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I am trying to show given $f_{n}\to f$ pointwise a.e. and $\int{f}<\infty$, it follows the sequence {$\int{f_{n}}$} has a limit. But I am not sure if extra condition is required. Can anyone give me a counter-example, or a simple proof?

Edit: a counter-example is already found. What if there is an extra condition $|f_{n}|\le g_{n}$ and $\int{g_{n}}\to \int{g} \le \infty$ ?

Edit2: It is already shown that this can be proven by Fatou's lemma. Thanks everyone who helped me.

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What kind of convergence are you assuming if you write $f_n\rightarrow f$? – user20266 May 27 '12 at 8:34
@Thomas Pointwise. I forgot to mention this. – Polymorpher May 27 '12 at 8:35
here I give you a example X=(0,1),$f_{n}=n^{2}\chi_{(0,1/n)}$, the claim in general is wrong – yaoxiao May 27 '12 at 8:38
@Polymorpher this will be true, just consider Fatou theorem, you will get what you want – yaoxiao May 27 '12 at 8:45
My point is, you need the extra condition. Since someone already gave a counter-example, I gave you one possible extra condition (and also I missed you edit by 6 minutes). – Najib Idrissi May 27 '12 at 8:47
up vote 2 down vote accepted

After precision have been brought here is a proof. We have $g_n-f_n\geq 0$ hence by Fatou lemma $$\int \liminf_n (g_n-f_n)\leq \liminf_n\int (g_n-f_n) $$ so $\int g-\int f\leq \int g+\liminf_n \int (-f_n)$ and $\limsup_n \int f_n\leq \int f$.

Since $g_n+f_n\geq 0$, we apply the previous job to $-f_n$ to get $-\liminf_n \int f_n\leq -\int f$ hence $\lim_{n\to +\infty}\int f_n=\int f$.

A shorter way suggested by @Sam L. is to apply Fatou lemma to $g+g_n-|f-f_n|$.

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It does not meet the condition $\int{g_{n}}\to\int{g}$ – Polymorpher May 27 '12 at 9:45
What is $g$ in this context? – Davide Giraudo May 27 '12 at 9:45
It is the limit of $g_{n}$. Sorry, I forgot to mention that... – Polymorpher May 27 '12 at 9:47
Then show $\limsup$ and $\liminf$ equals, from inequalities in both directions. – Polymorpher May 27 '12 at 10:03
I think it would be easier to consider $g+g_n - |f - f_n|\ge 0$ and apply Fatou to this. Then you'd get convergence $f_n\to f$ in $L^1$ in one go. – Sam May 27 '12 at 12:06

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