Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.