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I want so show the following: For $p \in [1, \infty)$ consider $f \in L^p(\mathbb{R})$ and the sequence $(f_n)_n$ such that $f_n \in L^p(\mathbb{R})$ for $n \in \mathbb{N}$. Show that, if $f_n$ converges to $f$ in $L^p$, then $(f_n)_n$ has a subsequence $(f_{n_k})_k$, such that $f_{n_k}(x) \rightarrow f(x)$ for $k \rightarrow \infty$ a.e on $x \in \mathbb{R}$.

My attempt:

Since $f_n \rightarrow f \in L^p$ we get $||f_n-f||^p_p \rightarrow 0$, i.e. $\int_{\mathbb{R}} |f_n(x)-f(x)| \mu(dx) \rightarrow 0$ a.e thus $|f_n(x)-f(x)| \rightarrow 0$, a.e. My problem is, that I am not really sure, if we showed that $\lim_{n \rightarrow \infty} ||f_n-f||_p=0 \Rightarrow \lim_{n \rightarrow \infty }f_n=f \text{ a.e } \text{ on } \mathbb{R}$.
I tried to prove this statement, but wasn't able to. So, my question is how to show that $\lim_{n \rightarrow \infty} ||f_n-f||_p=0 \Rightarrow \lim_{n \rightarrow \infty }f_n=f \text{ a.e } \text{ on } \mathbb{R}$. (If my approach is correct, else how do I show the statement in the beginning.)

Peter
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    See this https://math.stackexchange.com/q/138043 – psl2Z Jan 24 '24 at 22:48
  • The statement you're trying to prove is that convergence in $L^p$ implies a.e. convergence for a subsequence. What you're claiming in your proof of said statement is that convergence in $L^p$ implies a.e. convergence of the whole sequence. This is false (and also kind of a circular argument) – Michh Jan 24 '24 at 22:51

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