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If $\mu$ is a positive measure on a measurable space $(X,\mu )$ and $f, f_n \in L^p(\mu )$ for $1<p<\infty$, are such that $f_n \rightarrow f$ pointwise a.e., show that $||f_n-f||_p\rightarrow 0$ if and only if $||f_n||_p\rightarrow ||f||_p$.

I think that for the nontrivial implication, one approach could be to show somehow that $f_n$ converges weakly to $f$ and then using the reflexivity of the space and the boundedness of the sequence to conclude. But is $L^p$ reflexive for an arbitrary measure $\mu$, or it has to be $\sigma$-finite? Also is there a more elementary solution, probably using the Vitaly convergence theorem or things like that?

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  • See this for one approach. – David Mitra Jul 31 '15 at 10:28
  • see Corollary 7.15 https://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes_ch7.pdf for that $L^p$ is reflexive for $1<p<\infty$ for arbitrary measure space, and for $p=1$ it is reflexive if the space is $\sigma$ finite. – Svetoslav Jul 31 '15 at 11:48

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Generalized Lebesgue dominated convergence theorem works pretty well here. If we notice that $$ \underbrace{|f_n-f|^p}_{F_n}\le \underbrace{2^p(|f_n|^p+|f|^p)}_{G_n} $$ so a.e. $|F_n|\le G_n$, $F_n\to 0$, $G_n\to 2^{p+1}|f|^p=G\in L^1$ and $\int G_n\to \int G$. Thus $\|F_n\|_1\to 0$.

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A.Γ.
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  • I didn't know that the proof of this theorem is so simple by using the Generalized LDCT – Svetoslav Jul 31 '15 at 12:02
  • @A.G. Perfect! So we don't even need that p>1. I was misled by this assumption and think we somewhere need reflexivity. Thank you very much! – user163644 Jul 31 '15 at 13:14