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Lp space convergence

This is related to a previous question that I posted. Here is the problem:

Suppose that $\{f_n\}$ is a sequence of functions on $[0,1]$ such that $f_n(x) \rightarrow f(x)$ almost everywhere. Suppose also that $\sup{\lVert f_n\rVert_{L^4}} = M < \infty $.

Prove that $\lVert f_n - f\rVert_{L^3} \rightarrow 0$.

I could complete this proof if I knew that $\lVert f_n\rVert_{L^3} \rightarrow \lVert f\rVert_{L^3}$ by a problem that I proved from Royden's book. How can we recover this fact from the problem statement?

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    Apparently. And it wasn't solved there?2012-08-18
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    A sequence bounded in $L_p$, $p>1$, is uniformly integrable. Use this to show that $(f_n^3)$ is uniformly integrable (take $p=4/3$). You can then use [Vitali's Convergence Theorem](http://planetmath.org/VitaliConvergenceTheorem.html) to show your result.2012-08-18
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    @David Vitali would then say that $\int{|f_n^3 -f^3|} \rightarrow 0$, which is not $L^3$ convergence.2012-08-18
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    There are several versions of VCT. The one in my link is the more general theorem, which is applicable here.2012-08-18
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    You can also find this version of Vitali's Theorem in, e.g., Hewitt and Stromberg, *Real and Abstract Analysis*, Exercise 13.38, pg. 203.2012-08-18
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    Thanks David. Just a rant : This seems quite silly. This was on a qualifying exam, and I know for a fact the Vitali convergence theorem isn't even taught in this course. I'm just wondering how students are expected to solve this problem on a timed qualifying exam without knowing this? You would have to somehow arbitrarily create the definition of uniformly integrable, then arbitrarily create in your mind Vitali convergence theorem, then prove Vitali, then use it to prove this problem?2012-08-18
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    Actually, by Holder inequality, we can change the question to $L_4$. And by Egroff's Th, we can prove there is a tiny set $N_\delta$ with measure $\delta$, so that $||f_n-f||_{L^4([0,1]-N_\delta)}\rightarrow 0$. But unfortunately, i do not know how to deal with $N_\delta$ so far.2012-08-18
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    Here, i try to show $||f_n-f||_{L^4[0,1]}\rightarrow0$. By my comment above, we have for any $\epsilon_n\rightarrow0$, there are $N_{\delta_n}$ with measures $\delta_n\rightarrow 0$, and $N_n\in\mathbb{N}$ (increasing to $\infty$) so that $||f_n-f||_{L^4[0,1]-N_{\delta_n}}<\epsilon_n$ for each $n>N_n$. Moreover, we can assume $N_{\delta_{n+1}}\subset N_{\delta_n}$, for we can use Egroff's Th again on the $N_{\delta_n}$ to create $N_{\delta_{n+1}}$. So one have $||f_n-f||_{L^4[0,1]}\rightarrow0$, because $\cap N_{\delta_n}$ is a $0$ measure set and it doesn't affect the integration.2012-08-18
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    Remark. It should be noticed that $\sup ||f_n||_{L^4}<\infty$ is important in the proof sketch. Because, it leads to $||f||_4<\infty$ so that $||f_n-f||_4=\mbox{limit}_{n\rightarrow\infty}||f_n-f||_{L^4([0,1]-N_{\delta_n})}$. That is, it guarantees the uniform continuousness of integration with respect to integral domain. If $\sup ||f_n||_{L^4}<\infty$ fails, then one can easily give a counterexample to show $||f_n-f||_{L^4(N_\delta)}$ can be $\infty$ all the time, which is a lot greater than $||f_n-f||_{L^4([0,1]-N_\delta)}$.2012-08-18

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