Subsequence convergence in $L^p$ I recall a fact that for functions $f_1,f_2,\ldots\in L^1$ such that $\|f_n-f\|_1\rightarrow 0$ as $n\rightarrow\infty$, there exists a subsequence $f_{n_i}$ that converges to $f$ almost everywhere.
Is this still true if we replace $L^1$ by $L^p$, and $\|f_n-f\|_1$ by $\|f_n-f\|_p$?
 A: This is similar to what julien linked to, but I think it is worth sharing:
CLAIM Suppose that $f_n\in {\mathscr L}^p(X)$ is such that $\displaystyle \sum_{n\geqslant 1}\lVert f_n\rVert<+\infty$. Then $\displaystyle\sum_{n\geqslant 1}f_n $ converges a.e. in $X$, is in ${\mathscr L}^p(X)$ and $$\left\lVert\displaystyle\sum_{n\geqslant 1}f_n\right\rVert \leqslant \sum_{n\geqslant 1}\lVert f_n\rVert$$
P Let $M=\sum_{n\geqslant 1}\lVert f_n\rVert <+\infty$. Note $\left\lVert\displaystyle \sum_{k\leqslant n}|f_k|\right\rVert\leqslant \displaystyle \sum_{k\leqslant n}\lVert f_k \rVert $. 
Define $h_n=\displaystyle\left(\sum\limits_{k\leqslant n}|f_k|\right)^p$. Then $h_n\in {\mathscr L}^1(X)$, $h_n$ increases and $\lVert h_n\rVert_1\leqslant M^p$. By Levi's theorem, there is $h_0\in \mathscr L^1$ such that $h_n\to h_0$ a.e. $X$. It follows $h=\displaystyle\sum_{k\geqslant 1}f_k$ converges (absolutely) a.e. $X$. Let $g_n=\left|\displaystyle\sum_{k\leqslant n}f_k\right|^p$. Then each $g_n\in \mathscr L^1(X)$ and $g_n\to |h|^p$ a.e. on  $X$, and $g_n\leqslant h_n\leqslant h_0$ a.e. on $X$. By Lebesgue's dominated convergence theorem, $|h|^p\in\mathscr L(X)$, and $$\int_X |h|^p=\lim_{n\to\infty}\int_X \left|\displaystyle\sum_{k\leqslant n}f_k\right|^p=\lim_{n\to\infty}\left\lVert \displaystyle\sum_{k\leqslant n}f_k\right\rVert^p\leqslant\lim_{n\to\infty}\left(\sum_{k\leqslant n}\lVert f_k \rVert\right)^p= \left(\sum_{n\geqslant 1}\lVert f_n\rVert\right)^p$$
COR1 If $f_n\in {\mathscr L}^p(X)$ is such that $\displaystyle \sum_{n\geqslant 1}\lVert f_n\rVert<+\infty$, then $\displaystyle\sum_{n\geqslant 1}f_n $ converges in the ${\mathscr L}^p$ norm to some $f\in\mathscr L^p(X)$.
COR2 The space $\mathscr L^p(X)$ is complete, for 
PROP A normed space is complete iff $\displaystyle \sum_{k\geqslant 1}x_k$ converges whenever $\displaystyle \sum_{k\geqslant 1}\lVert x_k\rVert$ converges. 
