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Let $(X,\mathcal M, \mu)$ be an arbitrary measure space and $1\le p<\infty$. I am curious whether the following statement holds:

Let $\{f_n:X\to\mathbb{R}:n\in\mathbb{N}\}_n$ be a sequence in $L^p=L^p(X,\mathcal M, \mu)$. If $f_n\to f\in L^p$ (with respect to the $p$-norm) and $f_n\le f$ almost everywhere, then $f=\sup_{n\in\mathbb{N}}f_n$ almost everywhere.

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Every $L^p$ convergent sequence converges, up to subsequences, a.e. to the same limit. Thus for that subsequence your statement holds.

Wrong argument: However it is false in general. For example (taken from Terry Tao's blog) consider the "typewriter" sequence. Namely consider $L^1([0,1])$ with the usual Lebesgue measure and define each $f_n$ to be the indicator function of the interval $\left[\frac{n-2^k}{2^k},\frac{n+1-2^k}{2^k}\right]$ if $2^k<n\le 2^{k+1}$. These interval decreases in length as $n$ increases, thus $f_n\rightarrow 0$ in the $L^1$ norm. However they move through the $[0,1]$ interval, and thus for any $x\in[0,1]$ there are infinitely many $n$ such that $f_n(x)=1$.

EDIT: the previous example does not satisfy the hypotesis $f_n(x)\le f(x)$. In fact the statement is right.

Take any $f_n\rightarrow f$ in $L^p$ and you call $A$ the full measure set on which some $(f_{n_k})_k$ converges a.e. to $f$, then for any $x\in A$ you have that $\sup_n f_n(x)\le f(x)$ by hypothesis and that $\sup_n f_n(x)\ge f(x)$ since $f_{n_k}(x)\rightarrow f(x)$.

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Your example doesn't satisfy the condition $f_n\leq f$ almost everywhere. We have $f\geq \sup_kf_{n_k}$, where $f_{n_k}$ is a subsequence which converges almost everywhere. In this case $\sup_kf_{n_k}(x)=f(x)$. – Davide Giraudo Jul 18 '12 at 11:27
You are perfectly right, in fact the statement is correct. I proceed to fix. – Dario P. Jul 18 '12 at 16:02

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