# A subsequence of a sequence in $L^p$

Let $1\leq p \le \infty$. If $\{f_n\}$ is a sequence in $L^p$ that converges to $f$ in $L^p$, how can I show that there exists a subsequence, say, $f_{n_k}$ that converges pointwise a.e. to $f$. (without using convergence in measure)?

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Related – leo May 2 '12 at 3:02

You can assume that $f=0$, and $1\leq p<\infty$. We can construct a subsequence $\{f_{n_k}\}$, denoted $g_k$ such that $\lVert g_k\rVert\leq 2^{-k}$, so for a fixed integer $n$ $$\mu\{x:g_k(x)\geq n^{-1}\}\leq n^p2^{-kp},$$ and since the sets $\{x:g_k(x)\geq n^{-1}\}$ have a finite measure, and the series $\sum_k 2^{-kp}$ is convergent, we get that $\mu(\limsup_k\{x:g_k(x)\geq n^{-1}\}=0)$, so outside of a set $N_n$ of measure $0$, for each $x$, an integer $k(x)$ such that if $k\geq k(x)$, $g_k(x)\leq n^{-1}$. We can conclude since $\bigcup_{n\geq 1}N_n$ still have measure $0$.

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