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Here's what I really wanted to ask

Suppose that $f_n \rightarrow f$ a.e.

If we have $\int{\left|\lvert f_n\rvert^p - \lvert f\rvert^p\right|} \rightarrow 0$, is it true that $\|f_n\|_p \rightarrow \|f\|_p$?

What I want to do is let $g_n = \lvert f_n\rvert^p, g=\lvert f\rvert^p$ and then use the fact that $\int{\lvert g_n - g\rvert} \rightarrow 0$ if and only if $\int{\lvert g_n\rvert} \rightarrow \int{\lvert g\rvert}$. Then we get $\|f_n\|_p^p \rightarrow \|f\|_p^p$ and thus $\|f_n\|_p \rightarrow \|f\|_p$. Is this valid?

I'm kind of a beginner at this stuff.

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    Oh ok I get what you are $s$aying. Thank you.2012-08-20

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Still it is unnecessary to require $f_n\to f$ a.e.

To see this fix $\varepsilon>0$. Then exists $N_0(\varepsilon)\in\mathbb N$ such that for any $n>N_0$ $\int ||f_n|^p-|f|^p| < \varepsilon$ by your assumption.

Then it is just a matter of triangular inequality.

Indeed, for $n>N_0$, it follows

$\int |f|^p-\int |f_n|^p\leq \int ||f_n|^p-|f|^p|<\varepsilon. $

Moreover, reversing the arguments we get $\int|f_n|^p\leq \int|f|^p+\int||f_n|^p-|f|^p|$ which implies

$\int |f_n|^p-\int |f|^p\leq \int ||f_n|^p-|f|^p|<\varepsilon.$

We conclude by the arbitrariness of $\varepsilon>0$ and the coninuity of $x\mapsto \sqrt[p] x.$

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    @uforoboa For the "reversed" argument : You have added and subtracted $|f|^p$. But now I don't think it is true that |f_n + f - f|^p < |f|^p + ||f_n|^p - |f|^p| I think you need a $2^p$ factor.2012-08-20