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Let $\Omega \subset \mathbb{R}^n$ be bounded or unbounded. Suppose we have a sequence $\{f_n\} \in L^p_{loc}(\Omega)$ such that $f_n \rightarrow f$ in $L^p_{loc}(\Omega)$ for $f \in L^p_{loc}(\Omega)$.

Now suppose further that $f \in L^\infty(\Omega)$. Since $L^\infty(\Omega) \subset L^p_{loc}(\Omega)$, I am wondering if it is possible to always extract a subsequence of $\{f_n\}$, call it $\{f_{n_k}\}$, such that $f_{n_k} \rightarrow f$ in $L^\infty(\Omega)$. I think this is possible but I cannot convince myself that we can always choose a convergence subsequence such that all its elements belong to $L^\infty$.

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  • $\begingroup$ @AnneBauval Sorry I should have been more specific. I meant convergence in $L^\infty$ implies pointwise convergence a.e. $\endgroup$
    – CBBAM
    Feb 20 at 2:20
  • $\begingroup$ @AnneBauval I have corrected my post. $\endgroup$
    – CBBAM
    Feb 20 at 23:35

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Not possible. Take $f_n = \chi_{ [1/n,\ 1+1/n] }$.

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  • $\begingroup$ Doesn't each $f_n$ belong to $L^\infty$? $\endgroup$
    – CBBAM
    Feb 19 at 19:59
  • $\begingroup$ @AnneBauval If we take the sequence itself as a subsequence don't we have $f_n \rightarrow f$ in $L^\infty$? $\endgroup$
    – CBBAM
    Feb 20 at 2:19
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    $\begingroup$ No: $\|f_n-f\|_\infty=1$. $\endgroup$ Feb 20 at 6:40

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