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I'm trying to prove the following

Theorem: Let $\{f_n\}_{n\in\mathbb N}\subset L^p(\Omega)$, $f_n \rightharpoonup f$ in $L^p(\Omega)$ ($\Omega\subset\mathbb{R}^n$ is open and bounded, $1\leq p \leq \infty$) and $f_n \to \hat{f}$ almost everywhere. Then $f=\hat{f}$ almost everywhere.

Ideas for the proof:

We should prove that $$\int (f-\hat{f}) =0$$ We can write $$\int (f-\hat{f}) = \int (f-f_n)+\int (f_n-\hat{f})$$ Then, by weak convergence, the first integral tends to $0$ (because $1\in L^\infty\subseteq L^p$).

The problem is the limit of the second integral. We can write $$ \int (f_n-\hat{f})=\int_{\Omega \cap E} (f_n-\hat{f}) + \int_{\Omega\cap E^c} (f_n-\hat{f})$$ where $E=\{x: f_n(x) \not \to \hat{f}(x)\}$. Then $|E|=0$ and the first integral vanishes.

My question is: What should I do with the second integral? Is it enough if I use the Lebesgue Dominated Convergence Theorem, since by Hölder inequality I can easily dominate $\{f_n\}$ in $L^1$?

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  • $\begingroup$ I do, but $(f_n)$ does not converge to $f$ in $L^p$. We have only weak convergence. $\endgroup$ Commented May 20, 2014 at 20:38
  • $\begingroup$ I think the proof in second answer is also valid for $p=1,\infty$ since Egorov's Theorem requires that the sequence is measureable and pointwise a.e. convergent, which is satisfied in both cases anyway. Am I wrong? $\endgroup$ Commented May 20, 2014 at 21:09
  • $\begingroup$ The edit completely changed the problem statement. In the original, $(f_n)$ converges weakly to $f$. $\endgroup$ Commented Aug 18, 2021 at 10:52

1 Answer 1

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Let's subtract $f$ from everything, so we have weak convergence to zero. Suppose $\hat f$ is nonzero on a set of positive measure. Choosing small $\epsilon>0$ and possibly flipping the sign of $\hat f$, we get a set $E$ of finite positive measure on which $\hat f\ge \epsilon $. Replacing $E$ with a smaller set of positive measure, according to Egorov's theorem, we get uniform convergence $f_n\rightrightarrows \hat f$ on $E$. Then for all large $n$ $$\int f_n \chi_E \ge \frac{\epsilon}{2} |E|$$ contradicting the fact that $f_n\to 0$ weakly.

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    $\begingroup$ Nice and simple. $\endgroup$ Commented May 21, 2014 at 22:10
  • $\begingroup$ By replacing $E$ with smaller set you mean that integration is over new set $E\cap F$, where $F$ is set for which $f_n$ converges uniformly to $\hat{f}$, is that right? $\endgroup$ Commented May 24, 2014 at 21:14
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    $\begingroup$ @user2871983 You can argue like that. My thought process was to restrict attention to $E$ and apply Egorov's theorem there. Then it gives $F\subset E$ such that $|E\setminus F|$ is small. $\endgroup$ Commented May 24, 2014 at 21:37
  • $\begingroup$ Late question here but prompted by @DavidMitra in a similar question of mine, can the subtraction of $f$ be bypassed, while also working on an answer strictly involving $1< p < \infty$ ? It feels it will be similar enough. $\endgroup$ Commented May 18, 2019 at 12:21

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