I've been looking on Stack Exchange for a solution, but I cannot figure out how to apply the solution of similar statements to this problem, because the conditions seem to be weaker.
Assume $\int f > – \infty$, $f_n \geq f$ a.e. . Here, $f,f_n: X \rightarrow \mathbb{\bar{R}}$ and measurable. Now, show
$\liminf \int f_n \geq \int \liminf f_n $.
I know that I can now apply Fatou's Lemma on $f_n – f \geq 0$ to obtain
$\liminf \int f_n -f \geq \int \liminf f_n -f $
But I do not know how to proceed from here. I think I need to add $\int f$ to both sides and show that we can pull $f$ into the first integral. But how exactly does that work? It is not trivial that the integrals are linear here, because $f \not \in L^1$ and also $ f \not \geq 0$ (at least not necessarily).
Best Answer
If $\int f = +\infty$ then $\int f_n = +\infty$ for all $n$, and since $\liminf f_n\geq f$, $\int \liminf f_n=+\infty$ as well.
Thus we can assume that $-\infty <\int f < +\infty$ and in that case, $$\liminf \int f_n -f \geq \int \liminf f_n -f $$ imlpies that $$\liminf \int f_n \geq \int \liminf f_n$$