# Does convergence in $L^{p}$ implies convergence almost everywhere?

If I know $\|f_{n}(x) - f(x)\|_{L^{p}(\mathbb{R})} \rightarrow 0$ as $n \rightarrow \infty$, do I know $\lim_{n \rightarrow \infty}f_{n}(x) = f(x)$ for almost every $x$?

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For sure along a subsequence. This is the first step when you prove that $L^p$ is a complete normed space. See johndcook.com/modes_of_convergence.html –  Siminore Apr 28 '12 at 14:10
Choosing suitable intervals with length going to zero you can exhibit a sequence of L^p, but pointwise the sequence doesn't converges at any point! –  checkmath Apr 28 '12 at 14:25
Look at $\chi_{[0,1]}$, $\chi_{[0,1/2]}$, $\chi_{[1/2,1]}$, $\chi_{[0,1/4]}$, $\chi_{[1/4,1/2]}$, $\ldots\,$. –  David Mitra Apr 28 '12 at 14:36
If we add a condition: $(f_n)_{n=1}^\infty$ is a sequence of nonnegative functions and $f_{n+1}(x)\geq f(x)$, then will $L^p$-norm convergence imply pointwise convergence? –  RSQ May 26 '14 at 13:36

Let $\rho(x)=\chi_{[0,1]}$ be the characteristic function of the interval $[0,1]$. Then take the "dancing" sequence

$$f_n(x) = \rho(2^mx-k)$$

where $n=2^m+k$ with $0\leq k<2^m$. This sequence converges to $0$ in $L^p$ but for any $x\in(0,1)$ we have $f_n(x)$ is not convergent.

However, it is a general fact that one can always extract a subsequence converging almost everywhere to $f$.

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For $n$ integer and $0\leq k\leq 2^n-1$, denote $f_{n,k}:=n\chi_{[k2^{-n},(k+1)2^{-n}]}$. Consider $f_{1,0},f_{1,1},f_{2,0},f_{2,1},f_{2,2},f_{2,3},\ldots$ (put $g_n:=f_{\alpha_n,n-\alpha_n}$ where $\alpha$ is an integer such that $1+\ldots+2^{\alpha_n}\leq n<1+\ldots+2^{\alpha_n+1}$). Then $g_n$ doesn't converge to $0$ for any $x$, but converge to $0$ in $L^p$ for $1\leq p<\infty$. However, we can find an almost everywhere converging subsequence (here we can pick $g_{1+\ldots+2^n}$).

More generally, if $f_n\to 0$ in $L^p$, with $1\leq p<\infty$, we get can find $n_k$ increasing to $+\infty$ such that $\lVert f_{n_k}\rVert_{L^p}\leq 2^{-k}$. Let $g_k:=|f_{n_k}|$. Then $\mu\{x:g_k(x)\geq n^{-1}\}\leq \frac {2^{-kp}}{n^{-p}}$, so by Borel-Cantelli $\mu\{\limsup_k g_k\geq \frac 1n\}=0$ for each $n$ and $g_k\to 0$ almost everywhere.

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