Consider a measure space with an abstract measure $\mu$ and function $f_n$ that converges to $f$.

I have seen the proof of Fatou lemma lower bound: $\liminf_{n\to\infty}\int f_n \:d\mu\geqslant\int\liminf_{n\to\infty}f_n\:d\mu$

However I would like to prove:$\int f\:d\mu\leqslant\liminf_{n\to\infty}\int f_n\:d\mu$ ::This last equation seems contradictory to me. Thanks in advance!

  • $\begingroup$ Dear Pedro, could you kindly prove some more context? It does seem odd that you're being asked to prove something that is contradictory! $\endgroup$ – Kenny Wong Apr 13 '17 at 17:00
  • $\begingroup$ @Kenny Wong There was a mistake, I corrected it already. $\endgroup$ – Pedro Gomes Apr 13 '17 at 17:07
  • 1
    $\begingroup$ Just note that $$f = \lim f_n = \liminf f_n.$$ $\endgroup$ – saz Apr 13 '17 at 17:07
  • $\begingroup$ @saz Is that because the sequence converges from below, right? I mean the $\{f_n\}_{n\in\mathbb{N}}$ sequence. $\endgroup$ – Pedro Gomes Apr 13 '17 at 17:43
  • $\begingroup$ What exactly do you mean by "convergence from below"? If a sequence $(a_n)_n$ of real numbers converges to some number $a \in \mathbb{R}$ it holds that $$a=\lim_n a_n = \liminf_n a_n.$$ Apply this for $a_n := f_n(x)$. $\endgroup$ – saz Apr 13 '17 at 17:49

People have been commenting and I appreciate the insight to the point of coming up with an answer myself. I have been able to prove that

$\liminf_{n\in\mathbb{N}}\int f_n\:d\mu\geqslant\int\liminf_{n\in\mathbb{N}} f_n\:d\mu$

As we we know that the sequence of functions $f_n$ converge to $f$, we know that $\limsup_{n\in\mathbb{N}}\:f_n=\liminf_{n\in\mathbb{N}}\:f_n=f$



$\liminf_{n\in\mathbb{N}}\int f_n\:d\mu\geqslant\int\liminf_{n\in\mathbb{N}} f_n\:d\mu\Leftrightarrow\liminf_{n\in\mathbb{N}}\int f_n\:d\mu\geqslant\int f\:d\mu $

As we wanted.$\blacksquare$

If there is anything wrong with my proof, feel free to point out. Comments are welcomed after all it was I who answered it.


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