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Let $(X,\mathcal{B}, \mu)$ be a measure space and assume the sequence $(f_n)_n$ converges to $f$ in $L^p(\mu)$, where $1\leq p<\infty$. Show that there is a subsequence of $(f_n)_n$ that converges to $f$ almost everywhere.

Isn't it true that for all subsequence of $(f_n)_n$?

Attempt: Since $f_n\to f$ in $L^p$, for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\leq N$, $\|f_m-f_n\|_p<\epsilon /2$ or $\|f_n-f\|_p<\epsilon /2$

Let $(f_{n_k})_k$ be any subsequence of $(f_n)_n$. Then $$\|f_{n_k}-f\|_p\leq \|f_{n_k}-f_n\|_p+\|f_n-f\|_p< \epsilon /2+\epsilon /2=\epsilon.$$

I don't know what the wrong is here. Can anyone check my proof? Thanks!

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    $\begingroup$ The statement $||f_{n_k}(x) - f(x)||_p$ doesn't make any sense. $\endgroup$ – guest Jul 16 '16 at 13:46
  • $\begingroup$ Ok, I remove my last statement. But uniform convergence implies the convergence almost everywhere, by definition. Right? $\endgroup$ – Ergin Suer Jul 16 '16 at 13:54
  • $\begingroup$ Relevant: math.stackexchange.com/questions/611355/… $\endgroup$ – Clement C. Jul 16 '16 at 13:57
  • $\begingroup$ And note that what you actually proved in your question is that if $(f_n)_n$ converges to $f$ in $L_p$, then all subsequences of $(f_n)_n$ converge to $f$... in $L_p$. Which is (a) true, (b), sort of trivial (and not specific to $L_p$ convergence), and (c) not what you want to show. $\endgroup$ – Clement C. Jul 16 '16 at 13:58
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    $\begingroup$ The convergence here is not uniform. It's in $L^p$. The assumption is that $\int |f(x) - f_n(x)|^p dx \to 0$ as $n \to \infty$. What you need to show is that there exists a subsequence $f_{n_k}$ and a set $N$ of measure $0$ so that if $x \notin N$ then $|f_{n_k}(x) - f(x)| \to 0$. $\endgroup$ – guest Jul 16 '16 at 13:58
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What you proved is that if $f_n$ converges to $f$ in $L^p$, then also every subsequence $f_{n_k}$ converges to $f$ in $L^p$.

You need to prove almost everywhere convergence, that is $|f_n(x) - f(x)| \to 0$ for almost every $x$.

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Choose a subsequence $(\tilde f_k)_k = (f_{n_k})_{k}$ such that $\|\tilde f_{k+1} - \tilde f_k\|_p \le 2^{-k}$. Then $$g:= \sum_{k=1}^{\infty} |\tilde f_{k+1} -\tilde f_k| \in L^p(\mu)$$ since $$\|g\|_p \le \sum_{k=1}^{\infty} 2^{-k} \le 1.$$ Therefore the series $\sum_{k=1}^{n}|\tilde f_{k+1} - \tilde f_k|$ converges almost everywhere, since $|g|<\infty$ almost everywhere. Therefore also $$ \tilde f_{n+1} - \tilde f_1 =\sum_{k=1}^n \tilde f_{k+1} - \tilde f_k$$ converges almost everywhere, i.e. the subsequence converges almost everywhere. Since $$ |\tilde f_{n+1}| \le |\tilde f_1| + |g| \in L^p(\mu), $$ from the Lebesgue convergence theorem, the subsequence also converges in $L^p(\mu)$ towards its pointwise limit. As the subsequence also converges to $f$ in $L^p(\mu)$, the pointwise limit is $f$.

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  • $\begingroup$ Why does $\sum \limits_{k=1}^{n} |f_{k + 1} - f_{k}|$ converge a.e. if $|g| < \infty$? $\endgroup$ – layman Jul 16 '16 at 14:25
  • $\begingroup$ You're right. I just adjusted the definition of $g$ to account for that. Now everything is fine. $\endgroup$ – be5tan Jul 16 '16 at 14:35
  • $\begingroup$ The argument is that if it wouldn't converge, $g=\infty$ on a $\mu$-non-zero set which contradicts that $g \in L^p(\mu)$. $\endgroup$ – be5tan Jul 16 '16 at 14:38
  • $\begingroup$ This is Ok, but your choice of the name of $f_k$ for the subsequence of $f_n$ is not very adequate... $\endgroup$ – Kelenner Jul 16 '16 at 14:44
  • $\begingroup$ What do you precisely mean with "not very adequate"? $\endgroup$ – be5tan Jul 16 '16 at 14:50
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The standard counterexample to your claim that the pointwise convergence holds for every subsequence is the following. Set

$$ A_{n,m}:=[(n-1)/m, n/m] $$

Then

$$1_{A_{1,1}}, 1_{A_{1,2}}, 1_{A_{2,2}}, 1_{A_{1,3}}, \dots$$

converges in $L_p[0,1]$ to the zero function. But it does not converge pointwise to the zero function (in fact it diverges at every point).

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