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If $E\subset R^n$ be a Lebesgue measurable set and let $(f_k)$ be a sequence of non-negative Lebesgue measurable function on E such that $\lim f_k=f$ a.e. I want to prove that if $\int_E fd\lambda<\infty$,and$$\lim_k \int_E f_kd\lambda=\int_E fd\lambda,$$then$$\lim_k \int_A f_kd\lambda=\int_A fd\lambda$$ for every Lebesgue measurable set $A\subset E$.

Does it hold if $\displaystyle\int_E fd\lambda=\infty$?

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2 Answers 2

Here is a counter example in the case when $\int f = \infty$:

Take $E = \mathbb{R}$ and set $$f_k = 1_{(-\infty, 0)} + 1_{(k,k+1)} \rightarrow 1_{(-\infty, 0)}$$ And take $A = [0,\infty)$.

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For the first question, use Fatou's lemma and the sequence $g_k:=f_k+f-|f-f_k|\geqslant 0$.

For the second part, take $f_n:=\chi_{(n,n+1)}+\chi_{(-1,0)}\frac 1x$ and $f:=\chi_{(-1,0)}\frac 1x$, $A:=[0,+\infty)$.

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I don't understand the counterexample. –  Joe Dec 8 '12 at 18:02
    
It seems doesn't work. Since f=1 $\lim_k \int_A f_kd\lambda=\int_A fd\lambda=1$ –  Joe Dec 8 '12 at 18:08
    
I've edited, the first attempt didn't worked. –  Davide Giraudo Dec 8 '12 at 18:35

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