why is $\lim f_{n} = \lim \inf f$?

Question

I was reading the solution to this problem

And I did not understand why in the fifth line he substituted $\lim f_{n}$ by $\lim \inf f$ .... could anyone explain this for me?

Because if a sequence has a limit then this limit equals to the $\liminf$. — Mark, Oct 15 '19 at 12:37
but in my question the $\lim \inf$ is of $f$ but the limit is of $f_{n}$ @HansEngler — , Oct 15 '19 at 12:56
I know that $\lim \inf$ is the smallest accumulation point is this the reason of the correctness of your first equality ? Also why it is not equal $\lim \sup$? or it is also equal to it? @Mark — , Oct 15 '19 at 12:57
It also equals to the $\limsup$. But to use Fatou's lemma you need $\liminf$, that's why it was used in that answer. — Mark, Oct 15 '19 at 13:00

score 0 · Accepted Answer · answered Oct 15 '19 at 12:38

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The equality $\lim_nf_n=\liminf_nf_n$ is trivially true. And then one can apply Fatou's lemma:$$\int_E\liminf_nf_n\leqslant\liminf_n\int_Ef_n.$$

answered Oct 15 '19 at 12:38

I know that $\lim \inf$ is the smallest accumulation point is this the reason of the correctness of your first equality ? Also why it is not equal $\lim \sup$? or it is also equal to it? – Oct 15 '19 at 12:54
but in my question the $\lim \inf$ is of $f$ but the limit is of $f_{n}$ – Oct 15 '19 at 12:56
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For any convergent sequence $(a_n)_{n\in\mathbb N}$ of real numbers we have$$\lim_na_n=\liminf_na_n=\limsup_na_n.$$ – José Carlos Santos Oct 15 '19 at 12:56
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And in that answer $\liminf_nf$ is a typo. It should be $\liminf_nf_n$. – José Carlos Santos Oct 15 '19 at 12:58
got it! thank you very much! – Oct 15 '19 at 12:58

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