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If $f_n \rightarrow f$ in $L^1_{loc}(\Omega)$ then is it true that $f_n \rightarrow f$ a.e. in $\Omega$? Or is it just true for a subsequence? And why?(Here $\Omega$ is an open set in $\mathbb{R}^m$)

I have seen similar result for just $L^1$ without a proof. I have no clue about it when we replace it by $L^1_{loc}.$ Can somebody please help me to understand it?

Extremal
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1 Answers1

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It is just true along a subsequence.

Take a countable compact exhaustion $K_i, i\in\mathbb N$ of $\Omega$ with $K_i\subset K_{i+1}$.

On $K_1$ there is a subsequence of $f_n$ that converges a.e. on $K_1$, see, e.g. $L^1$ convergence gives a pointwise convergent subsequence

Now you can apply the same argument to $K_2$ and the resulting subsequence of $K_1$, so you have a subsequence of a subsequence that converges pointwise a.e. on $K_2$.

You can repeat this argument, and in the end you can choose a diagonal sequence out of these subsequences. Those converge pointwise a.e. on all $K_i$, and hence a.e. on $\Omega$.

supinf
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    You can give a counter-example : the typewriter sequence $$f_n(x) = \sqrt{n} \ 1_{\textstyle x \in \phi([\frac{n}{\pi}, \frac{n}{\pi}+1/n])}$$ where $\phi$ reduces an interval $[a,b]$ modulo $1$. This sequence doesn't converge pointwise for any $x \in [0,1)$ – reuns May 29 '17 at 15:57