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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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    You should edit your old question rather than asking a new one, using the "edit" button located under the tags.2012-08-20
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    You need some integrability condition here. Take $X=[0,1]$, and $f_n(x) = f(x) = \frac{1}{\sqrt[p]{x}}$. Then all your conditions are satisfied, but $f,f_n$ are not in $L^p(X)$.2012-08-20
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    Ok but what is wrong with my argument then? Denoting the $g$ and the like.2012-08-20
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    The argument is fine, but when you write $\|g\|_p$ there is an implication that $g \in L^P$. What is true is that $\int |f_n|^p \to \int |f|^p$. However to conclude that the norms converge you need to know that $f \in L^P$.2012-08-20
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    I never wrote $||g||_p$??2012-08-20
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    I don't know what to say. You conclude from $\int |f_n|^p \to \int |f|^p$ that $\|f_n\|_p \to \|f\|_p$. However, this conclusion is only valid if $f \in L^p$. Otherwise $\|f\|_p$ has no meaning.2012-08-20
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    Oh ok I get what you are saying. 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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    You are assuming that the individual integrals are finite.2012-08-20
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    Yes.. I think we can assume that.. I know it was not written.. I thought about it in any case..2012-08-20
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    Then it follows from the fact that $|\int g| \leq \int |g|$. In this case $g = |f_n|^p - |f|^p$.2012-08-20
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    I know.. it is what it is written..2012-08-20
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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