Convergence in $L^p_{loc}$ implies convergence of a subsequence in $L^\infty$

Question

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 $L^\infty$.

Answer 1

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