In generall, if $\;x_n\;$ is a sequence of real numbers, then $\;\forall \epsilon \gt 0\;\exists N\in \mathbb N:\;x_n \gt \liminf x_n -\epsilon\;\;\forall n\gt N\;$
as you can see here: https://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior
Taking the above inequality into account, if $\;x_n(t)\;$ is a sequence of real integrable functions and $\;I\subset \mathbb R\;$ then, is it true to claim that $\;\int_{I} x_n(t)\;dt \ge \liminf \int_{I} x_n(t)\;dt\;$?
I think it should be valid if I set $\;y_n=\int_{I} x_n(t)\;dt\;$. However I'm quite unsure of this argument.
I would really appreciate if somebody could confirm or fix the above inequality in order to hold.
Thanks in advance!
Your claim is false. Take $I=[0,1]$ and $x_n(t):=-1/n$ for all $t \in I$. Then $\int_I x_n=-1/n$ for all $n$, while $\liminf \int_I x_n=0$. The correct version is: for all $\varepsilon>0$, there exists $N$ such that $$ \forall n\ge N,\,\,\, \int_I x_n(t)\mathrm{d}t > \left(\liminf_{k\to \infty} \int_I x_k(t)\mathrm{d}t\right) -\varepsilon. $$
However, you can be probably interested in the Fatou's lemma.