I have a very short question: Let $\Omega\subseteq \mathbb R^d$ and $p\in[1,\infty[$.
Assuming $f_n\rightarrow f$ in $L^p(\Omega)$ as $n\to\infty$.
Can I then follow that $f_n \to f$ almost everywhere ?
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No. For simplicity, consider the interval $\Omega=[0,1]$ and construct a sequence of sets $A_n$ such that the measures $\lambda(A_n)$ tend to $0$ but every point belongs to infinitely many $A_n$. For example $A_1=[0,1/2]$, $A_2=[1/2,1]$, $A_3=[0,1/4],\ldots,A_6=[3/4,1]$, $A_7=[0,1/8],\ldots$. If $f_n$ is the indicator function of $A_n$, that is $f_n(x)=1$ if $x\in A_n$ and $f_n(x)=0$ else, then $f_n \to 0$ in all $L^p([0,1])$ because $\|f_n\|_p=\lambda(A_n)^{1/p}\to 0$ but there is no $x\in [0,1]$ with $f_n(x)\to 0$.
Jochen
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