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If $f_n\to f$ in $L^1(\mu)\implies f_n $ are uniformly integrable where $\mu $ is positive measure

My Attempt:

Uniformly Integrable family: $\{f_n\}_{n\in A}$ is said to be uniformly integrable if $\forall \epsilon>0$ $\exists \delta >0 $ such that $|\int_E f_n d\mu|$ wherever $\mu (E)<\delta$

Let $\epsilon>0$ given

As $f_n\to f$ in $L^1(\mu), \exists N$ that $\int |f_n-f|d\mu<\epsilon/2, \forall n>N$

As for $n<N$

I can find $\delta _i>0$ such that $\mu(E)<\delta_i$

$\int_{E} |f_i|d\mu<\epsilon/2$

For $n\geq N$

$\int |f_n|<\int |f-f_n|+\int |f|$

As Form assumption $\int |f-f_n|<\epsilon/2$ for any measurable set

And for f I can always find $\delta_a>0$ such that $\mu(E)<\delta_a$ $\int_E|f|d\mu<\epsilon/2 $

So we can choose $\delta=\min \left\{\delta_1,.....\delta_n,\delta_a\right\}$.

So done.

Am I correct?

I would be thankful if some one help me to find mistake in my proof if there is any.

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  • $\begingroup$ this is correct $\endgroup$ Apr 20 '19 at 12:35
  • $\begingroup$ Thanks A lot Sir $\endgroup$
    – Math Lover
    Apr 20 '19 at 12:39
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As pointed out by David Bowman, the attempt is correct. There is a small inaccuracy: the inequality $\int |f_n|<\int |f-f_n|+\int |f|$ may not be strict (for example if $f=0$ and $f_n\geqslant 0$).

Maybe it would have been nicer to add a step to check that the mentioned $\delta$ indeed work for all $f_n$: for $n\lt N$ it is by definition; for $n\geqslant N$, take $E$ such that $\mu(E)\lt\delta$. Then $$ \int_E |f_n|\leqslant \int_E |f-f_n|+\int_E |f|\leqslant \int |f-f_n|+\int_E |f| $$ and both terms are smaller than $\epsilon/2$.

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