1$ limit of ${f_n}$ by $\hat{f}$. By dominated convergence theorem, $$\int | \lim f_{N_m} - \hat{f} | d\mu=\lim \int | f_{N_m} - \hat{f} | d\mu = \lim | f_{N_m} - \hat{f} |=0,$$ where the dominating $L^1$ function is $g+ | \hat{f} |.$ So $\lim f_{N_m} = \hat{f}$ a.e. – Sam Wong Jul 25 '21 at 10:28
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    The sum in $f_{N_m} = f_{N_1} + \sum_{i=1}^{m} f_{N_{i+1}} - f_{N_i}$ should only be taken up to $m-1$. – MathematicsStudent1122 Jan 01 '22 at 02:25