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 $L^1$?
Best Answer
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.