Show that there is a subsequence of $(f_n)_n$ that converges to $f$ almost everywhere. 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!
 A: 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 $f$. 
A: 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 $x$. 
A: 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).
