# $f_n\to f$ in $L^1(\mu)\implies f_n$ are uniformly integrable

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

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.

• this is correct Apr 20 '19 at 12:35
• Thanks A lot Sir Apr 20 '19 at 12:39

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 $$\epsilon/2$$.