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I'm trying to prove the following

Theorem: Let $\{f_n\}_{n\in\mathbb N}\subset L^p(\Omega)$, $f_n \to f$ 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 $L^1$?

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 $L^1$?

I'm trying to prove the following theorem:

Let $\{f_n\}\subset L^p(\Omega)$, $f_n \rightharpoonup f$ in $L^p(\Omega)$ ($\Omega\subset\mathbb{R}^n$ is open and bounded, $1\leq p \leq \infty$) and $f_n \to \hat{f}$ a.e. then we have $f=\hat{f}$ a.e.

I think we should prove that $$\int (f-\hat{f}) =0$$ so I write $$\int (f-\hat{f}) = \int (f-f_n)+\int (f_n-\hat{f})$$ Then from weak convergence first integral tends to $0$ (because $1\in L^\infty\subseteq L^p$)

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

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

Thanks in advance, K.

I'm trying to prove the following

Theorem: Let $\{f_n\}_{n\in\mathbb N}\subset L^p(\Omega)$, $f_n \to f$ 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 $L^1$?