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!
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 $\limsup_{n\in\mathbb{N}}\:f_n=\liminf_{n\in\mathbb{N}}\:f_n=f$
Once:$\int\liminf_{n\in\mathbb{N}}\:f_n\:d\mu=\int\:f\:d\mu$
Therefore:
$\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.