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 $$\liminf \int f_n \geq \int \liminf f_n$$

  • $\begingroup$ 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. $\endgroup$ Aug 2 at 18:53
  • $\begingroup$ the condition $\int f > - \infty$ is an assumption and yes I mean -as I in fact wrote- that $-\infty < \int f_n < + \infty $ $\endgroup$
    – alphaomega
    Aug 2 at 19:36
  • 1
    $\begingroup$ Well, then I got it I think. Thanks a lot for the explanation :). I wouldn't have thought about that tbh! $\endgroup$ Aug 2 at 21:57

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