If $0\leq f_n$ and $f_n\rightarrow f$ a.e and $\lim\int_Xf_n=\int_X f$, p,rove or disprove that $\lim\int_Ef_n=\int_E f$ for all $E\in\mathcal{M}$.

I think it is true. It is easy to see $\lim\int_Ef_n\geq\int_E f$ using Fatou's Lemma, yet I could not show that $\lim\int_Ef_n\leq\int_E f$.

To disprove, each time I am either frustrated by either a.e convergence or non-negativity.

Thanks in advance,

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    $\begingroup$ Consider $\int_{X\setminus E} f_n$. $\endgroup$ – Daniel Fischer Jan 9 '14 at 19:38
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    $\begingroup$ Is there an assumption $\int_X f < \infty$ in the exercise? If so (and I expect that is so), you should not forget to put it in the question. $\endgroup$ – Daniel Fischer Jan 9 '14 at 20:17
  • $\begingroup$ Actually there is no any assumption $\int_X f<\infty$. If we put this assumption then considering $E$ and $E^c$ and using method of contradiction will give the result. $\endgroup$ – seriously divergent Jan 10 '14 at 2:49

Take $X= [0,\infty]$ then define take $f_n=n^2 \chi_{[0, \frac{1}{n}]}$ then you $\lim _{n\rightarrow \infty} \int _{X} f_n =\infty$ but $f = 0$ almost everywhere so $\int _{X} f =0$ to fix that define $\tilde{f}_n =n^2 \chi_{[0, \frac{1}{n}]} + \chi _{[1,\infty]} $. Now $\tilde {f} = \chi _{[1, \infty]}$ almost everywhere.

The hypothesis now is satisfied but if you restrict the integral on $[0,1]$ you do not have the desired limiting behavior.

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    $\begingroup$ thank you good flow of argument :) $\endgroup$ – seriously divergent Jan 10 '14 at 2:58

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