Extended Fatou's Lemma with weaker assumptions

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).

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 $$\liminf \int f_n \geq \int \liminf f_n$$
• Hi, thanks for your answer! However, I have a followup question: I understand now that if $\int f = \infty$, then Fatou's Lemma is trivially true. But I fail to understand your last step: How does $\int f < \infty$ imply $\liminf \int f_n \geq \int \liminf f_n$ ? Aren't the linearlity constrains required from both functions/integrals? Do you mean that $-\infty < \int f_n < + \infty$ ? Then I would understand your conclusion, because $\int f < + \infty$ follows and linearity applies. Aug 2 at 18:53
• the condition $\int f > - \infty$ is an assumption and yes I mean -as I in fact wrote- that $-\infty < \int f_n < + \infty$ Aug 2 at 19:36