Let the functions $f, f_n : X \to \bar{\Bbb R}$ be measurable and $f$ integrable:
I have to proof:

  • If $f_n \ge f, \: \mu -$almost everywhere $\forall n \in \Bbb N$ then $$\int_X \liminf_{n \to \infty}{f_n} \: d\mu \le \liminf_{n \to \infty} \int_X{f_n}d\mu$$
  • For $f_n \le f, \: \mu -$almost everywhere $\forall n \in \Bbb N$ then $$\int_X \limsup_{n \to \infty}{f_n}d\mu \ge \limsup_{n \to \infty} \int_X{f_n}d\mu$$

I also have to show why one can't give up the condition $f_n \ge f$ in a) respectively $f_n \le f$ in b)

I don't really know how to show all this, and i also don't get why the $f_n$ in case a) are integrable (in case b) it's clear since f is always bigger and integrable) Any ideas or tipps on how to show the two things above? Thanks in advance!

  • $\begingroup$ Also it isn't written anywhere that $f_n$ converges to $ f $. I don't know if that's implied or it isn't necessary. $\endgroup$
    – DeltaChief
    Dec 11, 2016 at 7:52

1 Answer 1


These are all (in their own way) different statements of Fatou's lemma.

If you know Fatou's lemma then (for the first) simply plug $g_n = f_n -f \geq 0$. $g_n$ are measurable (as sum of measurable functions) and the theorem follows (But wait, what is $\liminf f$? And can we just cancel them two terms in both sides of the equation?)

For the second, note that $h_n = -g_n \geq 0$ and we (again) meet the conditions for Fatou's lemma, only this time we have terms like $\limsup -f_n $, but this is easy (If you can connect $\limsup$ with $\liminf$).

  • $\begingroup$ Ah i see, thanks a lot. Could you also help me with showing that you can't leave out the condition $f _n \gt f$ in a) and the other on in b). I think one has to find a counterexample to show that one really needs it. $\endgroup$
    – DeltaChief
    Dec 11, 2016 at 8:38
  • $\begingroup$ @DeltaChief, I take you back to Fatou's: What if we drop the condition $f_n \geq 0$ (In Fatou's lemma)? In $\mathbb{R}$ take $f_n =-\frac{1}{n} \cdot 1_[0,n)$. All of them are negative, $\int_0^\infty f_n =1\quad (\forall n)$ But $\liminf f_n =0$. (To understand this, maybe after accumulating some more knowledge about measures): What went wrong? The signed measures $\nu _n (A) = \int _A f_n $ are not measures. $\endgroup$
    – Ranc
    Dec 11, 2016 at 10:58

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