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If $f_n \in L_2(\mu)$, $f_n\rightarrow f$ almost everywhere, this is not enough to conclude $f\in L_1(\mu)$.

But is it enough to conclude whether $f\in L_2(\mu)$ or $$\lim_{n \to \infty}\int_{R}{|f_n(x)-f(x)|}^2<\infty$$

What about the assumption change to $\sup\int_{R}{|f_n(x)|}^2d(\mu)<\infty$

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If you change the assumption to $$\sup_n \int_{\mathbb{R}} |f_n(x)|^2 dx < \infty$$ then you can at least conclude that $f\in L^2$ by Fatou's lemma. – Chris Janjigian Oct 27 '12 at 19:04

No, it is not true. $$f_n(x) = \begin{cases} n & x \in [0,1/n)\\ 0 & \text{else}\end{cases}$$

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  • Take $\Bbb R$ with Lebesgue measur, and $f_n=\chi_{(0,n)}$. Then $f_n\in L^2$ and $f_n\to \chi_{(0,+\infty)}$, which is not (square)-integrable.

  • The fact that $\sup_n\int_X f_n^2d\mu$ implies by Fatou's lemma that $f\in L^2$.

  • But even in this case, we don't have necessarily convergence in $L^2$. Taking $f_n:=\sqrt n \chi_{(n,n+n^{—1})}$, we can see that $\int_Xf_n^2=1$ for all $n$ and $f_n\to 0$ almost everywhere.
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Thanks, I am considering when u(x) is finite, is that convergence in $L^2$ right? – user46007 Oct 28 '12 at 1:00
I got that just need make a little change of your counterexample,when u(x) is finite,$f_n:=\sqrt{ n(n+1) }\chi_{(1/(n+1),1/n}$,consider [0,1]. – user46007 Oct 28 '12 at 1:08
Thank you!!!!!!!! – Bombyx mori Dec 25 '12 at 13:28

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