Extra condition (or proof) for $\int{f_{n}}$ converging to a limit, assuming $f_{n}\to f$ pointwise a.e .and $\int{f}<\infty$

Question

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.

What kind of convergence are you assuming if you write $f_n\rightarrow f$? — , May 27 '12 at 08:34
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 08:38
You might be interested in this: https://en.wikipedia.org/wiki/Dominated_convergence — Najib Idrissi, May 27 '12 at 08:38
@yaoxiao What if I put an extra condition: $|f_{n}|\le g_{n}$ and $\int{g_{n}}\to \int{g} \le \infty$ — aaronqli, May 27 '12 at 08:42
@Polymorpher this will be true, just consider Fatou theorem, you will get what you want — yaoxiao, May 27 '12 at 08:45
@N.I We do not have a single function to bound each $f_{n}$ therefore this theorem does not apply. — aaronqli, May 27 '12 at 08:45
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 08:47
@yaoxiao I know it is very easy to use Fatou's lemma to show $\int{f_{n}}=\int{f}$. I was trying to avoid that. But it looks like this cannot be avoided. Thanks! — aaronqli, May 27 '12 at 08:48

Davide Giraudo · Accepted Answer · 2012-05-27T12:07:53.903

2

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 $g+g_n-|f-f_n|$.

edited May 27 '12 at 12:07

answered May 27 '12 at 09:35

Davide Giraudo

172,925

It does not meet the condition $\int{g_{n}}\to\int{g}$ – aaronqli May 27 '12 at 09:45
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What is $g$ in this context? – Davide Giraudo May 27 '12 at 09:45
It is the limit of $g_{n}$. Sorry, I forgot to mention that... – aaronqli May 27 '12 at 09:47
Pointwise limit? – Davide Giraudo May 27 '12 at 09:52
Yes, a.e pointwise limit. I worked out a proof using Fatou's lemma to show this limit exists. Thank you for your help! I have edited my question. – aaronqli May 27 '12 at 09:55
To apply Fatou's lemma you need $f$ non-negative. Is it the case? – Davide Giraudo May 27 '12 at 09:56
It can be applied on $g-f$ – aaronqli May 27 '12 at 10:01
But it won't prove that $\int f_n$ is convergent. – Davide Giraudo May 27 '12 at 10:02
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Then show $\limsup$ and $\liminf$ equals, from inequalities in both directions. – aaronqli May 27 '12 at 10:03
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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
Indeed, thanks to that we just apply Fatou one time. – Davide Giraudo May 27 '12 at 12:06

Extra condition (or proof) for $\int{f_{n}}$ converging to a limit, assuming $f_{n}\to f$ pointwise a.e .and $\int{f}<\infty$

1 Answers1