Fix a real number $1\leq p<\infty$.

Is it true that if functions $f\in L^p(\mathbb{R})$ and $f_1,f_2,\ldots\in L^p(\mathbb{R})$ are such that $\|f_n-f\|_p\rightarrow 0$ as $n\rightarrow \infty$, then $f_n(y)\rightarrow f(y)$ as $n\rightarrow\infty$ for almost every $y\in\mathbb{R}$?

It really shouldn't be true, but what would be a counterexample?

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    $\begingroup$ Up to a subsequence it is true. It is usually proved as a by-product of the completeness of $L^p$. $\endgroup$ – Siminore Nov 15 '13 at 17:45
  • $\begingroup$ @Siminore Sure, I think I remember that. But what's a counterexample for my statement? $\endgroup$ – Kunal Nov 15 '13 at 17:46
  • $\begingroup$ See the typewriter sequence here terrytao.wordpress.com/2010/10/02/… $\endgroup$ – Chris Janjigian Nov 15 '13 at 17:56

Let $f_n(x) = \sin (2\pi nx) \cdot 1_{[0,2 \pi]}(x)$. $f_n \in L^p$, and $\int |f_n| \le \frac{2}{\pi n}$, hence $f_n \to 0$.

For $x \in [0, 2 \pi] \setminus \mathbb{Q}$, we have that $\{2\pi ( nx - \lfloor nx \rfloor ) \}_n$ is dense in $[0, 2 \pi]$, and hence $\{ f_n(x) \}_n$ is dense in $[-1,1]$. In particular, $f_n(x)$ does not have a limit, and $m ([0, 2 \pi] \setminus \mathbb{Q}) = 2 \pi >0$.

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